Circular_motion

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PHYSICS: Circular Motion

Circular MotionINTRODUCTION:

In many natural phenomena the motion is circular or nearly circular, e.g., satellite motion or motion of the eartharound the sun. A particle moving in a circular path illustrates many of the important features of the velocity andacceleration vectors in two dimensions.

We first consider circular motion with constant speed. In everyday usage, we would say that if the speed isconstant, there is no acceleration. However, we have defined acceleration as the rate of change of the velocityvector. Even when the speed is constant, a particle moving in a circle is accelerating because the direction of thevelocity is constantly changing.

UNIFORM CIRCULAR MOTION:

If a particle is moving in a circular path with constant speed, we define this motion as uniform circular motion.Now, we have to obtain the acceleration in uniform circular motion.

The situation is shown in figure 4.1(a). Let P be the position of the particle at the time t and 'P be its positionat the time t + ∆t. The velocity at P is 1,

!v a vector tangent to the curve at P. The velocity at 'P is 2!v , a vector

tangent to the curve at '.P As the speed of the particle is constant, the vectors 1!v and 2

!v are equal in magnitude,but

fig. 4.1(a)

Cθ

r

r

P

P’ v1!

v2!

fig. 4.1(b)

v2!

v1!

v 1!

v 2! �

θ r r

C

P’ Pv. t∆

θ

their directions are different. The length of the path traversed during ∆t is the arc length ',PP which is equal to,⋅ ∆v t where v is the constant speed of the particle.

Now, redraw the vectors 1!v and 2 ,!v as shown in figure 4.1(b) so that they originate at a common point. We are

free to do this as long as the magnitude and direction of each vector are the same as in figure 4.1(a). Figure4.2(b) enables us to see clearly the change in velocity as the particle moved from P to '.P This change,

2 1 ,− = ∆! ! !v v v is the vector which must be added to 1!v to get 2.

!v Notice that it points inward, approximatelytoward the centre of the circle.

Now the triangle fromed by 1 2,! !v v and ∆!v is similar to the triangle 'CCP formed by the chord 'PP and the

radii CP and '.CP This is so because both are isosceles triangles having the same vertex angle; the angle

θ between 1!v and 2

!v is the same as the angle 'PCP because 1!v is perpendicular to CP and 2

!v is perpendicularto the '.CP We can therefore write



| |∆ ⋅∆=!v v t

v r approximately,

The chord 'PP being taken equal to the arc length '.PP This relation becomes more nearly exact as ∆t isdiminished, since the chord and the arc approach each other. Notice also that ∆!v approaches closer andcloser to a direction perpendicular to 1

!v and 2!v as ∆t is diminished and therefore approaches closer and

closer to a direction pointing to the exact centre of the circle. It follows from this relation that2

,∆ =∆

!v vt r approximately,

and in the limit when ∆t → 0 this expression becomes exact. We therefore obtain

2

0lim

t

v vat r∆ →

∆= =∆

!...(4.1)

as the magnitude of the acceleration. The direction of !a is instantaneously along a radius inward toward thecentre of the circle. Figure 4.2 shows the instantaneous relation between !v and !a at various points ofmotion. The magnitude of !v is constant, but its direction changes continuously. This gives rise to an acceleration!a which is also constant in magnitude but continuously changing in direction. The velocity !v is always tangent

to the circle in the direction of motion; the acceleration !a is always directed radially inward. Because of this,!a is called a radial, or centripetal acceleration. Centripetal means �seeking a center�.

v!

v!a!

a!a!

fig. 4.2

v!

The acceleration resulting from a change in direction of a velocity is just as real and just as much an accelerationin every sense as that arising from a change in magnitude of a velocity. The velocity, being a vector, canchange in direction as well as magnitude. If a physical quantity is a vector, its directional aspects cannot beignored, for their effects will prove to be every bit as important and real as those produced by changes inmagnitude.

NONUNIFORM CIRCULAR MOTION:

If a particle moves in a circle with speed that is varying, there is a component of acceleration tangent to thecircle as well as the centripetal acceleration inward. The tangential component of acceleration is simply therate of change of speed dv/dt, whereas the radially inward component arises due to change in direction and

has a magnitude 2

.vr

Let us introduce the unit vector �t fixed to the moving particle and oriented along atangent to the path in the direction of the movement of the particle as shown in fig. 4.3. Therefore �t is avariable vector since it depends upon the position of the particle. The velocity vector !v of the particle isoriented along a tangent to the path and therefore can be represented as follows



O

l

A

v = v.t�!

fig. 4.3

�= ⋅!v v t

where speed , being the distance covered dlv ldt

= is the rate of covering distance.

Let us differentiate above equation with respect to the time:

� �( ) �⋅= = = ⋅ + ⋅!! dv d v t dv dta t v

dt dt dt dt

2� �� = ⋅ + ⋅ ⋅ = ⋅ + ⋅dv dt dl dv dtt v t vdt dl dt dt dl ...(4.2) using dl v

dt =

Let us examine the increment of the vector �t in the interval dl (see figure 4.4). It can be strictly shown thatwhen point 2 approaches point 1, the segment of the path between them tends to turn into an arc of a circle

fig. 4.4

�1

2dl

t1

�t2

d θdθ

�t2

�t1�dt

with centre at some point O.O is referred to as the centre of curvature of the path at the given point, and theradius r of the corresponding circle as the radius of curvature of the path at the same point. It is seen fromfigure 4.4, that the angle



1

� �1�θ = = =

dt dtdld r t

⇒� 1 ;=dt

dl r

At the same time, if 0,→dl then � �dt t . Introducing a unit vector �n alog the normal to the path at point 1directed toward the centre of curvature, we write the last equality in a vector form:

� �=dt n

dl r...(4.3)

Now let us substitute equation (4.3) into equation (4.2). Then we get,

2��= +! v dva n t

r dt ...(4.4)

Here the first term is called the centripetal (normal) acceleration na and the second term is called thetangential acceleration ta :

2

;=nvar

=tdvadt ...(4.5)

Here normal acceleration (which must be provided by a normal component of the net force acting on theparticle) is responsible for the change in direction of the velocity and the tangential acceleration (which must beprovided by a tangential component of the net force acting on the particle) is responsible for the change of thespeed.

Thus, the total acceleration a of a particle can be represented as the sum of the tangential and the normalaccelerations as shown in fig. 4.5. The magnitude of the total acceleration of the particle is

2 2= +n ta a a

2 22 = +

v dvr dt ...(4.6) r

v a =tdvdt

a =n vr

² a

O

fig. 4.5

Obviously this concept can also be applied to a particle moving on a circular path. At any point on the path,radius of the circle should be regarded as the radius of curvature of the path at that point and the centre of thecircle should be regarded as the centre of curvature of the path. Therefore, if equations (4.4), (4.5) and (4.6)are applied on a particle moving in a circular path, r denotes the radius of the circle.



ANGULAR VARIABLES:

It is often convenient to describe the motion of a particle moving in a circle in terms of angular variables.

If the particle is moving with constant speed then time required to complete one revolution, called the timeperiod T, is given by

2π= rTv

where r is the radius of the circle and v is the speed of the particle.

The reciprocal of the period is called the frequency f. We can write

12π

= = vfT r

The frequency is usually given in revolutions per second (rev/s). Like the radian, the revolution is also adimensionless unit.

Figure 4.6 shows a portion of a circle of arc length s andradius r. The angle θ swept by the radius vector when aparticle moves from P to 'P is θ = s/r. The θ is called theangular position of the particle with respect to its initialposition.

The rate of change of the angular position is called the angular velocity ω :

P

θ

r

r

s

P

fig. 4.6

'

( / )θω = =d d s rdt dt

1= dsr dt

Since ds/dt is the rate of covering distance, it is basically linear speed of the particle. Therefore, the linear andangular speeds are related by

ω = vr

⇒ =v ωr ...(4.7)

The rate of change of the angular velocity isdefined as the angular acceleration α :

( / )ωα = =d d v rdt dt

fig 4.7

αω

r

r

s

P

P

v

'at

v = dsdt = ωr

= dvdt = αrat

1= dvr dt

where is the tangential acceleration and / rate of change of speed.

tt

t

aaa dv dtr

= = =

Therefore, the tangential and the angular accelerations are related by

t =a αr ....(4.8)



NOTE: * At all points the acceleration of the particle has an inward radial component too.

* If angular acceleration α is constant then we can use the following equations:

( ) ;ω ω α= +int t ...(4.8)

21( ) ;2

θ θ ω α= + ⋅ +i int t t ...(4.9)

( )2 2( ) ( ) 2 .ω ω α θ= + ⋅ ⋅∆int ...(4.10)

A particle rotates in a circle according to the law 3,θ = −at bt where θ is the angular position of the particle at some

time �t�, 6 0= ⋅a rad/s and 2 0= ⋅b rad/s³. Find:

(a) the mean values of the angular velocity and angular acceleration averaged over the time intervalbetween t = 0 and the complete stop;

(b) the angular acceleration at the moment when the particle stops.

SOLUTION: If ω be the angular velocity and α be the angular acceleration of the particle, then

32( ) 3θω −= = = −d d at bt a bt

dt dt

and2( 3 ) 6ωα −= = = −d d a bt bt

dt dtIf the particle stops at t = t0, then

0(at ) 0ω = =t t

⇒ 203 0− =a bt

⇒ 0 3= at

b

Therefore, for the time interval [0, t0], the average angular velocity,

θ θθω−∆= =

∆ −f i

f it t t

0

0

( ) ( 0)0

θ θ= − ==−

t t tt

320 00

0

( ) (0)− −= = −at bt a btt

23 3

= − ⋅ =a aa bb

4 0= ⋅ rad/s;

EXAMPLE : 1



and the average angular acceleration,

0

0

( ) ( 0)0

ω ω ω ωωα− = − =∆= = =

∆ − −f i

f i

t t tt t t t

20

00

( 3 ) ( ) 3− −= = −a bt a btt

3 33

= − ⋅ = −ab abb

26 rad/s= −

At t = t0, the angular acceleration,

06α = − bt

6 2 33

= − = −ab abb

212 rad/s= −

A particle is moving in a circle of radius r with a varying angular speed ω, given by2ω β= t .

Find the net acceleration of the particle at some time t.

SOLUTION: We have

2ω β= t

∴ (let be the linear speed)ω=v r v

2β= r t

If at be the tangential acceleration and an be the normal/centripetal acceleration, then we have,

2 2 2 42 4β β= = =n

v r ta r tr r

and 2 β= =tdva r tdt

∴ Net acceleration, 2 2= +net t na a a

2 2 2 2 4 84 β β= +r t r t

2 64β β= +r t t

EXAMPLE : 2



1. A body is travelling in a circle at a constant speed. It

(a) has a constant velocity.(b) is not accelerated.(c) has an inward radial acceleration.(d) has an outward radial acceleration.

2. When a particle is in uniform circular motion it does not have

(a) radial velocity and radial acceleration(b) radial velocity and transverse acceleration(c) transverse velocity and radial acceleration(d) transverse velocity and transverse acceleration.

3. Assertion (A): If a body moving in a circular path has constant speed, then there is no force acting on it.

Reason (R) : The direction of the velocity vector of a body moving in a circular path is changing.

(a) both A and R are true but R is the correct explanation of A(b) both A and R are true but R is not the correct explanation of A(c) A is true but R is false(d) A is false but R is true.

4. A body is revolving with a constant speed along a circular path. If the direction of its velocity is reversed, keepingspeed unchanged, then

(a) the centripetal force does not suffer any change in magnitude and direction both(b) the centripetal force does not suffer any change in magnitude but its direction is reversed(c) the centripetal force disappears(d) centripetal force will be doubled.

5. An object follows a curved path. The following quantities may remain constant during the motion

(a) speed (b) velocity(c) acceleration (d) magnitude of acceleration.

TRY YOURSELF- I



ALTERNATE APPROACH FOR CIRCULAR MOTION:

Although we have already discussed both uniformand nonuniform circular motions, let us try tounderstand these concepts by a more generalapproach. In this method we will try to get thevelocity !v by differentiating the position vector !rand the acceleration !a by differentiating the velocity

.!v A particle is moving in a circle of radius r. Wehave assumed the centre as the origin of thereference frame, as shown in figure 4.8(a). At sometime the position vector !r is making an angle θwith the +ve-x direction. We define θ as the angularposition of the particle. We have also defined twovariable unit vectors, one along radially outwarddirection, �r and the other one, �,t along the tangentand coinciding with the direction of motion of theparticle. From figure 4.8(b), we have

� �� cos sinθ θ= +r i j ...(4.9)

θ

fig 4.8 (a)

Y

X

i

r

tj

O

ωα

r!

and � �� sin cosθ θ= − +t i j ...(4.10)

We have, � �� cos sinθ θ= +r r i r j

∴� �( cos sin )θ θ+= =

!! dr d r i r jvdt dt

θ

fig 4.8 (b)

i

tj

θ

r

� �( cos ) ( sin )θ θ= +d r i d r jdt dt

(cos ) (sin )� �d dri rjdt dt

θ θ= +

� �sin cosθ θθ θ = − ⋅ + ⋅ d dri rjdt dt

= � �sin cos [ angular velocity, / ]r i r j d dtω ω θ ω θ− ⋅ + ⋅ =∵

= � �[ sin cos ]ω θ θ− ⋅ + ⋅r i j

⇒ �⋅=!v ωr t [using equation 4.10]

Therefore, the velocity of the particle is along the tangent to the circle and the magnitude of the velocity is ωr.

Again, we have,

=!! dva

dt



sin� �sinω θθ ω= − ⋅ ⋅ − ⋅ ⋅d dr i r idt dt

cos� �cosω θθ ω+ +d dr j r jdt dt

2 2� � � �sin cos cos sin .[ angular acceleration / ]

α θ ω θ α θ ω θα ω

= − − + −=∵

r j r i r j r jd dt

2 � � � �(cos sin ) ( sin cos )ω θ θ α θ θ= − + + − +r i j r i j

⇒ 2 �� [using equations 4.9 and 4.10]a r r r tω α= − ⋅ + ⋅!

Therefore, the acceleration of the particle has a radially inward component of magnitude ω²r (= v²/r) and atangential component of magnitude α r. The magnitude of the net acceleration can be written as

2 2 2( ) ( )ω α= +neta r r

It is obvious that in the case of uniform circular motion, the angular acceleration α is zero and the particle hasonly radially inward component of the acceleration, which is often called centripetal acceleration. Theresults obtained by this method are in accordance with the previous results.

A point moves along a circle with a speed v = αt, where α is a positive constant. Find the magnitude of the accelerationat the moment when it covered the n-th (n = 0.10) fraction of the circle after the beginning of the motion at t = 0.

SOLUTION: Let r be the radius of the circle and s be the distance travelled along the circumference of theacceleration at the circle in the time interval [0, t], then we have

=ds vdt

⇒ α=ds tdt

⇒0 0

α= ⋅∫ ∫s t

ds t dt

⇒2

2α= ts ...(i)

When the point has covered nth (n = 0.10) fraction of the circle, we have,

(2 ) 0.10(2 ) 0.2( )s n r r rπ π π= = =

⇒2

0.2( ) [using (i)]2t rα π=

⇒ 2 0.4( )rt πα

= ...(ii)

EXAMPLE : 3



At this moment, the centripetal acceleration is

2 2 2

nv tar r

α= =

⇒ 0.4( )na πα= [using (ii)]

and the tangential acceleration is

( )α α= = =tdv d tadt dt

Therefore, the magnitude of the total acceleration is

2 2 2 2 2(0.16)n ta a a π α α= + = +

20.16 1α π= +

A point moves with deceleration along the circle of radius R so that at any moment of time its tangential and normalaccelerations are equal in moduli. At the initial moment t = 0 the velocity of the point equals v0. Find:

(a) the velocity of the point as a function of time and as a function of the distance covered s;

(b) the magnitude of the total acceleration of the point as a function of velocity and the distance covered.

SOLUTION :(a) Let at some time �t�, v be the speed of the point, then we have,

normal acceleration tangential acceleration=

⇒2

= −v dvR dt

⇒2

= −dv vdt R

...(i)

⇒0

20

= −∫ ∫v t

v

dv dtv R

⇒0

1 /− = −v

v

t Rv

⇒0

1 1 / − − − = −

t Rv v

⇒0

1 1= + tv v R

EXAMPLE : 4



⇒0

0

=+v Rv

R v t ...(ii)

From equation (i), again, we have

2

[ is distance covered]⋅ = −dv ds v sds dt R

⇒2

⋅ = −dv vvds R

⇒ = −dv vds R

⇒0 0

1= −∫ ∫v s

v

dv dsv R

⇒ 0ln( / ) /Rv v s= −

⇒ /0

s Rv v e−= ⋅ ...(iii)

(b) If an be the normal and at be the tangential acceleration, then the total acceleration,

2 2= +t na a a

2 [ ]= =∵n t na a a

2

2= vR

...(iv)

∴ 2 2 /02( ) [using (iii) and (iv)]

−

=s Rv ea s

R



1. In applying the equation for motion with uniform angular acceleration, ω = ω0 + αt, the radian measure

(a) must be used for both ω and α (b) may be used for both ω and α(c) may be used for ω but not for α (d) cannot be used for both ω and α.

2. The position vector of a particle in a circular motion about the origin sweeps out equal area in equal time. Its

(a) velocity remains constant (b) speed remains constant(c) acceleration remains constant (d) tangential acceleration remains constant.

3. A particle is going in a spiral path as shown in figure with constant speed.

(a) The velocity of the particle is constant(b) The acceleration of the particle is constant. (c) The magnitude of acceleration is constant.(d) The magnitude of acceleration is decreasing continuously.

4. If the equation for the displacement of a particle moving on a circular path is given by 32 0.5,tθ = + where θ is inradian and t in second, then the angular velocity of the particle after 2 s from its start is :

(a) 8 rad/s (b) 12 rad/s(c) 24 rad/s (d) 36 rad/s.

5. In the Bohr model of the hydrogen atom, the electron revolves in a circular orbit around the nucleus. If the radiusof the orbit is 115 3 10−⋅ × meter and the electron makes 156 6 10⋅ × rev/sec, find (a) the acceleration (magnitudeand direction) of the electron and (b) the centripetal force acting on the electron. (This force is due to the attractionbetween the positively charged nucleus and the negatively charged electron). The mass of the electron is 319 1 10−⋅ ×kg.

6. A particle is projected with a speed u at an angle θ with the horizontal. Consider a small part of its path near thehighest position and take it approximately to be a circular arc. What is the radius of this circle ? This radius is calledthe radius of curvature of the curve at that point.

7. What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point wherethe particle velocity makes an angle θ /2 with the horizontal ?

8. A wheel rotates around a stationary axis so that the rotation angle ϕ varies with time as ϕ = at2, where a = 0.20rad/s2. Find the total acceleration α of the point A at the rim at the moment t = 2.5 s if the linear velocity of the pointA at this moment v = 0.65 m/s.

9. (a) Write an expression for the position vector r! for a particle describing uniform circular motion, using rectangularcoordinates and the unit vectors �i and �j . (b) From (a) derive vector expressions for the velocity !v and theacceleration !a . (c) Prove that the acceleration is directed toward the center of the circular motion.

10. A point moves along an arc of a circle of radius R. Its velocity depends on the distance covered s as ,v a s=where a is a constant. Find the angle α between the vector of the total acceleration and the vector of velocity as afunction of s.

TRY YOURSELF- II



DYNAMICS OF CIRCULAR MOTION:

If we resolve the net force acting on a particle moving on a circular path along the motion of the particle(i.e., along the tangential direction) and along the radially inward direction (i.e., towards the centre), then wesee that the tangential component of the net force provides tangential acceleration and the radial component ofthe net force provides radial or centripetal acceleration. In the case of uniform circular motion, the tangentialcomponent of the net force is zero. The radial component of the net force is defined as centripetal force.

Therefore, we can write,

2

centripetal centripetal ;= = mvF mar

...(4.11)

tangential ;= =tdvF ma mdt ...(4.12)

and ( ) ( )2 2

net centripetal tangential= +F F F ...(4.13)

Here, I would like to emphasize that the centripetal force is not just another force which must be added withother forces present to get the net force. Centripetal force is just the radial component of the resultant of theforces already present. This concept would be more clear by the following examples.

Suppose we release a small ball of mass m from the highestpoint inside a smooth hemispherical bowl, as shown in figure4.9. When the position of the ball with respect to centre ofthe bowl, C, is making an angle θ, let it has acquired aspeed v. At this moment the ball has components ofacceleration along both radial and tangential directions. Theradial component of the acceleration or centripetalacceleration is being provided by the normal contact force,N, acting on the ball. Here N is balancing the radialcomponent of the weight of the ball and also providing thecentripetal acceleration to the ball, as shown in figure 4.10.

`

θN

C

vR

2

θ

m

v smooth

mg

R

fig. 4.9

Here it would be a mistake to think that there are followingthree force acting on the ball: (1) weight of the ball, (2)normal contact force on the ball, (3) centripetal force onthe ball. As we have already discussed before, centripetalforce is not an additional force acting on the ball; it is theradial component of the net force acting on the ball from itssurrounding or we can say it is the sum of the radialcomponents of the forces acting on the ball from itssurrounding. In this case the resultant of N and mg cosθshould be defined as the centripetal force.

θN

C

θdv/dt

smooth

mg

R

fig. 4.10

mg.sinθ

mg.cosθ

vR

2

As shown in figure 4.10, the tangential component of the weight of the ball, mg sinθ, is the net tangential forceon the ball. This force is responsible for the change in speed of the ball and provides the tangential acceleration

.dvdt It is quite obvious that in this case sinθ=dv g

dt and when ball reaches the lowermost point, the tangentialacceleration is zero.



v

mT

O vl

2

fig. 4.11 (a) (Top view)

N

mTO

vl

2

mg

fig. 4.11 (b) (side view)

Consider the case shown in figures 4.11(a) and (b). A small ball of mass m is being revolved with constantspeed v in a circular path on a smooth horizontal table by means of a thread of length l. Top view is shown infigure 4.11(a) and side view is shown in figure 4.11(b). Here, the ball is moving uniformly in a horizontal circle

of radius l with speed v, therefore, it has a centripetal acceleration of magnitude 2vl

which is pointing towardsthe centre of the circle, O. This acceleration is always in the horizontal plane only. From figure 4.11(b) it isclear that the tension force from the thread is providing the centripetal acceleration and hence the tensionforce is acting as centripetal force. Normal contact force from the surface of the table is balancing the weightof the ball. In this case tangential acceleration is zero.

A particle of mass m is suspended from a string of length L andmoves at constant speed in a horizontal circle of radius r. Thestring makes an angle θ given by sin ,θ = r L as shown in figure4.12. Find the tension in the string and the speed of the particle.(This configuration is known as a conical pendulum.)

`

θL

v

sin θ = rL

r

fig. 4.12SOLUTION: Let the particle is moving with speed v and tensionin the thread is T. The two forces acting on the particleare its weight mg, acting vertically downward, and thetension T, which acts along the string, as shown in figure4.13. In this situation we know that the acceleration istowards the centre of the horizontal circle and has a

magnitude 2

.vr

Thus the vertical component of the tension

must balance the weight mg. The horizontal componentof the tension is the centripetal force and is providing thecentripetal acceleration. Therefore, from figure 4.14, wehave

θL

Tv

mg

sin θ = rL

r

fig. 4.13

m

EXAMPLE : 5



T cos θ = mg ...(i)

and T sin θ = mv²/r ...(ii)

v r2/T sinθ

θ

mg

T cosθT

fig. 4.14

Solving equations (i) and (ii), we get,

tanθ=v gr and cosθ= mgT

A car travel on a horizontal road in a circle of radius 30 m. If the coefficient of static friction is 0 6,µ = ⋅s how fast canthe car travel without slipping?

SOLUTION: This situation is shown in figure 4.15. The normal contact force on the car from the road balances thedownward force due to gravity, mg. The only horizontal force is due to friction. As the car is moving in ahorizontal circle, its centripetal acceleration is always in the horizontal plane only. Therefore, in this case frictionalforce f is the centripetal force.

N

v

mg

f

r

vr

2

fig. 4.15

F.B.D. of the car is shown in figure 4.16. From this figure, we have.

N = mg m (car)

mg

Nv r2/

f

fig. 4.16and f = 2mv

rAs the car is not slipping on the road, frictional force must be static in nature.

Therefore, we have

limiting≤f f

⇒ µ≤ sf N

⇒2

µ≤ smv mg

r

⇒ 2 µ≤ zv gr

⇒ 0.6 9.8 30 13.3≤ = × × =sv µ gr m/s

EXAMPLE : 6



If the car travels at a speed greater than 13.3 m/s, the force of static friction will not be great enough to providethe acceleration needed for the car to travel in a circle of specified radius and the car would slide out away fromthe centre of the circle; i.e., it will tend to travel in a circle of larger radius. You can note that if a horizontal roadis smooth then car can not take a turn.

NOTE: If the road is not horizontal but banked, as shown in figure4.17, the normal force of the road will have a componentinward toward the centre of the circle, which will contributeto the centripetal force. The banking angle can be chosenin such a way that for a given speed no friction is neededfor the car to make the curve. θ

N

vr

2

θ

mg

fig. 4.17

A hemispherical bowl of radius R is set rotating about its axis of symmetry which is kept vertical. A small blockkept in the bowl rotates with the bowl without slipping on its surface. If the surface of the ball is smooth, and theangle made by the radius through the block with the vertical is θ, find the angular speed at which the bowl isrotating.

SOLUTION: Let the ball is at the point A on thehemispherical surface. As the ball is also rotatingwith the surface and there is no slipping betweenthe ball and the surface, ball would rotate in ahorizontal circle with same angular speed as that ofthe surface. If ω be angular speed of the bowl andr be the radius of the horizontal circle in which theball is rotating, then ball has a horizontalacceleration of magnitude ω2r, pointing towards thecentre of the circle, as shown in figure 4.18. As thesurface is smooth, there are only two forces actingon the ball : (1) normal contact force, N; (2) itsweight, mg. F.B.D of the block is shown in figure4.19. Vertical component of the normal contactforce is balancing the weight of the ball and itshorizontal component is providing the centripetalacceleration. Therefore, we have,

fig 4.18

R

AB

N

C

θ

ω

θ

mg

ω2r

fig 4.19

N

ω2r

mg

m

θ

N cos θ

mg

mN sin θ

ω2r

EXAMPLE : 7



2sinθ ω=N m r

( )2 sinω θ= m R [ ]sinr R θ=∵

2N m Rω⇒ = ...(i)

and N cos θ = mg

2 cosm R mgω θ⇒ = [using (i)]

ω⇒ = gRcosθ

A simple pendulum consists of a mass m suspended at the end of a rope of length l. When the rope is at an angle θ tothe vertical, the speed of the bob is v. Find :

(a) the radial and tangential components of the acceleration ;

(b) the tension in the thread.

SOLUTION: It is obvious that the path of the bob is a part of a vertical circle of radius l. When thread makes an angleθ with the vertical, the speed of the bob is given to be v. At this moment there are only two forces acting on theball : (1) tension force, T; (2) its weight, mg, as shown in figure 4.20 (a).

g sin θ

mg sin θ mg cos θ

θ

v

θ

mg

v2

l

fig 4.20(a)

m manet

g sin θ

F.B.D. of the ball is shown in figure 4.20(b). As the ball is moving on a circle of radius l, it has a centripetalacceleration of magnitude v2/l towards the centre. Therefore, applying Newton�s 2nd law along radial direction,we get

2

cos mvT mgl

θ− =

2

cosθ⇒ = + mvT mgl

From figure 4.20(b) it is clear that mg sin θ is the only tangential force on the bob and hence, tangentialacceleration of the bob is g sin θ . Both radial and tangential accelerations are shown in figure 4.20(c). Theirresultant, net acceleration, is also shown. You should note that when the bob reaches at the extreme position itsspeed becomes zero and hence at that position it has only tangential acceleration.

EXAMPLE : 8



In the previous example suppose the bob is given a horizontal speed u, when the thread is vertical. Find the speed andthe tension as a function of θ, where θ is angle between the vertical direction and the thread.

SOLUTION: From the previous example it is clear that atsome angle θ if the ball has a speed v then it has a

radial acceleration of magnitude 2vl

and a tangential

acceleration of magnitude gsin θ, as shown in figure4.21. We have,

sinθ= = −tdv a gdt

sinθ θθ

⇒ ⋅ = −dv d gd dt

θ

mg sin θ

mg cos θ

vv²l

g sin θ

T

fig. 4.21

l

sinω θθ

⇒ ⋅ = −dv gd

angular velocity /θω = = = d v ldt

sinθ θ⇒ ⋅ = − ⋅v dv g dl

0

sinθ

θ θ⇒ ⋅ = − ⋅∫ ∫v

u

v dv gl d

( )2 2

1 cos2

θ−⇒ = − −v u gl

( )2 2 2 1 cosθ⇒ = − −v u gl ...(i)

( )2 2 1 cosv u gl θ⇒ = − −

If T be the tension in the thread at this moment, then, using Newton�s 2nd law along the radial direction, we get,2

cosθ− = mvT mgl

2

cosθ⇒ = + mvT mgl

2

cos 2 2 cosθ θ= + − +mumg mg mgl

[using (i)]

2

3 cos 2θ= − + mumg mgl

EXAMPLE : 9



1. A car rounds an unbanked curve with radius of curvature 40m. The coefficient of friction between the tires andthe road is 0.6. What is the maximum speed the car can travel without slipping?

2. A curve of radius 30 m is banked so that a car can round the curve at 48 km/h even if the road is frictionless.Show in a force diagram that a component of the normal force exerted by the road on the car can provide thecentripetal force necessary and calculate the banking angle θ for these conditions.

3. A small coin is placed on a flat; horizontal turntable. The turntable is observed to make three revolutions in 3.14sec. (a) What is the speed of the coin when it rides without slipping at a distance 5.0 cm from the center of theturntable? (b) What is the acceleration (magnitude and direction) of the coin in part (a)? (c) What is the frictional-force acting on the coin in part (a) if the coin has a mass m? (d) What is the coefficient of static friction betweenthe coin and the turntable if the coin is observed to slide off the turntable when it is greater than 10 cm from thecenter of the turntable?

4. (a) What is the smallest radius of a circle at which a bicyclist can travel if his speed is 18 mi/hr and the coefficientof static friction between the tires and the road is 0.32? (b) Under these conditions what is the largest angle ofinclination to the vertical at which the bicyclist can ride without falling? (solution of part (b) requires the conceptof torque).

5. A rod of length l and mass m is rotated in a horizontal plane about its one end with constant angular velocity ω.The tension at the middle section of the rod is :

(a) 2m lω (b)2

2m lω

(c) 234

m lω (d) 238

m lω .

6. A car moves at a constant speed on a road with varying slope as shown in figure. The normal force by the roadon the car is NA and NB when it is at the points A and B respectively.

(a) NA = NB(b) NA > NB

BA

(c) NA < NB(d) insufficient information to decide the relation of NA and NB.

7. Let θ denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the bobis m, the tension in the string is mg cosθ

(a) always (b) never(c) at the extreme positions (d) at the mean position.

8. A simple pendulum is oscillating without damping. When the distancement of the bob is less then maximum, itsacceleration vector a is correctly shown in

(a)

!a

(b) !a

(c) !a (d)

!a

.

TRY YOURSELF- III



9. A car of mass m is moving on a horizontal circular path of radius r. At an instant its speed is v and is increasing ata rate a

(a) the acceleration of the car is towards the centre of the path

(b) the magnitude of the frictional force on the car is greater than 2mv

r(c) the friction coefficient between the ground and the car is not less than a/g

(d) the friction coefficient between the ground and the car is 2

1tan vrg

µ − =

.

10. A mass m on a frictionless table is attached to a hanging mass M by a cord through a hole in the table (Fig). Findthe conditions (v and r) with which m must spin for M to stay at rest.

m

M

11. Two blocks each of mass M are connected to the ends of a light frame as shown in figure. The frame is rotatedabout the vertical line of symmetry. The rod breaks if the tension in it exceeds T0. Find the maximum frequencywith which the frame may be rotated without breaking the rod.

l

M M



12. A block of mass m1 is attached to a cord of length L1, which is fixed at one end. The mass moves in a horizontalcircle supported by a frictionless table. A second block of mass m2 is attached to the first by a cord of length L2and also moves in a concentric circle, as shown in figure. If the angular speed is ω, find the tension in each cord.

m2m1

L1 L2

13. A small sphere of mass m suspended by a thread is first taken aside so that the thread forms the right angle withthe vertical and then released. Find :

(a) the total acceleration of the sphere and the thread tension as a function of θ, the angle of deflection of thethread from the vertical;

(b) the thread tension at the moment when the vertical component of the sphere�s velocity is maximum;

(c) the angle θ between the thread and the vertical at the moment when the total acceleration vector of thesphere is directed horizontally.

14. A small sphere of mass m suspended by a thread is first taken aside so that the thread forms the right angle withthe vertical and then released, then:

(a) total acceleration of sphere as a function of θ is 21 3cos θ+g

(b) thread tension as a function of θ is T = 3 mg cos θ

(c) the angle θ between the thread and the vertical at the moment when the total acceleration vector of the

sphere is directed horizontally is 1cos 1 3−

(d) the thread tension at the moment when the vertical component of the sphere�s velocity is maximum will be mg.

15. A ball suspended by a thread swings in a vertical plane so that its acceleration values in the extreme and thelowest position are equal. Find the thread deflection angle in the extreme position.

16. A block of mass m at the end of a string is whirled around in a vertical circle of radius R. Find the critical speedbelow which the string would become slack at the highest point. [Hint : when the string slacks, take T = 0, notv = 0]

17. A small body A starts sliding off the top of a smooth sphere of radius R. Find the angle θ (Fig.) corresponding tothe point at which the body breaks off the sphere, as well as the break-off speed of the body.

R

Aθ

[Hint : Find V(θ ) using tangential acceleration and find N(θ ) using Newton�s 2nd law along the radial direction.At the break off point take N = 0.]



CENTRIFUGAL FORCE:

In chapter 2(laws of motion) we discussed that the Newton�s laws of motion are not valid in a noninertial frame(a frame which is accelerating with respect to an inertial frame). If we insist to work from a noninertial frame,we need to apply a pseudo force to make the Newton�s 2nd law applicable in that frame. If the frame istranslating with an acceleration !a , then we apply ��m !a � as pseudo force where m is the mass of the body onwhich it is being applied. Now, we know that rotating frames are also accelerating frames. Therefore, we mustapply pseudo forces on the different bodies while making observations from a rotating frame of reference.

If a body is rotating with a constant angular speed ω and its distance from the axis of rotation is r then body hasa radially inward acceleration ω²r, as shown in figure 4.22(a). But if we observe the body from the framerotating with it, then we find that the body is at rest. To use the Newton�s 2nd law in its usual form, we applya pseudo force of magnitude mωωωωω²r in the radially outward direction, as shown in figure 4.22(b). Thispseudo force is defined as the centrifugal force. In Greek centrifugal means �centre fleeing�.

fig. 4.22(a)

mr

axis

ωω²r

fig. 4.22(b)

F = m rω²r

REST

It must be clear to you that centrifugal force is just a pseudo force and you need not to consider it if you aremaking your observations from an inertial frame. Consider the situation shown in figure 4.11. When we observefrom ground frame, we see that the ball is rotating with a constant angular speed ω in a circle of radius r andhence, it has centripetal acceleration ω²r, which is being provided by the tension force acting on it.Magnitude of the tension force is mω²r. But if we observe the ball from a frame rotating with it, we find that theball is at rest. If we do not apply any pseudo force then Newton�s 2nd law is not applicable in this frame,because the tension force is the only horizontal force on the ball and stillthe ball is at rest; as shown in figure 4.23(a).

Centre (= ² )m rω

mg

RESTm

N

(= ² )m rω

fig. 4.23(a)

Centre

mg

RESTm

N

Fm r

centrifugal

(= ² )ω

fig. 4.23(b)

T T

(violation of Newton's 2nd law)

To make the Newton�s 2nd law applicable in this frame, we must apply a pseudo force of magnitude mωωωωω²r inthe radially outward direction as shown in figure 4.23(b). This pseudo force is referred to as centrifugal force.



Let us consider the situation discussed in example 7 once again, but this time we would observe from theframe rotating with the ball, i.e., the frame attached with the bowl. F.B.D. of the ball for this frame is shown infigure 4.24. You should notice the difference between the figures 4.19 and 4.24. In 4.19 the ball has anacceleration ω²r towards the centre of the horizontal circle in which it is moving, while in 4.24 the ball is at restand a centrifugal force of magnitude mω²r is acting on it in the radially outward direction. From figure 4.24,we have,

fig. 4.24

²m rω

mg

REST

m

N

N sin θ

mg

REST

m

θ

N cos θ

²m rω

2sin ;θ ω=N m r

and cos .θ =N mg

Note that we have got the same mathematical equations as we had earlier while making observations fromground frame.Let us consider the situation when we take a turn whiledriving the car. At turnings we feel that we are being pushedin the radially outward direction with respect to the curvedroad. If the speed is high then the objects placed on seatsof the car also slip radially outward. For this generallycentrifugal force is held responsible. Same force is maderesponsible for the separation of water from the clothes inthe dryer of a washing machine. There are numerous cases where centrifugal force is held responsible, but you mustquestion that being a pseudo (unreal) force, how is itpossible for it to behave in such a manner. You will getyour answer when we will analyze these cases from aninertial frame. Suppose a box is placed on the back seatof a car which offers negligible friction. Initially the road isstraight then comes a turn. Let us analyze the motion ofthe box both from the car and the ground frames.

fig. 4.25: Motion of the box observed in the car frame

m



In the car frame the seat is at rest but the blockmoves in the radially outward direction, as shownin figure 4.25. If the car is moving with speed v andr be the radius of the circular turn, then we apply apseudo force mω²r along radially outward directionon the box while observing from the car frame andthis force is held responsible for the radially outwardmotion of the box.

fig. 4.26

Now, let us observe the same event from the groundframe. Top view of positions of the car and the boxat different moments are shown in figure 4.26. Youshould notice that as the car goes on a curved partof the road, the box continues to move along astraight line with respect to the road. Here there isenough friction between the tires of the car and theroad which provides the centripetal acceleration tothe car while taking the turn but due to negligiblefriction between the seat and the box, the box doesnot get centripetal force and hence it continues tomove along a straight path. So, the reason ofrelative slipping between the seat and the box isnot the centrifugal force, it is actually due to thelack of centripetal force. Therefore, instead ofsaying that the box is sliding radially outwards, weshould say the car is sliding inwards.

In a rotating frame a centrifugal force is sufficient pseudo force only if the body is at rest. If a body moves ina rotating frame, a second pseudo force which depends upon the velocity of the body must be introduced inorder to use =

! !netF ma in that frame. Called the Coriolis force, it is perpendicular to the velocity of the body

relative to the rotating frame and causes a sideways deflection.

Consider two persons standing along a radial line on a rotating platform and playing catch. If the ball is thrownradially outward, they will see it deflect to the right and the receiver will miss the ball as shown in figure 4.27(b).In an inertial frame the ball travels in a straight line (distance between the two persons is small, so effect ofgravity is neglected) after leaving the thrower and misses the receiver because the receiver is moving as shownin figure 4.27(a). The path of the ball relative to the rotating platform is the curved line shown in the figure. Theball must be thrown to the left of the receiver to take into account this sideways deflection.



v v

v

fig. 4.27(a)

fig. 4.27(b)

fig. 4.27(a)

OBSERVER OBSERVER

OBSERVEROBSERVER

These two pseudo forces, the centrifugal force and Coriolis forces for a rotating frame, have direct applicationto reference frame attached to the earth because of the earth�s rotation. In particular, Coriolis forces areimportant for understanding weather. For example, these forces are responsible for the fact that cyclones arecounterclockwise in the northern hemisphere and clockwise in the southern hemisphere.

At last I would like to remind you that you need not to apply any pseudo force if you are observing from aninertial frame of reference and even if you are observing from a rotating reference frame, you need not to applyCoriolis force if the body is at rest in that frame.



DIRECTIONS: For (Q.1 to Q4): Read the following questions and give answer using the following option (a, b, c and d):

(a) Statement-1 is true, Statement-2 is true: Statement-2 is correct explanation for Statement-1.(b) Statement- is true, Statement-1 is true; Statement-2 is not correct explanation for Statement-1.(c) Statement- is true, Statement-2 is false.(d) Statement- is false, Statement-2 is true.

1. Statement - 1: Force required to move a body uniformly along straight line is zero.

Statement - 2: The force required to move a body uniformly along a circle is zero.

2. Statement - 1: Cream gets separated out of milk when it is churned, it is due to centrifugal force only.

Statement - 2: Centrifugal and gravitational forces play significant role to separate cream from milk.

3. Statement - 1: As the frictional force increases, the safe velocity limit for taking a turn on an unbanked roadalso increases.

Statement - 2: Banking of roads will increase the value of limiting velocity.

4. A particle of mass m is observed from an inertial frame of reference and is found to move in a circle of radius rwith a uniform speed v. The centrifugal force on it is

(a)2mv

rtowards the centre (b)

2mvr

away from the centre

(c)2mv

ralong the tangent through the particle (d) zero.

5. A particle of mass m rotates with a uniform angular speed ω about the z-axis. It is viewed from a frame rotatingabout the z-axis with a uniform angular speed ω0 . The centrifugal force on the particle is

(a) m aω2 (b) 20m aω

(c)2

0

2m aω ω+

(d) 0 .m aω ω

6. A car some times over turns while taking a turn. When it overturns, it is :

(a) the inner wheel leaves the ground first(b) The outer wheel leaves the ground first(c) Both the wheels leave the ground simultaneously(d) Either wheel leaves the ground first.

7. A person stands on a spring balance at the equator.

(a) By what fraction is the balance reading less than his true weight ?(b) If the speed of earth�s rotation is increased by such an amount that the balance reading is half the true

weight, what will be the length of the day in this case?

[Hint : Use centrifugal force]

TRY YOURSELF- IV



8. A sleeve A can slide freely along a smooth rod bent in the shape ofa half-circle of radius R (fig.). The system is set in rotation with aconstant angular velocity ω about a vertical axis OO´. Find the angleθ corresponding to the steady position of the sleeve.

A

R

O´

O

9. A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block iskept in the bowl at a position where the radius makes an angle θ with the vertical. The block rotates with thebowl without any slipping. The friction coefficient between the block and the bowl surface is µ. Find the rangeof the angular speed for which the block will not slip.

[Hint : Make use of situation explained in figure 4.24]



MISCELLANEOUS EXAMPLES

A particle is projected with a speed u at an angle θ with the horizontal. What is the radius of curvature of the parabolatraced out by the projectile at the point where velocity is perpendicular to initial velocity of the projectile.

SOLUTION: Let the ball is projected from the point O. Path of the ball is shown in figure. Let us draw lines parallelto the direction of the initial velocity of the ball. We see that at the point P such a line is perpendicular to thevelocity of the ball at that point. It is clear from the figure that at the point P, the normal acceleration of the ballis g sinθ. Therefore, if v be the speed of the ball at the point P and r be the radius of curvature of the path at thesame point, then we have,

2

sinθ = vgr

vgcosθg sinθ

u

y

xO

P θ

θθθ θ

fig. 4.28

⇒2

sinvr

g θ= ...(i)

As there is no acceleration along the horizontal direction,

we have,

sin cosθ θ=v u

⇒ cotθ=v u ...(ii)

From (i) and (ii), we get,2 2cotsinθ

= urg

An automobile moves with a constant tangential acceleration ta in a horizontal plane circumscribing a circle of radius R.The coefficient of friction between the wheels of the automobile and the surface is µ. What distance will be covered bythe automobile without slipping if its initial speed is zero?

SOLUTION: Let the automobile starts from the position A and when it has covered a distance s, it is at the position Band has a speed v as shown in figure 4.29. At B the radial and the tangential components of the automobile�s

acceleration are 2v

R and ,ta respectively. If m be the mass of the automobile, then, the frictiton force acting on

the automobile is

= netf ma

222( )

= +

t

vm aR

Bv

at

A

S

R

vR

2

fig. 4.29

EXAMPLE : 10

EXAMPLE : 11



As the wheels are not slipping over the horizontal surface, we have,

f < µmg

⇒22

2( ) µ

+ ≤

tvm a mgR

⇒22

2 2 2( ) µ

+ ≤

tv a gR

As ta is tangential acceleration, v is speed and s is distance travelled, we have,

= tdv adt

⇒ ⋅ = tdv ds ads dt

⇒ ⋅ = ⋅tv dv a ds

⇒0 0

[ is constant]⋅ =∫ ∫ ∵v s

t tv dv a ds a

⇒2

Note that this expression is identical2 -(ii) with the expression we use for a uniformly

accelerated motion along a straight linetv a s

=

Using (i) and (ii), we have,

22 2 22 µ + ≤

tt

a s a gR

⇒2 2 2 2

2

4µ −≤ tg R as

⇒ 2 2 2 212

µ≤ − ts g R a

Therefore, maximum value of s without slipping over the horizontal surface is 2 2 2 2

.2

µ − tg R a



A particle moves uniformly with the speed u along a parabolic path 2 ,α=y x where α is a positive constant. Find theacceleration of the particle at x = 0. Also find the radius of curvature of the path at the same point.

SOLUTION: Let us differentiate twice the path equation with respecto to the time:

22 2

2 22 ; 2α = ⋅ = + ⋅

dy dx d y dx d xx a xdt dt dt dt dt

since, the particle moves with uniform speed, its acceleration at all points of the path is purely normal and at thepoint x = 0 its direction of acceleration coincides with the y-axis as shown in figure 4.30.

Therefore, at x = 0:

= =n ya a a

2

2= d ydt

2 2

22 = + ⋅

dx d xa xdt dt

a an=

y x= α 2

x

y

O dxdtu v= =x

fig. 4.30

keeping in mind that at x = 0, ,= =xdxu vdt therefore,

22=a au

If r be the radius of curvature of the path at x = 0, then

2

nuar

= = 22au

⇒ .r = 12a

A particle A moves in one direction along a given trajector with a tangential acceleration �,α= ⋅!ta t where α! is aconstant vecctor coinciding in the direction with the x axis, and �t is a unit vector coinciding in the direction with thevelocity vector at a given point, as shown in figure 4.31. Find how the speed of the particle depends on x provided thatits speed is negligible at the point x = 0.

A α

xO

!

fig. 4.31

�t

EXAMPLE : 13

EXAMPLE : 12



SOLUTION: Some arbitirary position of the partical is shown infigure 4.32. If ds be the magnitude ofdisplacement for thenext time interval dt and θ be the angle between the tangentand the x-axis, then we have, cos .θ =ds dx If v be thespeed of the particle, then, tangential acceleration,

=tdvadt

Aαθ

dx

vds

!dy

fig. 4.32

⇒ � cos [ is manitude of ]dv a tdt

α θ α α= ⋅ = ⋅ !!

⇒ cosα θ⋅ =dv dsds dt

⇒ cosα θ ⋅ = ⋅ = ∵ dsv dv ds v

dt

⇒ [ ]cosα θ⋅ = ⋅ =∵v dv dx ds dx

⇒0 0

⋅ = ⋅∫ ∫v x

v dv a dx

⇒2

2α=v x

⇒ 2α=v x .

A wire of mass m is bent in the shape of a circle of radius r. The wire lies on a smooth horizontal plane and is rotatedwith a constant angular velocity ω about the vertical axis through its centre. Find the tension in the wire.

SOLUTION: Let us single out a small element of the wire of massdm as shown in figure 4.33(a). This element moves alongthe circle due to a force which is sum of two vectors eachof which has the magnitude of the tension sought T, as shownin figure 4.33(b). Therefore, applying Newton�s second law,we get,

2 2 sin2θω⋅ = ⋅ ddm r T

dm

TT

ω²r

rr

dθ

4.33(a)when is very 2

small sin2θθ

θ θ

⋅

##

dT

EXAMPLE : 14



⇒2

2θ ω θ

π × ⋅ = ⋅

m d r T d

⇒ T =2

2mω rπ

T

2T.sin dθ2

T

dθ/2 dθ/2

4.33(b)

A block of mass M rests on a turntable which is rotating at constantangular speed ω. A smooth cord runs from the block through a holein the centre of the table down to a hanging block of mass m. Thecoefficient of friction between M and table is µ. Find the largest andsmallest value of radius r for which M will remain at rest relative tothe turntable.

M

m

r

fig. 4.34

SOLUTION: Let us analyze the given situation from the rotating frame.In this frame both the bodies will be at rest and we will have toapply a centrifugal force on the block of mass M. We will notapply centrifural force on the hanging block because its distancefrom the axis of rotation is zero. Except frictional force, differentforces acting on the two blocks are shown in figure 4.35(a). Mωωωωω²ris the centrifugal force acting on the block resting over thehorizontal surface. For a particular ω, this force is directlyproportional to r. When r is maximum, the block on thehorizontal surface is just about to slide radially outwards and hencethe frictional force acting on it would act radially inwards withmagnitude equal to its limiting value, as shown in figure 4.35(b).Obviously when r is minimum, the block on the horizontal surfacewould be just about to slide radially inwards and hence thefrictional force acting on it would act radially outwards withmagnitude equal to its limiting value because the body is just aboutto slide, as shown in figure 4.35(c).

fig. 4.35(a)

M

m

N

REST

M rω²

Mg

mg

T

T

REST

Calculation for max :r From figure 4.35(b), we have,

2maxω+ =T f M r

⇒ 2maxµ ω+ =T M r M

N

M rω²

Mg

T

f(= µN)

fig. 4.35(b)

⇒ 2maxµ ω+ =T mg M r

⇒ 2maxµ ω+ =mg Mg M r

⇒ max 2 µω

= + g mr

M

EXAMPLE : 15



Calculation for minr : From figure 4.34 (c), we have,

M

N

M rω² min

Mg

T

f (= µN)

fig. 4.35(c)

2minω + =M r f T

⇒ 2minω µ+ =M r N T

⇒ 2minω µ+ =M r Mg mg

⇒ min 2 µω

= − g mr

M .

The motion law for the point A of the rim of a wheel rolling uniformly along a horizontal path (the x-axis) has the form( sin ); (1 cos ),x a t t y a tω ω ω= − = − where a and ω are positive constants. Find the speed v of the point A, the

distance s which it traverses between two successive contacts with the roadbed, as well as the magnitude and thedirection of the acceleration a of the point A.

SOLUTION: We have,

( cos )ω ω ω= = −xdxv a tdt

(1 cos );ω ω= −a t

and sin ;ω ω= =ydyv a tdt

∴ speed, 2 2= +x yv v v

2(1 cos )ω ω= −a t

2 sin ;2

ωω = ta

Motion of the point A is shown in figure.4.36. Let it net is in contact with roadbed at t = 0 and next time itcomes in contact .at 1=t t . Therefore,

1( ) 0=y t

⇒ 1(1 cos ) 0ω− =a t

⇒ 1 2tω π= ⇒ 12= πtω

t = 0

2πa

fig. 4.36

v a = 2 ω

2a

RESTt t = 1

A

C aω

REST

EXAMPLE : 16



You are urged to notice that, when y = 0, vx and vy are also zero, i.e., when point A comes in contact with theroadbed, it is momentarily at rest. You should also notice that when y = 2a, i.e., when the point A is at thehighest point, the y-component of its velocity is zero and the x-component of its velocity is 2aω.

If s be the distance traversed between the moments t = 0 and t = t1, then

1 2 /

0 0

2 sin2

π ω ωω = ⋅ = ⋅ ∫ ∫t ts v dt a dt

2 /

0

22 sin sin is non negative for t 0,2 2

π ω ω ω πωω

= ⋅ ∈ ∫ ∵t ta dt

2 /

04 cos π ωω= −a t

= 8a

Again, we have,

2 sin ;ω ω= = −xn

dva a tdt

2 cos ;ω ω=ya a t

and magnitude 2 2= +x ya a

= 2ω a

Let us show that the acceleration vector, constant in its magnittude, is always directed towards the centre of thewheel, the point C (same as in the case of uniform circular motion). In fact in 'k frame fixed to the point C andtranslating uniformly in the horizontal direction relative to the roadbed the point A moves uniformly along acircle of radius a about the point C. Consequently its acceleration in the 'k frame is directed towards the centreof the wheel. And since the 'k frame moves uniformly, the acceleration vector is same relattive to the roadbed.



EXERCISE

1. In uniform circular motion

(a) both velocity and acceleration are constant

(b) acceleration and speed are constant but velocity changes

(c) both acceleration and velocity change

(d) both acceleration and speed are constant.

2. An aeroplane is taking a turn in a horizontal plane. While doing so it

(a) remians horizontal (b) inclines inwards

(c) inclines outwards (d) makes wings vertical.

3. Which of the following statements is false for a particle moving in a circle with a constant angular speed:

(a) The velocity vector is tangent to the circle

(b) The acceleration vector is tangent to the circle

(c) the acceleration vector points to the centre of the circle

(d) the velocity and acceleration vectors are perpendicular to each other.

4. A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixedcentre at an angular velocity ω0. If the length of the string and angular velocity are doubled, the tension in thestring which was initially T0 is now

(a) 0T (b) 0

2T

(c) 04T (d) 08T .

5. A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2π revolutions

per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is

(a) ML

(b) ML

(c) 4 ML

(d) 16 ML.

LT

R

θ

S

M

6. A stone of mass m tied to a string of length l is rotated in a circle with the other end of the string as the centre.The speed of the stone is v. If the string breaks, the stone will move

(a) towards the centre (b) away from the centre

(c) along a tangent (d) will stop.

OBJECTIVE



7. If the earth stops rotating, the apparent value of g on its surface will

(a) increase everywhere

(b) decrease everywhere

(c) remain the same everywhere some

(d) increase at some places and remain the same at other places.

8. Three identical cars A, B and C are moving at the same speed on three bridges. The car A goes on a planebridge, B on a bridge convex upward and C goes on a bridge concave upward. Let FA, FB and FC be thenormal forces exerted by the cars on the bridges when they are at the middle of bridges.

(a) FA is maximum of the three forces (b) FB is maximum of the three forces.

(c) FC is maximum of the three forces. (d) FA= FB =FC.

9. A train A runs from east to west and another train B of the same mass runs from west to east at the same speedalong the equator. A presses the track with a force F1 and B presses the track with a force F2.

(a) F1 > F2 (b) F1 < F2

(c) F1 = F2

(d) the information is insufficient to find the relation between F1 and F2.

10. A simple pendulum having a bob of mass m is suspended from the ceiling of a car used in a stunt film shooting.The car moves up along an inclined cliff at a speed v and makes a mump to leave the cliff and lands at somedistance. Let R be the maximum height of the car from the top of the cliff. The tension in the string when the caris in air is

(a) mg (b)2mvmg

R−

(c)2mvmg

R+ (d) zero.

11. A circular road of radius r is banked for a speed v = 40 km/hr. A car of mass m attempts to go on the circularroad. The friction coefficient between the tyre and the road is negligible.

(a) The car cannot make a turn without skidding.

(b) If the car turns at a speed less than 40 km/hr, it will slip down.

(c) If the car turns at the correct speed of 40 km/hr, the force by the road on the car is equal to 2

.mvr

(d) If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater than mg

as well as greater than 2mv

r.



12. On a circular table, A and B are moving on the circumference. Man A runs behing man B to catch him. A runswith constant angular speed ω1 with respect to table and B runs at constant tangential speed v2 with respectto ground. If it is found that the table rotates 30° in the opposite direction in every one second and the initialangular separation between A and B is 30°, then A catches B after: (Radius of table is 3 m).

(a) 0.5 sec, if ω1=56π

rad/s and v2 = 3.14 m/s

(b) 0.5 sec, if ω1 = 43π


(c) 0.5 sec, if ω1=43π


(d) A can not catch B within 0.5 s, if ω1 = 6π

rad/s and v2=6.28 m/s

13. A particle is fired from a point on the ground with speed u making an angle θ with the horizontal. Then

(a) the radius of curvature of the projectile at a point where the tangential acceleration becomes zero is2 2cos θu

g

(b) the radius of curvature of the projectile at a point where the tangential acceleration becomes zero is2 2cos θu

g

(c) At the point of projection tangential acceleration is sinθg

(d) At the point of projection tangential acceleration is cosθg .

14. A car is travelling with linear velocity v on a circular road of radius r. If it is increasing its speed at the rate of�a� metre/sec², then the resultant acceleration will be

(a)2

22

v ar

−

(b)

42

2

v ar

+

(c)4

22

v ar

−

(d)

22

2

v ar

+

.

15. A smooth hollow cone whose vertical angle is 2α, with it axis vertical and vertex downwards, revolves aboutits axis n times per second. find where a particle may be placed on the inner surface of cone so that it rotateswith same speed

(a) 2 2

cot4

απ

gn (b) 2 2

sin4

απ

gn

(c)2 24π ng (d) 2 24 sinπ α

gn .



16. A particle is moving in a circle of radius R in such a way that at any instant the normal and tangential componentsof its acceleration are equal. If its speed at t = 0 is v0, the time taken to complete the first revolution is

(a) R/v0 (b) v0/R

(c)2

0

(1 )π−−R ev (d)

2

0

π−R ev .

17. A train is moving with a speed v on a curved railway track of radius r. Aspring balance loaded with a block ofmass m is suspended from the roof of the train. the reading of spring balance is

(a) mg (b) mv²/r

(c) 2( / )+mg mv r (d) 2 2 2[( ) ( ) ]+mg mv r .

18. Two identical particles are attached at the ends of a light string which passes through a hole at the centre of atable. One of the particle is made to move in a circle on the table with angular velocity ω1 and the other is madeto move in a horizontal circle as a contact pendulum with angular velocity ω2. If l1 and l2 are the length of thestring over and under the table, then in order that particle under table neither moves down nor moves up the

ratio 1

2

ll is:

(a)1

2

ωω (b)

2

1

ωω

(c)2122

ωω (d)

2221

ωω .

l

l2

1

m

m

19. A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figureand whose equation is x² = 4ay. The wire frame is fixed and the bead can slide on it without friction. The beadis released from the point y = 4a on the wire frame from rest. The tangential acceleration of the bead when itreaches the position given by y = a is :

(a) 2g

(b)32

g

O x

y

m

(c) 2g

(d) 5g

.

20. A car driver going at some speed suddenly finds a wide wall at a distance r. To avoid hitting the wall he should

(a) apply the brakes

(b) should turn the car in a circle of radius r.

(c) apply the brakes and also turn the car in a circle of radius r.

(d) jump on the back seat.



21. A disc of radius R has a light pole fixed perpendicular to the disc at the circumference which in turn has apendulum of length R attached to its other end as shown in figure. The disc is rotated with a constant angularvelocity ω. The string is making an angle 30° with the rod. Then the angular velocity ω of disc is :

(a)

1 23

gR (b)

1 23

2

gR

30°R

ω

R

(c)1 2

3

gR

(d)1 2

23 3

gR

.

22. A particle is attached to an end of a rigid rod. The other end of the rod is hinged and the rod rotates alwaysremaining horizontal. It�s angular speed is increasing at constant rate. The mass of the particle is �m�. The forceexerted by the rod on the particle is

!F , then :

(a) F > mg

(b) F is constant

(c) The angle between !F and horizontal plane decreases

(d) The angle between !F and the rod decreases.



1. A circular road of radius 50 m has the angle of banking equal to 30º. At what speed should a vehicle go on thisroad so that the friction is not used?

2. The bob of a simple pendulum of length 1m has mass 100 g and a speed of 1.4 m/s at the lowest point in itspath. Find the tension in the string at this instant.

3. Suppose the bob of the previous problem has a speed of 1.4 m/s when the string makes an angle of 0.20 radianwith the vertical. Find the tension at this instant. You can use 2cos 1 / 2θ θ≈ − and sinθ θ≈ for small θ.

4. A pilot comes out of a vertical dive in a circular arc such that her upward acceleration is 9g. (a) If the mass ofthe pilot is 50 kg, what is the magnitude of the force exerted by the airplane seat on her at the bottom of thearc? (b) If the speed of the plane is 320 km/h, what is the radius of the circular arc?

5. Show that the angle made by the string with the vertical in a conical pendulum is given by cosθ = g/Lω², whereL is the length of the string and ω is the angular speed.

6. A small block of mass m slides on a frictionless circular track in a vertical circle of radius R, as shown in Figure.(a) Show that the speed of the block cannot be constant. (b) If the speed of the block at the top of the trackis v, find the force exerted on the block by the track. (c) What is the minimum value of v for the block to satyon the track? If the speed is less than this value, describe the path of the block.

R

v

7. An earth satellite moves in a circular orbit 400 miles above the earth�s surface. The time for one revolution (theperiod) is found to be 98 min. Find the acceleration of gravity at the orbit from these data.

8. Two blocks of mass m1 = 10 kg and m2 = 5 kg, connected to each other by a massless inextensible string oflength 0.3 m are placed along a diameter of a turn table. The coefficient of friction between the table and m1is 0.5 while there is no friction between m2 and table. The table is rotating with angular velocity 10 rad/secabout a vertical axis passing through its centre O. The masses are placed along the O such that the mass m1 isat a distance of 0.124 m from O. The masses are observed to be at rest with respect to an observer on the turntable. Calculate the frictional force on m1.

9. A particle travels with constant speed on a circle of radius3.0 meters and completes one revolution in 20 sec (Fig).Starting from the origin O, find (a) the magnitude anddirection of the displacement vector 5. sec, 7.5 sec, and10 sec later; (b) the magnitude and direction of thedisplacement in the 5.0 sec interval from the fifth to thetenth second; (c) the average velocity vector in this invertal;(d) the instantaneous velocity vector at the beginning andat the end of this interval; (e) the average accelerationvector in this interval; and (f) the instantaneous accelerationvector at the beginning and at the end of this interval.

45°

90°

S

Q

P

O

SUBJECTIVE



10. A particle moves in a plane according tox = R sinωt + ωRt ,y = R cos ωt + R,

where ω and R are constants. This curve, called a cycloid, is the parth traced out by a point on the rim of awheel which rolls without slipping along the x-axis. (a) Sketch the path. (b) Calculate the instantaneous velocityand acceleration when the particle is at its maximum and minimum value of y.

θr

M

11. A partticle of mass M = 0.305 kg moves counterclockwise in a horizonntalcircle of radius r = 2.63 meters with uniform speed v = 0.754 meter/secas in Fig. Determine at the instant θ = 322° (measured counterclockwisefrom the positive x-direction) the following quantities: the x-componentof the velocity; (b) the y-component of the acceleration; (c) the totalforce on the particle; (d) the component of total force on the particle inthe direction of its velocity.

12. A large mass M and a small mass m hangs at the two ends of the stringthat passes through a smooth tube as shown in the figure. The mass mmoves round in a circular path which lies in the horizontal plane. Thelength of the string from the mass m to the top of the tube is l, and θ is theangle this length makes with the vertical. What should be the frequency ofrotation of the mass m so that the mass M remain stationary.

mr

l

M13. A particle moves in a circle of radius 1.0 cm at a speed given by v = 2.0 t where v is in cm/s and t is seconds.

(a) Find the radial acceleration of the particle at t = 1s.(b) Find the tangential acceleration at t = 1s.(c) Find the magnitude of the acceleration at t = 1s.

15. A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate .dv adt

= The

friction coefficient between the road and the tyre is µ. Find the speed at which the car will skid.

14. A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is m.The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated bout thefixed end in the horizontal plane through the fixed end.

(a) What can the maximum angular speed be for which the block does not slip ?(b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration α, at what

angular speed will the block slip?

16. In a children�s park a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rodalways remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod . Let the mass ofeach kid be 15 kg, the distance between the points of the rod where the two kids hold it be 3.0 m and supposethat the rod rotates at the rate of 20 revolutions per minute. Find the force of friction exerted by the rod on oneof the kids.



17. A chain of mass m forming a circle of radius R is slipped on a smooth round cone with half-angle θ. Find thetension of the chain if it rotates with a constant angular velocity ω about a vertical axis coinciding with thesymmetry axis of the cone.

18. A car moves uniformly along a horizontal since curve y = a sin (x/α), where a and α are certain constants. Thecoefficient of friction between the wheels and the road is equal to k. At what velocity will the car ride withoutsliding?

19. A car moves with a constant tangential acceleration 20.62 /w m sτ = along a horizontal surface cir curcumscribinga circle of radius R = 40m. The coefficient of sliding friction between the wheeels of the car and the surface isk = 0.20. What distance will the car ride without sliding if at the initial moment of time its velocity is equal tozero.

20. A cyclist rides along the circumference of a circular horizontal plane of radius R, the friction coefficient being

dependent only on distance r from the centre O of the plane as ( )0 1 /k k r R= − , where k0 is a constant. Findthe radius of the circle with the centre at the point along which the cyclist can ride with the maximum velocity.What is this velocity?

21. A device (Fig.) consists of a smooth L-shaped rod located in a horizontal plane and a sleeve A of mass mattached by a weight less spring to a point B. The spring stiffness is equal to x. The whole system rotates witha constant angular velocity ω about a vertical axis passing through the point O. Find the fractional elongation ofthe spring. How is the result affected by the rotation direction?

A

BO

22. A particle moves along the plane trajectory y (x) with velocity v whose modulus is constant. Find the accelerationof the particle at the point x = 0 and the curvature radius of the trajectory at that point if the trajectory has theform of an ellipse (x/a)2 + (y/b)2 = 1; a and b are constants here.

23. A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity ω in acircular path of radius R . A smooth groove AB of length L (<<R) is made on the surface of the table. Thegroove makes an angle θ with the radius OA of the circle in which the cabin rotates. A small particle is kept atthe point A in the groove and is released to move along AB. Find the time taken by the particle to reach thepoint B.

OR

ABθ



24. A car moving at a speed of 36 km/hr is taking a turn on a circular road of radius 50 m. A small wooden plateis kept on the seat with its plane perpendicular to the radius of the circular road . A small block of mass 100 gis kept on the seat which rests against the plate. The friction coefficient between the block and the plate is m =0.58.

(a) Find the normal contact force exerted by the plane on the block.

(b) The plate is slowly turned so that the angle between the normal to the plate and the radius of the roadslowly increases. Find the angle at which the block will just starts sliding on the plane.

O

25. A particle moves in a plane under the action of a force which is always perpendicular to the particle�s velocityand depends on distance to a certain point on the plane as 1/rn, where n is a constant. At what value of n willthe motion of the particle along the circle be steady ?

26. A chain of length l is placed on a smooth spherical surface of radius R with one of its ends fixed at the top of thesphere. What will be the acceleration w of each element of the chain when its upper end is released ? It is

assumed that the length of the chain l 12

Rπ< .

27. A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor ofthe room on which tthe block moves is smooth but the friction coefficient between the wall and the block is µ.The block is given an initial speed v0. As a function of the speed v write (a) the normal force by teh wall on theblock, (b) the frictional force by the wall and (c) the tangential acceleration of the block. (d) Integrate the

tangential acceleration dv dvvdt ds

= to obtain the speed of the block after one revolution.

28. A particle moves in the plane xy with velocity � �,= +!v pi qxj where �i and �j are the unit vectors of the x andy axes and p and q are constants. At the initial moment of the time, the particle was located at the pointx = y = 0. Find(a) the equation of the particle trajectory y(x)(b) the curvature radius of trajectory as a function of x.

29. A very small cube of mass m is placed on the insideof a funnel (Fig.) rotating about a vertical axis at aconstant rate of v rev/sec. The wall of the funnelmakes an angle θ with the horizontal. If thecoefficient of static friction between the cube andthe funnel is µ and the center of the cube is a distancer from the axis of rotation, what are the largest andsmallest values of v for which the block will notmove with respect to the funnel?

r

axis of rotation

θ



30. A small disc A is placed on an inclined plane forming an angle α with the horizontal (Fig.) and is imparted aninitial velocity v0. Find how the velocity of the disc depends on the angle ϕ if the friction coefficient k = tan αand at the initial moment ϕ0 = π/2.

Aϕ

vα

x

31. A point moves in the plane so that its tangential acceleration ,w aτ = and its normal acceleration 4 ,nw bt=where a and b are positive constants, and t is time. At the moment t = 0 the point was at rest. Find how thecurvature radius R of the point�s trajectory and the total acceleration w depend on the distance covered s.

32. Block A has a mass of MA = 15 kg and B has a mass of MB = 45 kg. They are on a rotating surface and areconnected by a cord passing around the frictionless pulley as shown in figure. If the coefficient of frictionbetween the masses and the surface is µ = 0.25, determine the value of ω at which radial sliding will occur.

AB

30 cm.45 cm.

ω

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