Chapter 10Chapter 10

Rotational motionRotational motion

and Energyand Energy

Rotational MotionRotational Motion Up until now we have been looking at the Up until now we have been looking at the

kinematics and dynamics of kinematics and dynamics of translationaltranslational motion – that is, motion without rotation. Now motion – that is, motion without rotation. Now we will widen our view of the natural world to we will widen our view of the natural world to include objects that both rotate and translate.include objects that both rotate and translate.

We will develop descriptions (equations) that We will develop descriptions (equations) that describe rotational motiondescribe rotational motion

Now we can look at motion of bicycle wheels Now we can look at motion of bicycle wheels and even more!and even more!

II. Rotation with constant angular acceleration

III. Relation between linear and angular variables- Position, speed, acceleration

I. Rotational variables - Angular position, displacement, velocity, acceleration

IV. Kinetic energy of rotation

V. Rotational inertia

VI. Torque

VII. Newton’s second law for rotation

VIII. Work and rotational kinetic energy

Rotational kinematicsRotational kinematics In the kinematics of rotation we encounter

new kinematic quantities Angular displacement Angular speed Angular acceleration Rotational Inertia I Torque

All these quantities are defined relative to an axis of rotation

Angular displacement

Measured in radians or degreesMeasured in radians or degrees There is no dimensionThere is no dimension

= = ff – – i i CWCW

Axis of rotationAxis of rotation

i

fradif 3

2

Angular displacement and arc lengthAngular displacement and arc length

Arc length depends on the distance it is measured away from the axis of rotation

radif 3

2

Axis of rotation

Qsp

sq

Pri

fr

s

qq

pp

rS

rS

Angular SpeedAngular Speed

Angular speed is the rate of change of Angular speed is the rate of change of angular positionangular position

We can also define the We can also define the instantaneous angular speedinstantaneous angular speed

t

t lim

0

t

Average angular velocity and Average angular velocity and tangential speedtangential speed

Recall that speed is distance divided by Recall that speed is distance divided by time elapsedtime elapsed

Tangential speed is arc length divided Tangential speed is arc length divided by time elapsedby time elapsed

And because we can writeAnd because we can write

t

svt

r

v

tr

s

tT

1

r

s

Average Angular AccelerationAverage Angular Acceleration

Rate of change of angular velocityRate of change of angular velocity

Instantaneous angular accelerationInstantaneous angular accelerationtt

12

ttif

tt

0

lim

Angular acceleration and tangential Angular acceleration and tangential accelerationacceleration We can find a link between tangential We can find a link between tangential

acceleration acceleration aatt and angular acceleration and angular acceleration αα

SoSo

r

a

trv

r

v

ttT

if

if

r

aT

Centripetal acceleration

We have thatWe have that

But we also know thatBut we also know that So we can also say So we can also say

r

va Tc

2

222 rvT

rac2

Example: RotationExample: Rotation A dryer rotates at A dryer rotates at 120 rpm120 rpm. What distance do your . What distance do your

clothes travel during one half hour of drying time in clothes travel during one half hour of drying time in a a 70 cm70 cm diameter dryer? What angle is swept out? diameter dryer? What angle is swept out? Distance: Distance: s = s = rr and and = = //tt so so s = s = trtr s = 120 /min x 0.5 h x 60 min/h x 0.35 m s = 120 /min x 0.5 h x 60 min/h x 0.35 m

= 1.3 km= 1.3 km Angle: Angle: t = 120 r/min x 0.5 x 60 min t = 120 r/min x 0.5 x 60 min

= 120x2= 120x2r /min x 0.5 h x 60 min/h = 2.3 x 10r /min x 0.5 h x 60 min/h = 2.3 x 1044 r r

Rotational motion with constant Rotational motion with constant angular accelerationangular acceleration

We will consider cases where We will consider cases where is constant is constant

Definitions of rotational and translational Definitions of rotational and translational

quantities look similarquantities look similar

The kinematic equations describing rotational The kinematic equations describing rotational

motion also look similarmotion also look similar

Each of the translational kinematic equations Each of the translational kinematic equations

has a rotational analoguehas a rotational analogue

Rotational and Translational Rotational and Translational Kinematic EquationsKinematic Equations

2

22

22

1

fi

2f

if

vvv

xavv

avx

avv

i

i tt

t

Constant Constant motion motion

What is the angular acceleration of a car’s wheels What is the angular acceleration of a car’s wheels (radius 25 cm) when a car accelerates from 2 m/s to (radius 25 cm) when a car accelerates from 2 m/s to 5 m/s in 8 seconds?5 m/s in 8 seconds?

srads

srad

t

sradm

sm

r

v

sradm

sm

r

v

if

ff

ii

/5.18

/)820(

/2025.0

/5

/825.0

/2

Example: Centripetal AccelerationExample: Centripetal Acceleration

A 1000 kg car goes around a bend that has a A 1000 kg car goes around a bend that has a radius of 100 m, travelling at 50 km/h. What is the radius of 100 m, travelling at 50 km/h. What is the centripetal force? What keeps the car on the bend? centripetal force? What keeps the car on the bend? [What keeps the skater in the arc?][What keeps the skater in the arc?]

Friction keeps the car and skater on the bendFriction keeps the car and skater on the bend

Nmhkm

kg

r

mvFc 1929

10036001000

5010002

2

Car rounding a bendCar rounding a bend Frictional force of road on tires supplies Frictional force of road on tires supplies

centripetal forcecentripetal force If If ss between road and tires is lowered then between road and tires is lowered then

frictional force may not be enough to provide frictional force may not be enough to provide centripetal force…car will slidecentripetal force…car will slide

Locking wheels makes things worse asLocking wheels makes things worse as

k < s

Banking of roads at corners reduces the risk Banking of roads at corners reduces the risk of skidding…of skidding…

Car rounding a bendCar rounding a bend Horizontal component of the normal force of Horizontal component of the normal force of

the road on the car can provide the the road on the car can provide the centripetal forcecentripetal force

If

then no friction is requiredthen no friction is required

Fg

NNcos

Nsin

r

mvN

2

sin

Rotational DynamicsRotational Dynamics

Easier to move door at Easier to move door at AA than at than at BB using the using the same force same force FF

More More torquetorque is exerted at is exerted at AA than at than at BB

A B

hinge

TorqueTorque

Torque is the rotational analogue of ForceTorque is the rotational analogue of Force

Torque, Torque, , is defined to be, is defined to be

Where Where FF is the force applied is the force applied tangent to the tangent to the rotationrotation and and rr is the distance from the axis of is the distance from the axis of rotationrotation

r

F

= Fr

Torque A general definition of torque isA general definition of torque is

Units of torque are NmUnits of torque are Nm

Sign convention used with torqueSign convention used with torque Torque is positive if object tends to rotate CCWTorque is positive if object tends to rotate CCW Torque is negative if object tends to rotate CWTorque is negative if object tends to rotate CW

r

F = Fsin r

Condition for Equilibrium

We know that if an object is in (translational) We know that if an object is in (translational) equilibrium then it does not accelerate. We equilibrium then it does not accelerate. We can say that can say that F = 0F = 0

An object in rotational equilibrium does not An object in rotational equilibrium does not change its rotational speed. In this case we change its rotational speed. In this case we can say that there is no net torque or in other can say that there is no net torque or in other words that:words that:

= 0= 0

An unbalanced torque (An unbalanced torque () gives rise to an ) gives rise to an angular acceleration (angular acceleration ())

We can find an expression analogous to We can find an expression analogous to F = maF = ma that relates that relates and and

We can see thatWe can see that

FFtt = ma = matt

and and FFttr = mar = mattr = mrr = mr22(since (since aatt = r = r

Therefore Therefore

Torque and angular accelerationTorque and angular acceleration

mrFt

= mr2

Torque and Angular Acceleration

Angular acceleration is directly proportional to the net torque, but the constant of proportionality has to do with both the mass of the object and the distance of the object from the axis of rotation – in this case the constant is mr2

This constant is called the moment of inertia. Its symbol is I, and its units are kgm2

I depends on the arrangement of the rotating system. It might be different when the same mass is rotating about a different axis

Newton’s Second Law for Rotation

Now we have

Where I is a constant related to the distribution of mass in the rotating system

This is a new version of Newton’s second law that applies to rotation

= I

Angular Acceleration and IAngular Acceleration and I

The angular acceleration reached by a The angular acceleration reached by a rotating object depends on, rotating object depends on, MM, , rr, (their , (their distribution) and distribution) and TT

When objects are rolling under the influence When objects are rolling under the influence of gravity, only the mass distribution and the of gravity, only the mass distribution and the radius are importantradius are important

T

Moments of Inertia for Rotating Moments of Inertia for Rotating ObjectsObjects

II for a small mass for a small mass mm rotating about a point a rotating about a point a distance distance rr away is away is mrmr22

What is the moment of inertia for an object that What is the moment of inertia for an object that is rotating –such as a rolling object?is rotating –such as a rolling object?

Disc?Sphere?Hoop?Cylinder?

Moments of Inertia for Rotating Moments of Inertia for Rotating ObjectsObjects

The total torque on a rotating system is the sum of the torques acting on all particles of the system about the axis of rotation –

and since is the same for all particles:

I mr2 = m1r12+ m2r2

2+ m3r32+…

Axis of rotation

Continuous ObjectsContinuous Objects

To calculate the moment of inertia for To calculate the moment of inertia for continuous objects, we imagine the object to continuous objects, we imagine the object to consist of a continuum of very small mass consist of a continuum of very small mass elements elements dmdm. Thus the finite sum . Thus the finite sum ΣmΣmi i rr22

ii

becomes the integralbecomes the integral

dmrI 2

Moment of Inertia of a Uniform RodMoment of Inertia of a Uniform Rod

L

Lets find the moment of inertia of a uniform Lets find the moment of inertia of a uniform rod of length rod of length LL and mass and mass MM about an axis about an axis perpendicular to the rod and through one end. perpendicular to the rod and through one end. Assume that the rod has negligible thickness. Assume that the rod has negligible thickness.

Moment of Inertia of a Uniform RodMoment of Inertia of a Uniform Rod We choose a mass element We choose a mass element dmdm at a distance at a distance xx from the axes. The mass per unit length (linear from the axes. The mass per unit length (linear mass density) is mass density) is λ = M / Lλ = M / L

Moment of Inertia of a Uniform RodMoment of Inertia of a Uniform Rod

dm = λ dx

dxL

Mdm

L L L

dxxL

Mdx

L

MxdmxI

0 0 0

222

23

0

3

3

1

33

1ML

L

L

Mx

L

M L

Example:Example:Moment of Inertia of a DumbbellMoment of Inertia of a Dumbbell

A dumbbell consist of point masses A dumbbell consist of point masses 2kg2kg and and 1kg1kg attached by a rigid massless rod of length attached by a rigid massless rod of length 0.6m0.6m. . Calculate the rotational inertia of the dumbbell Calculate the rotational inertia of the dumbbell (a) about the axis going through the center of the (a) about the axis going through the center of the mass and (b) going through the 2kg mass. mass and (b) going through the 2kg mass.

Example:Example:Moment of Inertia of a DumbbellMoment of Inertia of a Dumbbell

mkgkg

mkgkg

mm

xmxmX 2.0

12

)6.0)(1()0)(2(

21

2211

222

222

211

24.0)2.06.0)(1()2.00)(2(

)()()(

kgmmmkgmmkg

XxmXxmIa CMCMCM

Example:Example:Moment of Inertia of a DumbbellMoment of Inertia of a Dumbbell

2222 4.0)6.0)(1()( kgmmkgLmIb

Moment of Inertia of a Uniform HoopMoment of Inertia of a Uniform Hoop

R

dmAll mass of the hoop M is at distance r = R from the axis

2222 MRdmRdmRdmrI

Moment of Inertia of a Uniform DiscMoment of Inertia of a Uniform Disc

R

dr

Each mass element is a hoop of radius r and thickness dr. Mass per unit area

σ = M / A = M /πR2

r

We expect that I will be smaller than MR2 since the mass is uniformly distributed from r = 0 to r = R rather than being concentrated at r=R as it is in the hoop.

Moment of Inertia of a Uniform Moment of Inertia of a Uniform DiscDisc

R

dr

r

dAdm

R R

drrrdrrdmrI0 0

322 22

242

4

2

1

24

2MRR

R

MR

A

MI

A

M

Moments of inertia Moments of inertia II for Different Mass for Different Mass ArrangementsArrangements

Moments of inertia Moments of inertia II for Different Mass for Different Mass ArrangementsArrangements

Kinetic energy of rotationKinetic energy of rotation

Reminder:Reminder: Angular velocityAngular velocity, , ωω is the same for all is the same for all particles within the rotating body. particles within the rotating body.

Linear velocityLinear velocity,, vv of a particle within the rigid body of a particle within the rigid body depends on the particle’s distance to the rotation axis (r).depends on the particle’s distance to the rotation axis (r).

2222

233

222

211

2

1)(

2

1

2

1

...2

1

2

1

2

1

i

ii

i

ii

i

ii rmrmvm

vmvmvmK

Moment of InertiaMoment of Inertia

Kinetic energy of a body in pure rotation

Kinetic energy of a body in pure translationKinetic energy of a body in pure translation

Which one will win?Which one will win?

A hoop, disc and sphere are all rolled down an inclined plane. Which one will win?

1. Hoop I = MR2

2. Disc I = ½MR2

3. Sphere I = 2/5MR2

= / I

1. 1 = / MR2

2. α2= 2(/ MR2)

. = 2.5(/ MR2)

I. Rotational variablesI. Rotational variables

Rigid body: body that can rotate with all its parts locked body that can rotate with all its parts locked together and without shape changes.together and without shape changes.

r

s

radius

lengtharc

Rotation axis: every point of a body moves in a circle whose every point of a body moves in a circle whose center lies on the rotation axis. Every point moves through center lies on the rotation axis. Every point moves through the same angle during a particular time interval. the same angle during a particular time interval.

Angular position:Angular position: the angle of the reference line relative to the angle of the reference line relative to the positive direction of the x-axis.the positive direction of the x-axis.

Units: Units: radians (rad)radians (rad)

Reference line:Reference line: fixed in the body, perpendicular to the fixed in the body, perpendicular to the rotation axis and rotating with the body.rotation axis and rotating with the body.

revrad

radr

rrev

159.03.571

22

3601

Note:Note: we do not reset we do not reset θθ to to zero with each complete zero with each complete rotation of the reference line rotation of the reference line about the rotation axis. 2 about the rotation axis. 2 turns turns θθ =4=4ππ

Translation:Translation: body’s movement described by x(t).body’s movement described by x(t).

Rotation:Rotation: body’s movement given by body’s movement given by θθ(t) = angular position (t) = angular position of the body’s reference line as function of time.of the body’s reference line as function of time.

Angular displacement:Angular displacement: body’s rotation about its axis body’s rotation about its axis changing the angular position from changing the angular position from θθ11 to to θθ22..

12 Clockwise rotation Clockwise rotation negative negativeCounterclockwise rotation Counterclockwise rotation positive positive

Angular velocity:

tttavg

12

12

dt

d

tt

0

lim

Average:Average:

Instantaneous:Instantaneous:

Units:Units: rad/s or rad/s or rev/srev/s

These equations hold not only for the rotating rigid body These equations hold not only for the rotating rigid body as a whole but also for every particle of that body as a whole but also for every particle of that body because they are all locked together.because they are all locked together.

Angular speed (Angular speed (ωω)):: magnitude of the angular velocity.magnitude of the angular velocity.

Angular acceleration:

tttavg

12

12

dt

d

tt

0

lim

Average:

Instantaneous:

Angular quantities are “normally” vector Angular quantities are “normally” vector quantities quantities right hand rule. right hand rule.

Object rotates around the direction of Object rotates around the direction of the vectorthe vector a vector defines an axis a vector defines an axis of rotation not the direction in which of rotation not the direction in which something is moving.something is moving.

Examples:Examples: angular velocity, angular velocity, angular accelerationangular acceleration

Exception: angular displacements

II. Rotation with constant angular II. Rotation with constant angular accelerationacceleration

20

00

020

2

200

0

2

1

)(2

1

)(2

2

1

atvtxx

tvvxx

xxavv

attvxx

atvv

Linear equationsLinear equations Angular equations Angular equations

20

00

020

2

200

0

2

1

)(2

1

)(2

2

1

tt

t

tt

at

III. Relation between linear and angular variablesIII. Relation between linear and angular variables

Position:Position: rs θθ always in radians always in radians

Speed:Speed: rvdt

dr

dt

ds

ωω in rad/s in rad/s

Since all points within a rigid body have the Since all points within a rigid body have the same angular speed same angular speed ωω, points with greater , points with greater radius have greater linear speed, radius have greater linear speed, vv. .

vv is tangent to the circle in which a point is tangent to the circle in which a point movesmoves

If If ωω=cte =cte vv=cte =cte each point within the body undergoes each point within the body undergoes uniform circular motionuniform circular motion

Period of revolution:Period of revolution: 22

v

rT

Acceleration:Acceleration:

rarrdt

d

dt

dvt

Tangential component Tangential component of linear accelerationof linear acceleration

Radial component ofRadial component oflinear acceleration:linear acceleration:

rr

r

r

var 2

222

Responsible for changes in the direction of the linear velocity vector v

Units:Units: m/sm/s22

Kinetic energy of rotationKinetic energy of rotation

Reminder:Reminder: Angular velocityAngular velocity, , ωω is the same for all is the same for all particles within the rotating body. particles within the rotating body.

Linear velocityLinear velocity,, vv of a particle within the rigid body of a particle within the rigid body depends on the particle’s distance to the rotation axis (r).depends on the particle’s distance to the rotation axis (r).

2222

233

222

211

2

1)(

2

1

2

1

...2

1

2

1

2

1

i

ii

i

ii

i

ii rmrmvm

vmvmvmK

Moment of InertiaMoment of Inertia

Rotational inertia = Moment of inertia, I:Rotational inertia = Moment of inertia, I:

i

iirmI 2

Indicates how the mass of the rotating body is Indicates how the mass of the rotating body is distributed about its axis of rotation.distributed about its axis of rotation.

The moment of inertia is a constant for a particular The moment of inertia is a constant for a particular rigid body and a particular rotation axis.rigid body and a particular rotation axis.

Units:Units: kg mkg m22

Example:Example: long metal rod.long metal rod.

Smaller rotational inertia in Smaller rotational inertia in (a) (a) easier to rotate. easier to rotate.

Kinetic energy of a body in pure rotation

Kinetic energy of a body in pure translationKinetic energy of a body in pure translation

Discrete rigid body I =∑miri

2

Continuous rigid body I = ∫r2 dm

Parallel axis theoremParallel axis theorem

2MhII COM

Proof:Proof:

Rotational inertia about a given axisRotational inertia about a given axis = = Rotational Inertia about a parallel axis Rotational Inertia about a parallel axis that extends trough body’s Center of that extends trough body’s Center of Mass + MhMass + Mh22

h = perpendicular distance between the given axis and axis h = perpendicular distance between the given axis and axis through COM.through COM.

dmbaydmbxdmadmyx

dmbyaxdmrI

)(22)(

)()(

2222

222

R

222 22 MhIMhbMyaMxdmRI COMCOMCOM

FrFrFrFr t )sin()sin(

Units:Units: NmNm

Tangential component, FTangential component, Ftt:: does cause rotation does cause rotation pulling a pulling a

door perpendicular to its plane. door perpendicular to its plane. FFtt= F sin= F sinφφ

Radial component, Fr : does not cause rotation pulling a door parallel to door’s plane.

TorqueTorque

Torque: Torque:

Twist “Turning action of force F ”.

rr┴┴ : Moment arm of F : Moment arm of F

r : Moment arm of Fr : Moment arm of Ftt

Sign:Sign: Torque >0 if body rotates counterclockwise. Torque >0 if body rotates counterclockwise. Torque <0 if clockwise rotation.Torque <0 if clockwise rotation.

Superposition principle:Superposition principle: When several torques act on When several torques act on a body, the net torque is the sum of the individual a body, the net torque is the sum of the individual torquestorques

Vector quantityVector quantity

Newton’s second law for rotationNewton’s second law for rotation

ImaF

Proof:Proof:

Particle can move only along the circular path only the tangential component of the force Ft (tangent to the circular path) can accelerate the particle along the path.

ImrrrmrmarF

maF

tt

tt

)()( 2

Inet

VII. Work and Rotational kinetic energy

Translation Rotation

WmvmvKKK ifif 22

2

1

2

1WIIKKK ifif 22

2

1

2

1

f

i

x

x

FdxW f

i

dW

Work-kinetic energy Theorem

Work, rotation about fixed axis

dFW )( ifW Work, constant torque

vFdt

dWP

dt

dWP Power, rotation about

fixed axis

Proof:

2222222222

2

1

2

1)(

2

1)(

2

1)(

2

1)(

2

1

2

1

2

1ififififif IImrmrrmrmmvmvKKKW

f

i

dWddrFdsFdW tt

dt

d

dt

dWP

Rotational Kinetic EnergyRotational Kinetic Energy

We must rewrite our statements of conservation of We must rewrite our statements of conservation of mechanical energy to include KEmechanical energy to include KErr

Must now allow that (in general):Must now allow that (in general):

½ mv2+mgh+ ½ I2 = constant

Could also add in e.g. spring PECould also add in e.g. spring PE

Example - Rotational KE

What is the linear speed of a ball with radius 1 cm when it reaches the end of a 2.0 m high 30o incline?

mgh+ ½ mv2+ ½ I2 = constant Is there enough information?

2 m

Example - Rotational KE

So we have that

The velocity of the centre of mass and the

tangential speed of the sphere are the same,

so we can say that:

Rearranging for vf:

2

5

2MRISphere

222

222

5

1

2

15

2

2

1

2

1

ffi

ffi

Rvgh

MRMvMgh

R

v

R

vt

R

v

R

vt

smgh

v

vvvgh

if

fffi

/3.57.0

28.9

7.0

7.05

1

2

1 222

Example: Conservation of KEExample: Conservation of KErr

A boy of mass 30 kg is flung off the edge of a A boy of mass 30 kg is flung off the edge of a roundabout (m = 400 kg, r = 1 m) that is travelling at roundabout (m = 400 kg, r = 1 m) that is travelling at 2 rpm. What is the speed of the roundabout after he 2 rpm. What is the speed of the roundabout after he falls off?falls off?

Roundabout is a disk:Roundabout is a disk: 222 200)1()400(5.02

1kgmmkgMRIR

222 30)1)(30( kgmmkgRMI BB 222 23030200 kgmkgmkgmIII BRtotal

srad

KEKEkgmIKE

JkgmIKE

f

fiffff

iii

/22.02005.0

5

)200(2

1

2

1

560

22)230(5.0

2

1

222

222

Boy has

10.1 10.1 During a certain period of time, the angular position During a certain period of time, the angular position of a swinging door is described by of a swinging door is described by

θθ= 5.00 + 10.0= 5.00 + 10.0tt + 2.00 + 2.00tt22

where where θθ is in radians and is in radians and tt is in seconds. Determine the is in seconds. Determine the angular position, angular speed, and angular angular position, angular speed, and angular acceleration of the door (a) at acceleration of the door (a) at tt = 0 = 0 and (b) at and (b) at tt = 3.00 s = 3.00 s..

Solution:Solution:

0 5.00 radt

0 00

20

0

10.0 4.00 10.0 rad s

4.00 rad s

t tt

tt

dt

dt

ddt

(a)(a)

During a certain period of time, the angular position of a During a certain period of time, the angular position of a swinging door is described by swinging door is described by

θθ= 5.00 + 10.0= 5.00 + 10.0tt + 2.00 + 2.00tt22

where where θθ is in radians and is in radians and tt is in seconds. Determine the is in seconds. Determine the angular position, angular speed, and angular angular position, angular speed, and angular acceleration of the door (a) at acceleration of the door (a) at tt = 0 = 0 and (b) at and (b) at tt = 3.00 s = 3.00 s..

Solution:Solution:

(b)(b)

3.00 s 5.00 30.0 18.0 53.0 radt

3.00 s 3.00 s3.00 s

23.00 s

3.00 s

10.0 4.00 22.0 rad s

4.00 rad s

t tt

tt

dt

dt

ddt

10.8.10.8. A rotating wheel requires A rotating wheel requires 3.00 s3.00 s to rotate through to rotate through 37.0 revolutions37.0 revolutions. Its angular speed at the end of the . Its angular speed at the end of the 3.00-s3.00-s interval is interval is 98.0 rad/s98.0 rad/s. What is the constant . What is the constant angular acceleration of the wheel?angular acceleration of the wheel?

212fi it t

and fi t are two equations in two unknowns

i

2 21 12 2fi ff t t t t t

22 rad 137.0 rev 98.0 rad s 3.00 s 3.00 s

1 rev 2

2232 rad 294 rad 4.50 s

22

61.5 rad13.7 rad s

4.50 s

10.21.10.21. The four particles are The four particles are connected by rigid rods of connected by rigid rods of negligible mass. The origin is at negligible mass. The origin is at the center of the rectangle. If the center of the rectangle. If the system rotates in the the system rotates in the xyxy plane about the plane about the zz axis with an axis with an angular speed of angular speed of 6.00 rad/s6.00 rad/s, , calculate (a) the moment of calculate (a) the moment of inertia of the system about the inertia of the system about the zz axis and (b) the rotational axis and (b) the rotational kinetic energy of the system.kinetic energy of the system.

The four particles are connected by The four particles are connected by rigid rods of negligible mass. The rigid rods of negligible mass. The origin is at the center of the rectangle. origin is at the center of the rectangle. If the system rotates in the If the system rotates in the xyxy plane plane about the about the zz axis with an angular axis with an angular speed of speed of 6.00 rad/s6.00 rad/s, calculate (a) the , calculate (a) the moment of inertia of the system about moment of inertia of the system about the the zz axis and (b) the rotational axis and (b) the rotational kinetic energy of the system.kinetic energy of the system.

(a)(a)2

j jj

I mr

1 2 3 4

2 2

2

2

3.00 m 2.00 m 13.0 m

13.0 m 3.00 2.00 2.00 4.00 kg

143 kg m

r r r r

r

I

The four particles are connected by The four particles are connected by rigid rods of negligible mass. The rigid rods of negligible mass. The origin is at the center of the rectangle. origin is at the center of the rectangle. If the system rotates in the If the system rotates in the xyxy plane plane about the about the zz axis with an angular axis with an angular speed of speed of 6.00 rad/s6.00 rad/s, calculate (a) the , calculate (a) the moment of inertia of the system about moment of inertia of the system about the the zz axis and (b) the rotational axis and (b) the rotational kinetic energy of the system.kinetic energy of the system.

In this case,

(b(b)) 22 21 1

143 kg m 6.00 rad s2 2RK I

32.57 10 J

29.29. Many machines employ cams for various purposes, such Many machines employ cams for various purposes, such as opening and closing valves. In Figure P10.29, the cam is a as opening and closing valves. In Figure P10.29, the cam is a circular disk rotating on a shaft that does not pass through the circular disk rotating on a shaft that does not pass through the center of the disk. In the manufacture of the cam, a uniform solid center of the disk. In the manufacture of the cam, a uniform solid cylinder of radius cylinder of radius RR is first machined. Then an off-center hole of is first machined. Then an off-center hole of radius radius RR/2 is drilled, parallel to the axis of the cylinder, and /2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance centered at a point a distance RR/2 from the center of the cylinder. /2 from the center of the cylinder. The cam, of mass The cam, of mass MM, is then slipped onto the circular shaft and , is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the cam when it welded into place. What is the kinetic energy of the cam when it is rotating with angular speed about the axis of the shaft?is rotating with angular speed about the axis of the shaft?

We consider the cam as the superposition of the original solid disk and a disk of negative mass cut from it. With half the radius, the cut-away part has one-quarter the face area and one-quarter the volume and one-quarter the mass of the original solid cylinder:

By the parallel-axis theorem, the original cylinder had moment of inertia

The negative mass portion has :

The whole cam has:

and

0 014

M M M 043

M M

2 22 2

CM 0 0 0 01 3

2 2 4 4R R

I M M R M M R

2 20

01 12 4 2 32

M RRI M

22 2 2 20

0 03 23 23 4 234 32 32 32 3 24

M RI M R M R MR MR

2 2 2 2 21 1 23 232 2 24 48

K I MR MR

10.31.10.31. Find the net torque on the wheel in Figure Find the net torque on the wheel in Figure about the axle through about the axle through OO if if aa = 10.0 cm and = 10.0 cm and bb = 25.0 = 25.0 cm.cm.

0.100 m 12.0 N 0.250 m 9.00 N 0.250 m 10.0 N 3.55 N m

The thirty-degree angle is unnecessary information.

37.37. A block of mass m1 = 2.00 kg and a block of mass m2 = 6.00 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. These blocks are allowed to move on a fixed block-wedge of angle = 30.0 as in Figure P10.37. The coefficient of kinetic friction is 0.360 for both blocks. Draw free-body diagrams of both blocks and of the pulley. Determine (a) the acceleration of the two blocks, and (b) the tensions in the string on both sides of the pulley.

For m1: y yF ma1 0n m g

1 1 19.6 Nn m g

1 1 7.06 Nk kf n

x xF ma 17.06 N 2.00 kgT a

For pulley, I 2

1 212

aT R T R MR

R

1 21

10.0 kg2

T T a 1 2 5.00 kgT T a

For m2:

2 2 cos 0n m g

22 6.00 kg 9.80 m s cos30.0

50.9 N

n

2 2k kf n 18.3 N

2 2 218.3 N sinT m m a

218.3 N 29.4 N 6.00 kgT a

(a) Add equations for m1, m2, and for the pulley:

17.06 N 2.00 kgT a

1 2 5.00 kgT T a

218.3 N 29.4 N 6.00 kgT a

2

7.06 N 18.3 N 29.4 N 13.0 kg

4.01 N0.309 m s

13.0 kg

a

a

(b)

21 2.00 kg 0.309 m s 7.06 N 7.67 NT

22 7.67 N 5.00 kg 0.309 m s 9.22 NT