Ch. 3: Kinetics of Particles - Chulalongkorn University...

Ch. 3: Kinetics of Particles

3.0 Outline

IntroductionNewton’s Second LawEquations of MotionRectilinear MotionCurvilinear Motion

3.0 Outline


3.1 Introduction

3.1 Introduction

Kinetics is the study of the relations between the forcesand the motion. Here we will not seriously concernwhether the forces cause the motion or the motiongenerates the forces (causality).

In this chapter, the focus is on the particles. That is thebody whose physical dimensions are so small comparedwith the radius of curvature of its path.

There are at least 3 approaches toe the solution ofkinetic problems: (a) Newton’s second law (b) work andenergy method (c) impulse and momentum method.


3.2 Newton’s Second Law

3.2 Newton’s Second Law

For most engineering problems on earth, the accelerationmeasured w.r.t. reference frame fixed to the earth’ssurface may be treated as absolute. And Newton’s 2nd

law of motion holds.

Newton’s 2nd law breaks when the velocities of the orderof the speed of light are involved theory of relativity

mm mass (resistance to rate of change of velocity) of the

acting on the particleresulting acceleration measured in a of reference

particleresultant force

nonaccelerating frame

=F = a

F =a =


3.3 Equation of Motion and Solution of Problems

3.3 Equation of Motion and Solution

Two problems of dynamics(1) specified kinematic conditions, find forces

straightforward application of Newton’s law asalgebraic equations

(2) specified forces, find motion Difficulty depends on the form of force function(t, s, v, a), as the solutions are found by solvinga system of differential equations.For simple functions, we can find closed form solutionsof motion as in rectilinear motion (sec. 2.2).

m --- equation of motionscalar components decomposition according to a specified coordinate∑F = a



Unconstrained motionMotion of the particle is determined by its initial motion andthe forces from external sources. It is free of constraintsand so has three degrees of freedom to specifythe position. Three scalar equations of motion wouldhave to be applied and integrated to obtain the motion.

Constrained motionMotion of the particle is partially or totally determined byrestraining guides, other than its initial motion and theforces from external sources. Therefore, all forces, bothapplied and reactive, that act on the particle must beaccounted for in Newton’s law. The number of d.o.f. and equations are reduced regarding to the type of constraints.



Free body diagramAll forces acting on the particle needed to be accountedin the equations of motion. Free body diagram unveilsevery force that acts on the isolated particle. Only afterthe FBD has been completed should the equations ofmotion be written. The appropriate coordinate axes anddirections should be indicated and consistently usedthroughout the problem.

Treatment of the body as particle is valid when the forcesmay be treated as concurrent through the mass center.


3.4 Rectilinear Motion


x x y z

If the x-axis is the direction of the rectilinear motion,F ma F 0 F 0

If we are not free to choose a coordinate direction along the motion,the nonzero acceleration component will be sh

= = =∑ ∑ ∑

x x y y z z

A B A

own up in all equations:F ma F ma F ma

Other coordinate system such as n-t or r-

may be determined via the use of relative motionFor pure translating moving reference frame

θ

= = =∑ ∑ ∑

a

a = a + a /B



P. 3/17 The coefficient of static friction between the flatbed of the truck and the crate it carries is 0.30.Determine the minimum stopping distance s thatthe truck can have from a speed of 70 km/h withconstant deceleration if the crate is not toslip forward.



P. 3/17

If the crate is not to slip, crate and truck must have same acceleration.If the crate is not to slip, friction static friction at impending status.Minimum stopping distance when the deceleration is th

=

( ) ( )

x x x x

22 2

o o

e max allowable value.

F ma 0.3mg ma , a 0.3g constant for minimum distance

10v v 2a s s 0 70 2 0.3g s, s 64.2 m36

⎡ ⎤= − = = −⎣ ⎦

⎛ ⎞⎡ ⎤= + − = × + − =⎜ ⎟⎣ ⎦ ⎝ ⎠

∑

mg

N

F < 0.3N

+x



P. 3/18 If the truck of Prob. 3/17 comes to stop from an initial forwardspeed of 70 km/h in a distance of 50 m with uniform deceleration,determine whether or not the crate strikes the wall at the forwardend of the flat bed. If the crate does strike the wall, calculate itsspeed relative to the truck as the impact occurs. Use the frictioncoefficients μs = 0.3 and μk = 0.25.



P. 3/18

( )

( )

22 2 2

o o truck truck

o o stop

stopping distance 50 m, which is less than minimum value 64.2 mthe crate slips

10v v 2a s s 0 70 2a 50, a 3.781 m/s36

10v v a t t 0 70 3.781 t, t 5.36

=∴

⎛ ⎞⎡ ⎤= + − = × + × = −⎜ ⎟⎣ ⎦ ⎝ ⎠⎛ ⎞= + − = × − × =⎡ ⎤ ⎜ ⎟⎣ ⎦ ⎝ ⎠

( )

s k

truck crate

x x s

k

k crate c

14 s

Friction force: F 0.3mg 2.943m and F 0.25mg 2.45mAssume crate and truck go together a a

F ma F m 3.781 required friction 3.781m F

the crate slips and F F

F ma , a

= = = =→ =

⎡ ⎤= − = − → = >⎣ ⎦∴ =

− =

∑

[ ] ( )

( ) ( )

2rate

2crate/truck crate truck crate/truck

2 2o o o o

2.45 m/s

a a a a 2.45 3.781 1.331 m/sthe crate slips forward but will it strike the wall?

1s s v t t a t t relative motion calculation2

13 1.3312

= −

= − = − − − =

∴

⎡ ⎤= + − + −⎢ ⎥⎣ ⎦

= ×

( )

2strike stop

o o

crate/truck

t , t 2.123 s t crate will strike the wall before the truck stops

v v a t t relative motion calculation

v 0 1.331 2.123 2.826 m/s

× = < ∴

= + −⎡ ⎤⎣ ⎦= + × =

N

F < 0.3N

+x



P. 3/23 If the coefficients of static and kinetic frictionbetween the 20-kg block A and the 100-kg cart Bare both essentially the same value of 0.50,determine the acceleration of each part for(a) P = 60 N and (b) P = 40 N.



P. 3/23

A max A

2x x A A

2B B

max

x

(a) N 20g, F 0.5N 98.1 N 120 Nblock A moves forward relative to B

F ma 120 98.1 20a , a 1.095 m/s

98.1 100a , a 0.981 m/s

(b) F 80 N block A does not move relative to B

F

= = = <∴

⎡ ⎤= − = =⎣ ⎦= =

> ∴

=

∑

∑ x

2

max

ma A & B move together

80 120a, a 0.667 m/sFind developed friction by isolated FBD at A or B80 F 20a, F 66.67 N F assumption is validF 100a, F 66.67 N

⎡ ⎤⎣ ⎦= =

− = = < ∴= =

2P20g

100g

NB

NA NAF

F



P. 3/24 A simple pendulum is pivoted at O and is free toswing in the vertical plane of the plate. If theplate is given a constant acceleration a up theincline θ, write an expression for the steadyangle β assumed by the pendulum after allinitial start-up oscillations have ceased. Neglectthe mass of the slender supporting rod.



P. 3/24

y

x x

1

F 0 Tcos mgcos 0

F ma Tsin mgsin ma

a gsintangcos

β θ

β θ

θβθ

−

⎡ ⎤= − =⎣ ⎦⎡ ⎤= − =⎣ ⎦

⎛ ⎞+= ⎜ ⎟

⎝ ⎠

∑∑

T

mgθ

βx

y



P. 3/28 For the friction coefficients μs = 0.25 andμk = 0.20, calculate the acceleration of eachbody and the tension T in the cable.



P. 3/28

A B A B

max s

max

max

s 2s c a 2a 0N 60gcos30, F N 127.4 NAssume motion impends at block A F F and equilibrium

F 0 60gsin30 F T 0, T 166.9 N

but cylinder B will not be in equilibrium 20g 2T 0 move

lμ

+ + = → + == = =

→ =

⎡ ⎤= − − = =⎣ ⎦− < →

∑( )

k A B

2 2B B A

upAssum block A slides down and block B moves up

F ma 60gsin30 F T 60a 120a

20g 2T 20a , T 105.35 N, a 0.725 m/s , a 1.45 m/s

⎡ ⎤= − − = = −⎣ ⎦− = = = − =

∑

T

2T

20g

60g

NF



P. 3/35 A bar of length l and negligible mass connects the cart of massM and the particle of mass m. If the cart is subjected to a constantacceleration a to the right, what is the resulting steady-stateangle θthat the freely pivoting bar makes with the vertical?Determine the net force P (not shown) that must be applied tothe cart to cause the specified acceleration.



P. 3/35

y

1x x

x x

From the given statements, pendulum and cart have same accelerationAt the pendulum,

F 0 Tcos mg 0, T mg/cos

aF ma Tsin ma, tang

At the cart,

F ma P Tsin Ma, P

θ θ

θ θ

θ

−

⎡ ⎤= − = =⎣ ⎦⎛ ⎞

⎡ ⎤= = = ⎜ ⎟⎣ ⎦⎝ ⎠

⎡ ⎤= − =⎣ ⎦

∑

∑

∑ ( )m M gtanθ= +

x

y

mg

Mg

TTθ

P

N



P. 3/36 Determine the accelerations of bodies A and Band the tension in the cable due to theapplication of the 250 N force. Neglect all frictionand the masses of the pulleys.



P. 3/36

A B A B

x x A B

2 2A B

2s 3s c 2a 3a 0

F ma 2T 70a and 300 3T 35a

a 2.34 m/s , a 1.56 m/s , T 81.8 N

l+ + = → + =

⎡ ⎤= − = − =⎣ ⎦= − = =

∑

sA sB

2T 3T

70g 35g

300 N

NA NB



P. 3/44 The sliders A and B are connected by a lightrigid bar and move with negligible friction in theslots, both of which lie in a horizontal plane.For the position shown, the velocity of A is0.4 m/s to the right. Determine the accelerationof each slider and the force in the barat this instant.



P. 3/44

( )

A B A B A B2 2 2

A B A B

A A B B A B B A

A B2 2A A A B B B A B B

Kinematics: triangle OABs s and 0.5 s cos15 s cos15, s s 0.2588 m

s s 2s s cos150diff: 0 2s v 2s v 2cos150 s v s vgiven: v 0.4 m/s v 0.4 m

diff: 0 v s a v s a cos150 s a s a

l

= = + = =

= + −

= + − +

= → = −

= + + + − +( )( )

( )

A A B

A B

A B

2 2A B

2v v

0 0.04287 0.4829a 0.4829a 1Kinetics:

F ma 40 Tcos15 2a and Tcos15 3a into 1

a 7.95 m/s , a 8.04 m/s , T 25.0 N

+

= + +

⎡ ⎤= − = − =⎣ ⎦= = − =

∑

sB

sA

NB

NAT

T

40 N



P. 3/46 With the blocks initially at rest, the force P isincreased slowly from zero to 260 N. Plot theaccelerations of both masses as functions of P.



P. 3/46

max max

k k

A B A

A A B B

A A B B

N 35g, N N 42g 77gF 0.2N 68.67 N, F 0.15N 113.3 N

F 0.15N 51.5 N, F 0.10N 75.54 N

Three possible situations: no motion, B & A move together, and B & A move separately.Two impossibl

= = + == = = =

= = = =

( )max

max

A

A

B B

e situations: B moves alone then F will 0 A will move eventuallyand A moves alone P is applied at block B and force P is increased slowly from zero

not jump right to F

1) 0 P F : F will be

≠ →

≤ ≤

∵∵

A A B

developed to cancel with the applied P,

F will stay zero, and so there is no motion a 0 & a 0→ = =

35g

42g

NA

FA

FANA

NB

FB

P



P. 3/46

( ) ( )

max k

k

max

max

A A B B

A A B

2min B A

A A

2) assume both A and B go together in this phase F F and F F

F ma F 35a & P F F 42a

at P P F increased slowly , a 0.49 m/s and F 17.16 N jumping

at F F about to slip r

∴ ≤ =

⎡ ⎤= = − − =⎣ ⎦= = = =

=

∑

( )

( )

k

max

2

B

2B A B

elative to each other , P 226.6 N and a 1.962 m/s

P Fbetween these extremum values, a : linear function of P

77F P 226.6 : a a which varies linearly from 0.49 to 1.962 m/s

3) A slide backward

= =

−=

∴ < ≤ =

k

k k k

A A

A A A B B

2A B

relative to B increasing P makes B accelerates more and moreP 226.6 N makes A slips F F

F ma F 35a & P F F 42a

P 127.04a 1.47 m/s constant and a : linear function of P42

226.6

> → =

⎡ ⎤= = − − =⎣ ⎦−

= =

∴

∑

∵

( )2 2A BP 260.0 : a 1.47 m/s constant and 2.37 a 3.166 m/s jumping< ≤ = < ≤



P. 3/46



P. 3/48 The system is released from rest in the positionshown. Calculate the tension T in the cord andthe acceleration a of the 30 kg block. The smallpulley attached to the block has negligible massand friction. (Suggestion: First establish thekinematic relationship between the accelerationsof the two bodies.)



P. 3/48

( )( )

2 2 2

2 2

Kinematics: b x and b y

diff: bb xx and b y 0

b bb x xx and b y 0 1

at this instant: x/b 4 / 5, x 0, b 0 initially restassume cylinder moves down,

c l= + + =

= + =

+ = + + =

= = =

hence block moves to the leftKinetics: for 30 kg block

F ma T 3/5 T 30g N 0, N 30g 2T/5

F T 4/5 30xassume the block moves F 0.25Nfor 15 kg cylinder

15g T 15y 15b

r

⎡ ⎤= × − − + = = +⎣ ⎦− × =

→ =

− = = −

∑

( ) ( )

2

T/15 g 30x 4 becall 1 , , T 137.9 Nb 5 x 7.5g 0.7T

7.5g 0.7Tx 0.766 m/s30

− ×= = = =

−−

= = −

c

+b

+x +y

30g

15g

T

T

T

FN


3.5 Curvilinear Motion

3.5 Curvilinear MotionChoose appropriate coordinate system (x-y, n-t, or r-θ)for the given problem. Determine the motion alongthose axes. Then set up the Newton’s lawalong those axes. The positive sense of the force andacceleration must be consistent.

( ) ( )( ) ( )

x y

2 2n t

2r

x-y system: F mx F my

n-t system: F m m v / F mv

r- system: F m r r F m r 2rθ

ρβ ρ

θ θ θ θ

= =

= = =

= − = +

∑ ∑∑ ∑∑ ∑



P. 3/55 The member OA rotates about a horizontal axis through Owith a constant counterclockwise velocity ω= 3 rad/s. As itpasses the position θ= 0, a small block of mass m is placedon it at a radial distance r = 450 mm. If the block is observedto slip at θ= 50°, determine the coefficient of static frictionμs between the block and the member.



P. 3/55

( )

( )( )

t t

2n n

s s

use n-t coordinate systemgiven: 0.45 m, 50 , 0 no slip until 50 ,

3 rad/s, 0

F ma N mgcos50 m , N mgcos50

F ma mgsin50 F m

At 50 , F F N and directs u

ρ β ρ β

β β

ρβ ρβ

ρβ

μ

= = ° = = °

= =

⎡ ⎤= − = + =⎣ ⎦

⎡ ⎤= − =⎣ ⎦° = =

∑∑

( )

2

2

pward because gsin50which means bar OA rotates too slow than required to keepthe block stays on the bar. The friction will develop to resist

the block from sliding down or to match F with .

I

ρβ

ρβ

>

∑

( )2 2

2

f the bar rotates very very slow, friction force cannot make

F to match F cannot be reduced any more, F .

And then the block will slide down, hence decreases, to the positionwhere v / lar

ρβ ρβ

ρ

ρ

>∑ ∑ ∑

( )2s s

ge enough to match F (i.e., to satisfy Newton's law)

mgsin50 mgcos50 m , 0.549μ ρβ μ− = =

∑

mg

NF

n

t



P. 3/67 A 2 kg sphere S is being moved in a vertical plane by a roboticarm. When the arm angle θis 30°, its angular velocity abouta horizontal axis through O is 50 deg/s CW and its angularacceleration is 200 deg/s2 CCW. In addition, the hydraulicelement is being shortened at the constant rate of 500 mm/s.Determine the necessary minimum gripping force P if thecoefficient of static friction between the sphere and the grippingsurfaces is 0.5. Compare P to the minimum gripping force Psrequired to hold the sphere in static equilibrium in the 30˚ position.



P. 3/67

( )( )

r r

2

2r r f f

f f

given: m 1 kg, r 1 m, r 0.5 m/s, r 0

30 , 50 0.873 rad/s, 200 3.49 rad/s180 180

F ma mgsin 2F m r r , F 4.143 N

F ma 2F mgcos m r 2r , F 12.859θ θθ θ

π πθ θ θ

θ θ

θ θ θ

= = = − =

= ° = − × = − = × =

⎡ ⎤= − + = − =⎣ ⎦

⎡ ⎤= − = + =⎣ ⎦

∑∑

r

2 2f f f s

s s s s

N

F F F 13.51 P P 27.02 N

static equilibrium: 2F 2 P mg P 19.62 Nθ

μ

μ

= + = = → =

= = → =

rθ mg

2Fr

2Fθ

mg

2Fs



P. 3/69 A flatbed truck going 100 km/h rounds ahorizontal curve of 300 m radius inwardly bankedat 10°. The coefficient of static friction betweenthe truck bed and the 200 kg crate it carries is0.70. Calculate the friction force F acting onthe crate.



P. 3/69

2

n n

y

assume the crate tends to slide up the truck bedfriction directs downslope

the crate has absolute curve motion into the paper on the horizontal plane

m 10F ma Nsin10 Fcos10 100300 36

F

∴

⎛ ⎞⎡ ⎤= + = ×⎜ ⎟⎣ ⎦ ⎝ ⎠

=

∑

max

0 mg Ncos10 Fsin10 0

N 2021.52 N and F 165.9 Ncheck if this friction can be providedF 0.7N 1415 N F

the crate tends to slide up due to high speed curved motionbut still too far from slidi

⎡ ⎤ − + − =⎣ ⎦= =

= = >∴

∑

ng up (can increase the truck speedyet the crate does not move relative to the truck bed)

mg

N

F

n

y



P. 3/69 The flatbed truck starts from rest on a road whose constantradius of curvature is 30 m and whose bank angle is 10°. Ifthe constant forward acceleration of the truck is 2 m/s2,determine the time t after the start of motion at which the crateon the bed begins to slide. The coefficient of static frictionbetween the crate and truck bed is μs = 0.3, and the truckmotion occurs in a horizontal plane.



P. 3/69

static motion

200g

FN N

Fsn

Fst(inward)

200g

n

y



P. 3/69s s s

s

s

s

Static case: N 200gcos10 1932.2 N, F 0.3N 579.66 NF 200gsin10 340.7 N upward to prevent sliding down the incline, and F

Slipping when friction F but in what direction?F can be divided in two co

= = = =

= = <

=

t

n

s t

s n

s

mponents: along n- and t-axisF points in positive t (inward the paper) to match the positive a

F points down the incline to match the component of a down the incline

When the truck moves, N N to >

[ ]( )

n

n

n

2t

t t

y s

2

n n s

match the positive component of a up the truck bed

given: 30 m, 0, 0, a 2 m/sa v v a t 2t

F 0 Ncos10 200g F sin10 0 1

4tF ma F cos10 Nsin10 2003

ρ ρ ρ= = = =

= = =

⎡ ⎤= − − =⎣ ⎦

⎡ ⎤= + = ×⎣ ⎦

∑

∑ ( )

( )

( )

t

n t n t

n

n

t t s

22 2 2 2 2s s s s s

2s

s

s

20

F ma F 200 2 400 N

F F F F F 0.3N

F 0.09N 160000 and substitute into 1

N 2076.47, 1919.24 N but 1919.24 N which is impossibleN 2076.47 N, F

⎛ ⎞⎜ ⎟⎝ ⎠

⎡ ⎤= = × =⎣ ⎦

⎡ ⎤+ = + =⎣ ⎦

= −

= <

∴ = =

∑

ts477.55 N, F 400 N, t 5.58 s= =



P. 3/72 The small object is placed on the inner surface of the conicaldish at the radius shown. If the coefficient of static frictionbetween the object and the conical surface is 0.30, for whatrange of angular velocities ωabout the vertical axis will the blockremain on the dish without slipping? Assume that speed changesare made slowly so that any angular acceleration may be neglected.



P. 3/72

min n s n

max n s n

min y

n n

given: 0, 0.2 m, 0, 0 causes small a F upward to reduce F

causes large a F downward to increase F

: F 0 Ncos30 0.3Nsin30 mg 0

F ma Nsin30 0.3Ncos30 m 0

ω ρ ρ ρω

ω

ω

= = = =

→

→

⎡ ⎤= + − =⎣ ⎦⎡ ⎤= − =⎣ ⎦

∑∑

∑∑ ( )

( )

2min min

max y

2n n max max

.2 , 3.405 rad/s

: F 0 Ncos30 0.3Nsin30 mg 0

F ma Nsin30 0.3Ncos30 m 0.2 , 7.214 rad/s

3.405 7.214 rad/s

ω ω

ω

ω ω

ω

=

⎡ ⎤= − − =⎣ ⎦⎡ ⎤= + = =⎣ ⎦∴ < <

∑∑

mg

NFs

ωmin

mg

N

Fsωmax



P. 3/74 The 2 kg slider fits loosely in the smooth slot of the disk, whichrotates about a vertical axis through point O. The slider is freeto move slightly along the slot before one of the wires becomestaut. If the disk starts from rest at time t = 0 and has a constantclockwise angular acceleration of 0.5 rad/s2, plot the tensionsin wires 1 and 2 and the magnitude N of the force normal tothe slot as functions of time t for the interval 0<=t<=5 s.



P. 3/74

2 2

r r

given: 0.5 rad/s constant 0.5t, 0.25tr 0.1 m & free to move move with the disk r 0, r 0assume N and T to be in the indicated direction and use r- coordinate

F ma Nc

slightlyθ θ θ

θ

= → = == ≈ → = =

⎡ ⎤= −⎣ ⎦∑ ( )( )( )

2 2

2 2

1

os45 Tcos45 2 0.1 0.5t 0.05t

F ma Nsin45 Tsin45 2 0.1 0.5 0.1

0.05t 0.1 0.05t 0.1N T2 2

N is always positive the assumed direction is correctT will be negative for t 1.414 s

0, 0T

θ θ

− = × − × = −

⎡ ⎤= − = × × =⎣ ⎦+ −

= =

∴<

∴ =

∑

2

22

t 1.414 s 0.1 0.05t , 0 t 1.414 s and T 20.05t 0.1, t 1.414 s

0, t 1.414 s2

≤ ≤⎧ ⎧ −≤ <⎪ ⎪=−⎨ ⎨

>⎪ ⎪ ≥⎩ ⎩

r

θ

45°

N

Tθ



P. 3/74



P. 3/86 A small rocket-propelled vehicle of mass m travels down thecircular path of effective radius r under the action of its weightand a constant thrust T from its rocket motor. If the vehiclestarts from rest at A, determine its speed v when it reaches Band the magnitude N of the force exerted by the guide on thewheels just prior to reaching B. Neglect any friction and anyloss of mass of the rocket.



P. 3/86

[ ] ( )

2n n

t t t t

2 2t t

0

= /2 = /2

F ma N mgsin mv / r

T mgcosF ma T mgcos ma , am

Tvdv a ds v / 2 a rd , v 2r gsinm

N 3mgsin 2T

Tv r 2 N 3mg Tm

g

θ

θ π θ π

θ

θθ

θθ θ

θ θ

π π

⎡ ⎤= − =⎣ ⎦+⎡ ⎤= + = =⎣ ⎦

⎛ ⎞= = = +⎜ ⎟⎝ ⎠

= +

⎛ ⎞= + = +⎜ ⎟⎝ ⎠

∑

∑

∫

mg

n

t

T

N



P. 3/97 A hollow tube rotates about the horizontal axisthrough point O with constant angular velocityωo . A particle of mass m is introduced withzero relative velocity at r = 0 when θ= 0 andslides outward through the smooth tube.Determine r as a function of θ.



P. 3/97

( )( )

( )( )

( )

o

o

2r r

2o o

p h

2p o o

p

given: , 0at t 0, r 0, r 0, 0 t t

F ma mgsin m r r

r r gsin t differential equation of r t

r t r r

particular solution r is a solution of r r gsin t

r t force

θ ω θθ θ ω

θ θ

ω ω

ω ω

= =

= = = = ∴ =

⎡ ⎤= = −⎣ ⎦− = ←

= +

− =

=

∑

( ) ( )

o o

2 2o o o o o 2

o2

h ost

h

2 s

d response of gsin t gsin t

1sub. into diff. eq. gsin t gsin t gsin t2

homogeneous solution r is a solution of r r 0

r t free natural response

sub. into diff. eq. s

C

C C C

Ae

A e

ω ω

ω ω ω ω ωω

ω

=

− − = → = −

− =

= =

( )

( )

( ) ( )

( )

o o

o o

t 2 sto o o

t th

t tp h o2

o

o oo

2 2 2o o o

0, s ,

r t1r t r r gsin t, which must satisfy i.c.

2gr 0 0 and r 0 0

2g g g, r sinh sin

4 4 2

A e

Ae Be

Ae Be

A B A B

A B

ω ω

ω ω

ω ω ω

ωω

ω ωω

θ θω ω ω

−

−

− = = −

∴ = +

= + = + −

= = + = = − −

→ = = − ∴ = −

rθ

N mg



P. 3/98 The small pendulum of mass m is suspended from a trolleythat runs on a horizontal rail. The trolley and pendulum areinitially at rest with θ= 0. If the trolley is given a constantacceleration a = g, determine the maximum angle θmaxthrough which the pendulum swings. Also find the tension Tin the cord in terms of θ.



P. 3/98

[ ]

( )( )

P C P/C

2C P/C n t

t t

use n-t coordinate to avoid unknown T in t-direction translating axes attached to the cart to observe pendulum

g

F ma mgsin m gcos

g cos sin a

l l

l

l

θ θ

θ θ θ

θ θ θ

= +

= = +

⎡ ⎤= − = − +⎣ ⎦

= −

∑

a a a

a i a e e

( ) ( )

( )( )

2 2

0

max min max

2n n

s function of

g gd d / 2 cos sin d , 2 sin cos 1

or when 0 sin cos 1 / 2

F ma T mgcos m gsin

T mg 3sin 3cos 2

l l

l

θ

θ

θ θ θ θ θ θ θ θ θ θ θ

θ θ θ θ θ θ π

θ θ θ

θ θ

⎡ ⎤= = − = + −⎣ ⎦

= → + = ∴ =

⎡ ⎤= − = +⎣ ⎦= + −

∫

∑

t

n

T

mg

2nlθ e

tlθe

giaP



P. 3/99 A small object is released from rest at A and slides with frictiondown the circular path. If the coefficient of friction is 0.2,determine the velocity of the object as it passes B. (Hint: Writethe equations of motion in the n- and t- directions, eliminate N,and substitute vdv = atrdθ. The resulting equation is a linearnonhomogeneous differential equation of the form

, the solution of which is well known.)( ) ( )dy/dx f x y g x+ =



P. 3/99( )

( )( )

( ){ } ( )( ) ( ) ( )

2n n

t t

2

2 2

22 2

F ma N mgsin m 3

F ma mgcos 0.2N m 3

eliminate N: gcos 0.2 gsin 3 3

1 1d d d gcos 0.2 gsin 3 d d3 2

d 20.4 g cos 0.2sin , as a fund 3

θ θ

θ θ

θ θ θ θ

θ θ θ θ θ θ θ θ θ θ θ

θθ θ θ θ

θ

⎡ ⎤= − =⎣ ⎦

⎡ ⎤= − =⎣ ⎦

− × + =

⎡ ⎤= = − × + =⎣ ⎦

+ = −

∑∑

( ) ( )

( ) ( )

( ) ( )

2

p h

p

ction of

let u and to solve the differential equation for uu u u

2u forced response of g cos 0.2sin cos sin3

2sub. into diff. eq. sin cos 0.4 cos sin g cos 0.2sin3

match the c

A B

A B A B

θ

θ θ θ

θ θ θ θ θ

θ θ θ θ θ θ

=

= +

= − = +

− + + + = −

( )

( ) ( )

sh

s s

0.4

1.2 2 0.48oeff. of sin and cos : g g3.48 3 3.48

u solution of the homogeneous equation

s 0.4 0, s 0.41.2 2 0.48u gcos gsin with u 0 03.48 3 3.48

1.2 g 0 u3.48

A B

Ce

C e Ce

Ce

C

θ

θ θ

θ

θ θ

θ

θ θ θ −

⎛ ⎞= = −⎜ ⎟⎝ ⎠

= =

+ × = = −

⎛ ⎞∴ = + − + =⎜ ⎟⎝ ⎠

+ = → ( ) 2 0.4

2B

1.2 2 0.48 1.2gcos gsin g3.48 3 3.48 3.48

at / 2, 3.382 v r 5.52 m/sReal world where friction exists makes the phenomena difficult

e θθ θ θ θ

θ π θ θ

−⎛ ⎞= = + − −⎜ ⎟⎝ ⎠

= = → = =

n

t

F=0.2N

N

mg



P. 3/100 A small collar of mass m is given an initialvelocity of magnitude vo on the horizontalcircular track fabricated from a slender rod.If the coefficient of kinetic friciton is μk,determine the distance traveled before thecollar comes to rest. (Hint: Recognize thatthe friction force depends on the net normalforce.)



P. 3/100

[ ]

( )o

v h

y v

2

n n h

2 2t t k v h t

2 2 2 2 4kt

2 4 2 20 s2o o

22 2 2 kv 0k

Normal force has component N and N

F 0 N mg

vF ma N mr

F ma F N N ma

vdv a ds vdv r m g m v dsmr

v v r grdv rds, s ln2 rg2 r g v

μ

μ

μμ

⎡ ⎤= =⎣ ⎦

⎡ ⎤= =⎣ ⎦

⎡ ⎤= − = − + =⎣ ⎦

= = − +

⎛ + +− ⎜= =+ ⎝

∑

∑

∑

∫ ∫⎞⎟

⎜ ⎟⎠

y

t

nmg

Nv

Nh

F



P. 3/101 The slotted arm OB rotates in a horizontal plane about point Oof the fixed circular cam with constant angular velocity =15 rad/s. The spring has a stiffness of 5 kN/m and isuncompressed when θ= 0. The smooth roller A has a massof 0.5 kg. Determine the normal force N that the cam exertson A and also the force R exerted on A by the sides of the slotwhen θ= 45°. All surfaces are smooth. Neglect the smalldiameter of the roller.

θ



P. 3/101

2 2 2

2

2

Kinematics: 0.2 0.1 r 0.2rcosdiff: 0 2rr 0.2rcos 0.2r sin

0 2r 2rr 0.2rcos 0.2r sin

0.2r sin 0.2r sin 0.2r cos

given: / 4, 15 rad/s, 0r 0.

θ

θ θ θ

θ θ θ

θ θ θ θ θ θ

θ π θ θ

= + +

= + −

= + + −

− − −

= = =

∴ =

( )( )

( )

2

2r r

1164 m, r 0.66 m/s, r 15.05 m/s0.2 0.1 , 20.7

sin135 sinKinetics: spring force at / 4 : F 5000 r 0.1 compressed

F ma F Ncos20.7 m r r

F ma R Nsin20.7 m r 2r

N 81.7θ θ

ββ

θ π

θ

θ θ

= =

= = °

= = × −

⎡ ⎤= − + = −⎣ ⎦

⎡ ⎤= − = +⎣ ⎦=

∑∑

N R 38.7 N=

0.2

0.1

rθβ

N

R

F

r

θ

β



P. 3/102 The small cart is nudged with negligiblevelocity from its horizontal position at A ontothe parabolic path that lies in a vertical plane.Neglect friction and show that the cartmaintains contact with the path for all valuesof k.



P. 3/102

( )( )

3/ 222

22

3/ 22 2

2

n n

2

If the cart maintains contact, N 0use n-t coordinate since N aligns with the n-axis

1 ' dy d y y kx k 2x tan 2k'' dx dx

1 4k x2k

vF ma N mgcos m

1 tan

y

yρ θ

ρ

θρ

θ

>

⎡ ⎤⎡ ⎤+⎢ ⎥⎣ ⎦= = = = =⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤+⎣ ⎦=

⎡ ⎤= − + =⎣ ⎦

+

∑

[ ]

2

2 2

t t t

2 2t

23/ 2 3/ 22 2 2 2 2 2

1sec cos1 4k x

F ma mgsin ma

vdv a ds vdv gsin ds gdy, v 2gy 2kgxmg 2k mgN 2mkgx 0

1 4k x 1 4k x 1 4k x

θ θ

θ

θ

⎡ ⎤= =⎣ ⎦ +⎡ ⎤= =⎣ ⎦

= = = = =

∴ = − × = >+ ⎡ ⎤ ⎡ ⎤+ +⎣ ⎦ ⎣ ⎦

∑

dx

dydsθ

mg

N

n

t

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