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Mechanics Problem: Rigid Bodies in Motion, Summaries of Acting

Solutions to various mechanics problems involving rigid bodies in motion. Topics include kinematics equations, moment of inertia, and energy conservation. The problems involve calculating angular acceleration, center of mass, and moment of inertia for different shapes and configurations.

Typology: Summaries

2021/2022

Uploaded on 09/27/2022

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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
ω
f
=
ω
i
+
α
(t
f
– t
i
) becomes, after using
ω
i
= 2400 rpm = (2400)(2
π
)/60 = 251.3
rad/s, t
f
t
i
= 2.5 s – 0 s = 2.5 s, and
ω
f
= 0 rad/s,
0 rad = 251.3 rad/s +
α
(2.5 s)
2
100.5 rad/s
α
=
(b)
Applying the kinematic equation for angular position yields:
2
f i i f i f i
2 2
1
( ) ( )
2
1
0 rad (251.3 rad/s)(2.5 s 0 s) ( 100.5 rad/s )(2.
5 s 0 s)
2
314.2 rad 50.0 rev
t t t t
θ θ ω α
= + +
= + +
= =
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
xm m m
m y m y m y
ym m m
+ + + +
= = =
+ + + +
+ + + +
= = =
+ + + +
pf3
pf4
pf5
pf8
pf9

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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 Tm 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).