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PHYS1100 Practice problem set, Chapter 13: 7, 10, 14, 19, 24, 27, 30, 34, 56, 59, 60, 68, 72
13.7. Model: The drill is a rigid rotating body.
Visualize:
The figure shows the drill’s motion from the top. Solve: (a) The kinematic equation ωf = ωi + α( t f – t i) becomes, after using ωi = 2400 rpm = (2400)(2 π)/60 = 251. rad/s, t f – t i = 2.5 s – 0 s = 2.5 s, and ωf = 0 rad/s,
0 rad = 251.3 rad/s + α(2.5 s) ⇒ α= − 100.5 rad/s^2
(b) Applying the kinematic equation for angular position yields: 2 f i i f i f i 2 2
0 rad (251.3 rad/s)(2.5 s 0 s) ( 100.5 rad/s )(2.5 s 0 s)
314.2 rad 50.0 rev
θ = θ + ω t − t + α t − t
13.10. Visualize: Please refer to Figure Ex13.10. The coordinates of the three masses m A, m B, and m C are (0 cm,
0 cm), (10 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are
A A B B C C cm A B C A A B B C C cm A B C
(100 g)(0 cm) (200 g)(10 cm) (300 g)(10 cm) 8.33 cm (100 g 200 g 300 g) (100 g)(0 cm) (200 g)(10 cm) (300 g)(0 cm) 3.33 cm (100 g 200 g 300 g)
m x m x m x x m m m m y m y m y y m m m
13.14. Model: The disk is a rotating rigid body.
Visualize:
The radius of the disk is 10 cm and the disk rotates on an axle through its center. Solve: The net torque on the axle is
τ = F A r A sin φA + F B r B sin φB + F C r C sin φC + F D r D sin φD
= (30 N)(0.10 m) sin (− 90 °) + (20 N)(0.05 m) sin 90° + (30 N)(0.05 m) sin 135° + (20 N)(0.10 m) sin 0° = −3 N m + 1 N m + 1.0607 N m = −0.939 N m Assess: A negative torque means a clockwise rotation of the disk.
13.19. Model: The three masses connected by massless rigid rods is a rigid body.
Visualize: Please refer to Figure Ex13.19.
Solve: (a) (^) cm
(0.100 kg)(0 m) (0.200 kg)(0.06 m) (0.100 kg)(0.12 m) 0.060 m 0.100 kg 0.200 kg 0.100 kg
i i i
m x x m
2 2
cm
(0.100 kg)(0 m) (0.200 kg) (0.10 m) (0.06 m) (0.100 kg)(0 m) 0.040 m 0.100 kg 0.200 kg 0.100 kg
i i i
m y y m
(b) The moment of inertia about an axis through A and perpendicular to the page is
2 2 2 2 2 2
I A = ∑ m ri i = m B (0.10 m) + m C(0.10 m) = (0.100 kg)[(0.10 m) + (0.10 m) ] =0.0020 kg m
(c) The moment of inertia about an axis that passes through B and C is
2 2 2 2 I BC (^) = m A (0.10 m) − (0.06 m) =0.00128 kg m
Assess: Note that mass m A does not contribute to I A, and the masses m B and m C do not contribute to I BC.
13.30. Model: The disk is a rigid body rotating about an axis through its center.
Visualize:
Solve: The speed of the point on the rim is given by v rim = R ω. The angular velocity ω of the disk can be
determined from its rotational kinetic energy which is K = 12 I ω^2 = 0.15 J. The moment of inertia I of the disk
about its center and perpendicular to the plane of the disk is given by
2 2 5 2
2 5 2
(0.10 kg)(0.040 m) 8.0 10 kg m
2(0.15 J) 0.30 J
61.237 rad/s
8.0 10 kg m
I MR
I
ω ω
−
−
= = = ×
×
Now, we can go back to the first equation to find v rim. We get v rim = R ω = (0.040 m)(61.237 rad/s) = 2.45 m/s.
13.34. Model: The sphere is a rigid body rolling down the incline without slipping.
Visualize:
The initial gravitational potential energy of the sphere is transformed into kinetic energy as it rolls down. Solve: (a) If we choose the bottom of the incline as the zero of potential energy, the energy conservation equation will be K f = U i. The kinetic energy consists of both translational and rotational energy. This means
2 2 2 2 2 f cm cm
2 2
10 10 7 7 2 2
(2.1 m)sin 25 10 (2.1 m)(sin 25 ) (2.1 m)(sin25 ) 88.1 rad/s (0.04 m)
K I Mv Mgh MR M R Mgh
MR Mg
g g R
(b) From part (a)
2 2 2 2 2 2 2 2 2 total cm cm rot cm
1 2 2 rot 5 7 2 2 total 10
and 2 2 10 2 2 5 5 1 10
5 7
K I Mv MR K I MR MR
K MR
K MR
= + = = = ^ =
⇒ = = × =
13.56. Model: The beam is a rigid body of length 3.0 m and the student is a particle.
Visualize:
Solve: To stay in place, the beam must be in both translational equilibrium ( F net = 0 N)
r r
and rotational equilibrium
( τ net= 0 Nm). The first condition is
∑ Fy^ = −^ w beam^ −^ w student^ +^ F 1^ +^ F 2 =0 N
⇒ F 1 + F 2 = w beam + w student = (100 kg + 80 kg)(9.80 m/s^2 ) = 1764 N
Taking the torques about the left end of the beam, the second condition is
− w beam (1.5 m) – w student (2.0 m) + F 2 (3.0 m) = 0 N m − (100 kg)(9.8 m/s^2 )(1.5 m) – (80 kg)(9.8 m/s^2 )(2.0 m) + F 2 (3.0 m) = 0 N m
⇒ F 2 = 1013 N
From F 1 + F 2 = 1764 N, we get F 1 = 1764 N – 1013 N = 751 N. Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any point on the body of interest.
13.60. Model: The ladder is a rigid rod of length L. To not slip, it must be in both translational equilibrium
( F net = 0 N)
r r
and rotational equilibrium ( τ net= 0 N m).We also apply the model of static friction.
Visualize:
Since the wall is frictionless, the only force from the wall on the ladder is the normal force n 2.
r
On the other hand,
the floor exerts both the normal force n 1
r
and the static frictional force f s.
r
The weight w
r
of the ladder acts through
the center of mass of the ladder.
Solve: The x - and y -components of F net = 0 N
r r
are
∑ Fx^ =^ n 2^ −^ f s^ =^ 0 N^ ⇒^ f s^ =^ n 2^ ∑ Fy^ =^ n 1^ −^ w^ =^ 0 N⇒^ n 1^ = w
The minimum angle occurs when the static friction is at its maximum value f s max = μs n 1. Thus we have n 2 = f s =
μs n 1 = μs w. We choose the bottom corner of the ladder as a pivot point to obtain τnet, because two forces pass
through this point and have no torque about it. The net torque about the bottom corner is
τnet = d 1 w – d 2 n 2 = (0.5 L cos θmin) mg – ( L sin θmin) μs mg = 0 N m
min s min min min s
0.5cos sin tan 1.25 51.
θ μ θ θ θ μ
13.68. Model: Assume the string does not slip on the pulley.
Visualize:
The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward force on the block m 2 , but a downward force on the outer edge of the pulley. Similarly the string exerts a force on block m 1 to the right, but a leftward force on the outer edge of the pulley. Solve: (a) Newton’s second law for m 1 and m 2 is T = m 1 a 1 and T – m 2 g = m 2 a 2_._ Using the constraint – a 2 = + a 1 = a , we have T = m 1 a and − T + m 2 g = m 2 a. Adding these equations, we get m 2 g = ( m 1 + m 2 ) a, or
2 1 2 1 1 2 1 2
m g m m g
a T m a
m m m m
(b) When the pulley has mass m , the tensions ( T 1 and T 2 ) in the upper and lower portions of the string are different. Newton’s second law for m 1 and the pulley are:
T 1 = m 1 a and T 1 R – T 2 R = − I α
We are using the minus sign with α because the pulley accelerates clockwise. Also, a = R α. Thus, T 1 = m 1 a and
(^2 1 )
I a aI T T R R R
Adding these two equations gives
(^2 1 )
I
T a m
R
Newton’s second law for m 2 is T 2 – m 2 g = m 2 a 2 = − m 2 a. Using the above expression for T 2 ,
2 (^1 2 2 )
I m g
a m m a m g a
R m m I R
Since I = 12 m R p^2 for a disk about its center,
2 1 2 1 p 2
m g
a
m m m
With this value for a we can now find T 1 and T 2 :
( )
1 2 2 2 2 (^1 12 p) 1 1 1 2 1 1 1 p 1 1 2 2 p (^1 2 2) p 1 2 2 p
m m g m g^ m^ m^ m^ g
T m a T a m I R m m
m m m m m m m m m
^ +
Assess: For m = 0 kg, the equations for a , T 1 and T 2 of part (b) simplify to
2 1 2 1 2 1 2 1 2 1 2 1 2
and and
m g m m g m m g
a T T
m m m m m m
These agree with the results of part (a).
