Ch. 8: Kinetics of Particles
8.0 Outline 415 Introduction 416 Newton’s Second Law 417 Equations of Motion 418 Rectilinear Motion 421 Curvilinear Motion 444
8.0 Outline
415
Ch. 8: Kinetics of Particles
8.1 Introduction
8.1 Introduction
Kinetics is the study of the relations between the forces and the motion. Here we will not seriously concern whether the forces cause the motion or the motion generates the forces (causality). In this chapter, the focus is on the particles. That is the body whose physical dimensions are so small compared with the radius of curvature of its path. There are at least 3 approaches toe the solution of kinetic problems: (a) Newton’s second law (b) work and energy method (c) impulse and momentum method.
416
Ch. 8: Kinetics of Particles
8.2 Newton’s Second Law
8.2 Newton’s Second Law
For most engineering problems on earth, the acceleration measured w.r.t. reference frame fixed to the earth’s surface may be treated as absolute. And Newton’s 2nd law of motion holds. Newton’s 2nd law breaks when the velocities of the order of 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 =
417
Ch. 8: Kinetics of Particles
8.3 Equation of Motion and Solution of Problems
8.3 Equation of Motion and Solution
Two problems of dynamics (1) specified kinematic conditions, find forces straightforward application of Newton’s law as algebraic equations (2) specified forces, find motion Difficulty depends on the form of force function (t, s, v, a), as the solutions are found by solving a system of differential equations. For simple functions, we can find closed form solutions of motion as in rectilinear motion (sec. 2.2).
m --- equation of motionscalar components decomposition according to a specified coordinate∑F = a
418
Ch. 8: Kinetics of Particles
8.3 Equation of Motion and Solution
Unconstrained motion Motion of the particle is determined by its initial motion and the forces from external sources. It is free of constraints and so has three degrees of freedom to specify the position. Three scalar equations of motion would have to be applied and integrated to obtain the motion. Constrained motion Motion of the particle is partially or totally determined by restraining guides, other than its initial motion and the forces from external sources. Therefore, all forces, both applied and reactive, that act on the particle must be accounted for in Newton’s law. The number of d.o.f. and equations are reduced regarding to the type of constraints.
419
Ch. 8: Kinetics of Particles
8.3 Equation of Motion and Solution
Free body diagram All forces acting on the particle needed to be accounted in the equations of motion. Free body diagram unveils every force that acts on the isolated particle. Only after the FBD has been completed should the equations of motion be written. The appropriate coordinate axes and directions should be indicated and consistently used throughout the problem. Treatment of the body as particle is valid when the forces may be treated as concurrent through the mass center.
420
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
8.4 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
421
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/1 The coefficient of static friction between the flat bed of the truck and the crate it carries is 0.30. Determine the minimum stopping distance s that the truck can have from a speed of 70 km/h with constant deceleration if the crate is not to slip forward.
422
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/1
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
423
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/2 If the truck of Prob. 3/17 comes to stop from an initial forward speed of 70 km/h in a distance of 50 m with uniform deceleration, determine whether or not the crate strikes the wall at the forward end of the flat bed. If the crate does strike the wall, calculate its speed relative to the truck as the impact occurs. Use the friction coefficients μs = 0.3 and μk = 0.25.
424
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/2
( )
( )
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
425
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/3 If the coefficients of static and kinetic friction between the 20-kg block A and the 100-kg cart B are both essentially the same value of 0.50, determine the acceleration of each part for (a) P = 60 N and (b) P = 40 N.
426
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/3
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
= =
− = = < ∴= =
2P 20g
100g
NB
NA NA F
F
427
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/4 A simple pendulum is pivoted at O and is free to swing in the vertical plane of the plate. If the plate is given a constant acceleration a up the incline θ, write an expression for the steady angle β assumed by the pendulum after all initial start-up oscillations have ceased. Neglect the mass of the slender supporting rod.
428
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/4
y
x x
1
F 0 Tcos mgcos 0
F ma Tsin mgsin ma
a gsintangcos
β θ
β θ
θβθ
−
= − = = − =
+=
∑∑
T
mg θ
β x
y
429
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/5 For the friction coefficients μs = 0.25 and μk = 0.20, calculate the acceleration of each body and the tension T in the cable.
430
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/5
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
N F
431
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/6 A bar of length l and negligible mass connects the cart of mass M and the particle of mass m. If the cart is subjected to a constant acceleration a to the right, what is the resulting steady-state angle θthat the freely pivoting bar makes with the vertical? Determine the net force P (not shown) that must be applied to the cart to cause the specified acceleration.
432
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/6
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
T T θ
P
N
433
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/7 Determine the accelerations of bodies A and B and the tension in the cable due to the application of the 250 N force. Neglect all friction and the masses of the pulleys.
434
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/7
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
435
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/8 The sliders A and B are connected by a light rigid bar and move with negligible friction in the slots, both of which lie in a horizontal plane. For the position shown, the velocity of A is 0.4 m/s to the right. Determine the acceleration of each slider and the force in the bar at this instant.
436
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/8
( )
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
NA T
T
40 N
437
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/9 With the blocks initially at rest, the force P is increased slowly from zero to 260 N. Plot the accelerations of both masses as functions of P.
438
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/9
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
FA NA
NB FB
P
439
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/9
( ) ( )
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< ≤ = < ≤
440
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/9
441
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/10 The system is released from rest in the position shown. Calculate the tension T in the cord and the acceleration a of the 30 kg block. The small pulley attached to the block has negligible mass and friction. (Suggestion: First establish the kinematic relationship between the accelerations of the two bodies.)
442
Ch. 8: Kinetics of Particles
8.4 Rectilinear Motion
P. 8/10
( )( )
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
F N
443
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
8.5 Curvilinear Motion Choose appropriate coordinate system (x-y, n-t, or r-θ) for the given problem. Determine the motion along those axes. Then set up the Newton’s law along those axes. The positive sense of the force and acceleration 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
= =
= = =
= − = +
∑ ∑∑ ∑∑ ∑
444
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/11 The member OA rotates about a horizontal axis through O with a constant counterclockwise velocity ω= 3 rad/s. As it passes the position θ= 0, a small block of mass m is placed on it at a radial distance r = 450 mm. If the block is observed to slip at θ= 50°, determine the coefficient of static friction μs between the block and the member.
445
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/11
( )
( )( )
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
N F
n
t
446
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/12 A 2 kg sphere S is being moved in a vertical plane by a robotic arm. When the arm angle θis 30°, its angular velocity about a horizontal axis through O is 50 deg/s CW and its angular acceleration is 200 deg/s2 CCW. In addition, the hydraulic element is being shortened at the constant rate of 500 mm/s. Determine the necessary minimum gripping force P if the coefficient of static friction between the sphere and the gripping surfaces is 0.5. Compare P to the minimum gripping force Ps required to hold the sphere in static equilibrium in the 30˚ position.
447
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/12
( )( )
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
= + = = → =
= = → =
r θ mg
2Fr
2Fθ
mg
2Fs
448
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/13 A flatbed truck going 100 km/h rounds a horizontal curve of 300 m radius inwardly banked at 10°. The coefficient of static friction between the truck bed and the 200 kg crate it carries is 0.70. Calculate the friction force F acting on the crate.
449
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/13
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
450
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/14 The flatbed truck starts from rest on a road whose constant radius of curvature is 30 m and whose bank angle is 10°. If the constant forward acceleration of the truck is 2 m/s2, determine the time t after the start of motion at which the crate on the bed begins to slide. The coefficient of static friction between the crate and truck bed is μs = 0.3, and the truck motion occurs in a horizontal plane.
451
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/14
static motion
200g
F N N
Fsn
Fst (inward)
200g
n
y
452
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/14 s 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= =
453
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/15 The small object is placed on the inner surface of the conical dish at the radius shown. If the coefficient of static friction between the object and the conical surface is 0.30, for what range of angular velocities ωabout the vertical axis will the block remain on the dish without slipping? Assume that speed changes are made slowly so that any angular acceleration may be neglected.
454
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/15
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
N Fs
ωmin
mg
N
Fs ωmax
455
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/16 The 2 kg slider fits loosely in the smooth slot of the disk, which rotates about a vertical axis through point O. The slider is free to move slightly along the slot before one of the wires becomes taut. If the disk starts from rest at time t = 0 and has a constant clockwise angular acceleration of 0.5 rad/s2, plot the tensions in wires 1 and 2 and the magnitude N of the force normal to the slot as functions of time t for the interval 0<=t<=5 s.
456
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/16
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
T θ
457
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/16
458
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/17 A small rocket-propelled vehicle of mass m travels down the circular path of effective radius r under the action of its weight and a constant thrust T from its rocket motor. If the vehicle starts from rest at A, determine its speed v when it reaches B and the magnitude N of the force exerted by the guide on the wheels just prior to reaching B. Neglect any friction and any loss of mass of the rocket.
459
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/17
[ ] ( )
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
460
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/18 A hollow tube rotates about the horizontal axis through point O with constant angular velocity ωo . A particle of mass m is introduced with zero relative velocity at r = 0 when θ= 0 and slides outward through the smooth tube. Determine r as a function of θ.
461
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/18
( )( )
( )( )
( )
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
−
−
− = = −
∴ = +
= + = + −
= = + = = − −
→ = = − ∴ = −
r θ
N mg
462
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/19 The small pendulum of mass m is suspended from a trolley that runs on a horizontal rail. The trolley and pendulum are initially at rest with θ= 0. If the trolley is given a constant acceleration a = g, determine the maximum angle θmax through which the pendulum swings. Also find the tension T in the cord in terms of θ.
463
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/19
[ ]
( )( )
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
464
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/20 A small object is released from rest at A and slides with friction down the circular path. If the coefficient of friction is 0.2, determine the velocity of the object as it passes B. (Hint: Write the equations of motion in the n- and t- directions, eliminate N, and substitute vdv = atrdθ. The resulting equation is a linear nonhomogeneous differential equation of the form , the solution of which is well known.) ( ) ( )dy/dx f x y g x+ =
465
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/20 ( )
( )( )
( ){ } ( )( ) ( ) ( )
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
466
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/21 A small collar of mass m is given an initial velocity of magnitude vo on the horizontal circular track fabricated from a slender rod. If the coefficient of kinetic friciton is μk, determine the distance traveled before the collar comes to rest. (Hint: Recognize that the friction force depends on the net normal force.)
467
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/21
[ ]
( )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
n mg
Nv
Nh
F
468
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/22 The slotted arm OB rotates in a horizontal plane about point O of the fixed circular cam with constant angular velocity = 15 rad/s. The spring has a stiffness of 5 kN/m and is uncompressed when θ= 0. The smooth roller A has a mass of 0.5 kg. Determine the normal force N that the cam exerts on A and also the force R exerted on A by the sides of the slot when θ= 45°. All surfaces are smooth. Neglect the small diameter of the roller.
θ
469
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/22
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 cosgiven: / 4, 15 rad/s, 0
r 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
r θ β
N
R
F
r
θ
β
470
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/23 The small cart is nudged with negligible velocity from its horizontal position at A onto the parabolic path that lies in a vertical plane. Neglect friction and show that the cart maintains contact with the path for all values of k.
471
Ch. 8: Kinetics of Particles
8.5 Curvilinear Motion
P. 8/23
( )( )
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
dy ds θ
mg
N
n
t
472
Ch. 8: Kinetics of Particles
8.6 Work and Energy
8.6 Work and Energy
473
Many of the dynamic problems need to determine the velocity or the displacement quantities. This requires integrating the equations formulated from the Newton’s second law with the use of kinematical analysis for the acceleration term. However, this integration step can be performed beforehand. If we integrate the Newton’s law directly w.r.t. the displacement, i.e.
∫ ∫
−==
2
21
22 vvmmvdvFds
Ch. 8: Kinetics of Particles
8.6 Work and Energy
474
The left hand side is defined to be the work done on the particle. The right hand side corresponds to the change in the kinetic energy. This integrated equation of motion is called work-energy equation. On the other hand, if we integrate the Newton’s law directly w.r.t. time, i.e. For this one, the left hand side quantity is called the impulse applied to the particle, while the right hand one is the change of the momentum of the particle. This integrated equation of motion is then called impulse-momentum equation.
( )∫ ∫ −== 12 vvmmdvFdt
Ch. 8: Kinetics of Particles
8.6 Work and Energy
475
Definition of Work If the force acting on a particle at point A of which its position is described by the position vector , then
is the work done by the force during the differential displacement , of which its magnitude is equal to the force times the displacement in the direction of force. Alternatively, the magnitude will also be equal to the displacement times the force in the direction of the displacement. In other words, components of force in the direction normal to the displacement vector, , will do no work.
Fr
rF ddU ⋅=F
rd
nF
Ch. 8: Kinetics of Particles
8.6 Work and Energy
476
Ch. 8: Kinetics of Particles
8.6 Work and Energy
477
Calculation of Work When the force , which may not be constant, has been applied to the particle during a finite movement, the work performed by this force on the particle will be which may be further elaborated as if the x-y coordinate frame is used, or
F
∫ ⋅=2
1
rF dU
( )∫ +=2
1
dyFdxFU yx
Ch. 8: Kinetics of Particles
8.6 Work and Energy
478
if the n-t coordinate frame is used. Moreover, if the function of the force is not known explicitly, the numerical integration approach for approximating the area under graph may be used.
∫=2
1
s
stdsFU
sFt −
Ch. 8: Kinetics of Particles
8.6 Work and Energy
479
Work done by spring force From the figure, the spring force exerted on the particle when it is displaced from the unstretch by is . Therefore the work done by spring as the particle undergoes an arbitrary displacement from to would be
xiF kx−=
1x2x
( ) ( )22
2121 2
12
1
xxkdxkxUx
x
−=⋅−= ∫− ii
Ch. 8: Kinetics of Particles
8.6 Work and Energy
480
Ch. 8: Kinetics of Particles
8.6 Work and Energy
481
Work done by gravity force a) g = const.
b) g != const.
( ) ( ) ( )21
2
121 yymgdydxmgU −=+⋅−= ∫− jij
( )
−=⋅
−= ∫−
12
2
1221
11rr
mGmdrr
mGmU erre ee
Ch. 8: Kinetics of Particles
8.6 Work and Energy
482
Ch. 8: Kinetics of Particles
8.6 Work and Energy
483
Work done by force lead us to define the kinetic energy of the particle, which can be interpret as the amount of work given to the particle to bring it from rest to a velocity . Hence the above equation may be rewritten as
( )∫ ∫∫∫ −===⋅=⋅=−
2
1
2
1
21
22
2
1
2
121 2
1 vvmmvdvdsmadmdU trarF
F
2
21 mvT =
v
TTTU ∆=−=− 1221
Ch. 8: Kinetics of Particles
8.6 Work and Energy
484
Ch. 8: Kinetics of Particles
8.6 Work and Energy
485
This equation is known as work-energy equation for the particle. It says that the total work done by all forces acting on the particle as it moves from 1 2 equals to the change in kinetic energy of the particle. Alternatively, saying that the final kinetic energy equals the initial kinetic energy plus the work done on the particle during that period. The equation also applies to a system of particles of which their relative distances are kept unchanged.
2211 TUT =+ −
Ch. 8: Kinetics of Particles
8.6 Work and Energy
486
However, would be the net work done on the system by external forces and would be the total kinetic energy of the system. Before applying the work-energy equation, it is required to make clear the isolated system under consideration by drawing the free-body diagram of it, showing all externally applied forces which do work (active forces).
21−UT
Ch. 8: Kinetics of Particles
8.6 Work and Energy
487
Potential Energy Potential energy is here to treat the work, especially those done by gravity forces and by spring forces. Potential energy may be converted into kinetic energy and do work later. Gravitational Potential Energy is the work done against the gravitational force to elevate the particle a distance above some reference datum, where the potential energy is taken to be zero
h
mghVg =
Ch. 8: Kinetics of Particles
8.6 Work and Energy
488
The change in potential energy when the particle goes from the level to the new level is which is the negative of the work done by the gravitational force on the particle . If the level change is significant compared to the radius of the earth, the gravity force is not equal to the simple , rather it will be . Accordingly, the potential energy is the work done against this gravity force which, in this case, may be calculated from
1h
( ) hmghhmgVg ∆=−=∆ 12
2h
hmg∆−
mg 222 // rmgRrGmme =
rmgRVg
2
−=
Ch. 8: Kinetics of Particles
8.6 Work and Energy
489
where the reference datum for zero potential is taken to be at . The change in potential energy when the particle goes from to is which is the negative of the work done by the gravitational force on the particle .
∞→r
−=∆
21
2 11rr
mgRVg
1r 2r
( )122 /1/1 rrmgR −
Ch. 8: Kinetics of Particles
8.6 Work and Energy
490
Ch. 8: Kinetics of Particles
8.6 Work and Energy
491
Elastic Potential Energy is the work done against the spring force to deform the spring. Therefore, the potential energy of the spring when it is deformed by the amount of from the unstretched posture is The change in potential energy when the spring undergoes the deformation from to is which is the negative of the work done by the spring force.
x
2
0 21 kxdkV
x
e == ∫ δδ
1x 2x
( )21
222
1 xxkVe −=∆
Ch. 8: Kinetics of Particles
8.6 Work and Energy
492
Work-Energy Equation Revisited The original work-energy equation may be modified to account for the potential energy explicitly as where now the scope of the system is changed to include the spring and the gravity field. is then the work done by all external forces on the new system. Alternatively
1221 TTU −=−
'21−U
( ) ( ) 12'
21 TTVVU eg −=∆−+∆−+−
22'
2111 VTUVT +=++ −
Ch. 8: Kinetics of Particles
8.6 Work and Energy
493
P. 8/24 A constant horizontal force P = 700 N is applied to the linkage shown. With the 14-kg ball initially at rest on its support with , calculate the velocity of the ball as approaches zero where the ball reaches its highest position.
60=θθ