Q: A mechanic uses a wrench to loosen a bolt. The wrench is 0.30 m long, and the mechanic applies a force of 120 N at an angle of 45° to the handle. What is the magnitude of the torque produced?

$$T = Fr \sin \theta = (120\ \text{N})(0.30\ \text{m})\sin 45^\circ = 36\ \text{N-m}$$

Q: A bolt requires a torque of 40 N-m to loosen. If you can only exert a force of 100 N, what is the minimum length of the wrench needed if you pull perpendicular to the handle?

$$r = \frac{T}{F} = \frac{40\ \text{N-m}}{100\ \text{N}} = 0.40\ \text{m}$$

Q: A seesaw is 2.0 m long and massless. Child A (30 kg) sits at one end. Where should Child B (40 kg) sit to balance the seesaw?

$$r_B = \frac{m_A r_A}{m_B} = \frac{30\ \text{kg} \times 2.0\ \text{m}}{40\ \text{kg}} = 1.5\ \text{m}$$

Q: A solid wheel of mass 10 kg and radius 0.20 m is accelerated from rest to 10 rad/s in 5 s. What torque is required?

$$I = \frac{1}{2} m r^2$ = \frac{1}{2} \times 10\ \text{kg} \times (0.20\ \text{m})^2$ = 0.2\ \text{kg-m}^2$$ $$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{10\ \text{rad/s}}{5\ \text{s}} = 2\ \text{rad/s}^2$$ $$T = I \alpha = 0.2\ \text{kg-m}^2 \times 2\ \text{rad/s}^2 = 0.4\ \text{N-m}$$

Q: A thin hoop and a solid sphere both have mass $$m$$ and radius $$r. $$Which has the greater moment of inertia about their central axis?

The thin hoop, because $$I_{\text{hoop}} = mr^2$ and I$$_{\text{sphere}} = \frac{2}{5}$$mr^2$$

Q: A pulley of radius 0.15 m is accelerated from rest to 14 rev/min in 4.5 s by a force of 13.0 N. What is the moment of inertia of the pulley?

$$\omega = 14\ \text{rev/min} \times \frac{2\pi}{60} = 1.47\ \text{rad/s}$$ $$\alpha = \frac{1.47\ \text{rad/s}}{4.5\ \text{s}} = 0.33\ \text{rad/s}^2$$ $$T = F r = 13.0\ \text{N} \times 0.15\ \text{m} = 1.95\ \text{N-m}$$ $$I = \frac{T}{\alpha} = \frac{1.95}{0.33} = 5.9\ \text{kg-m}^2$$

Q: A baton consists of two masses (0.30 kg each) at the ends of a 0.66 m rod. What is the moment of inertia about the center?

$$I = 2 \times m r^2$ = 2 \times 0.30\ \text{kg} \times (0.33\ \text{m})^2$ = 0.066\ \text{kg-m}^2$$

Question 1

A mechanic uses a wrench to loosen a bolt. The wrench is 0.30 m long, and the mechanic applies a force of 120 N at an angle of 45° to the handle. What is the magnitude of the torque produced?

Accepted Answer

$$T = Fr \sin 	heta = (120\ 	ext{N})(0.30\ 	ext{m})\sin 45^\circ = 36\ 	ext{N-m}$$

Question 2

A door is 0.90 m wide. To open it with the least force, where and in what direction should you apply the force?

Accepted Answer

At the doorknob, perpendicular to the door

Question 3

A bolt requires a torque of 40 N-m to loosen. If you can only exert a force of 100 N, what is the minimum length of the wrench needed if you pull perpendicular to the handle?

Accepted Answer

$$r = \frac{T}{F} = \frac{40\ 	ext{N-m}}{100\ 	ext{N}} = 0.40\ 	ext{m}$$

Question 4

A seesaw is 2.0 m long and massless. Child A (30 kg) sits at one end. Where should Child B (40 kg) sit to balance the seesaw?

Accepted Answer

$$r_B = \frac{m_A r_A}{m_B} = \frac{30\ 	ext{kg} 	imes 2.0\ 	ext{m}}{40\ 	ext{kg}} = 1.5\ 	ext{m}$$

Question 5

A solid wheel of mass 10 kg and radius 0.20 m is accelerated from rest to 10 rad/s in 5 s. What torque is required?

Accepted Answer

$$I = \frac{1}{2} m r^2$ = \frac{1}{2} 	imes 10\ 	ext{kg} 	imes (0.20\ 	ext{m})^2$ = 0.2\ 	ext{kg-m}^2$$ 
 $$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{10\ 	ext{rad/s}}{5\ 	ext{s}} = 2\ 	ext{rad/s}^2$$ 
 $$T = I \alpha = 0.2\ 	ext{kg-m}^2 	imes 2\ 	ext{rad/s}^2 = 0.4\ 	ext{N-m}$$

Question 6

Which of the following statements about the moment of inertia is correct?

Accepted Answer

The moment of inertia depends on both the mass and its distribution relative to the axis of rotation.

Question 7

A thin hoop and a solid sphere both have mass $$m$$ and radius $$r. $$Which has the greater moment of inertia about their central axis?

Accepted Answer

The thin hoop, because $$I_{	ext{hoop}} = mr^2$ and I$$_{	ext{sphere}} = \frac{2}{5}$$mr^2$$

Question 8

A pulley of radius 0.15 m is accelerated from rest to 14 rev/min in 4.5 s by a force of 13.0 N. What is the moment of inertia of the pulley?

Accepted Answer

$$\omega = 14\ 	ext{rev/min} 	imes \frac{2\pi}{60} = 1.47\ 	ext{rad/s}$$ 
 $$\alpha = \frac{1.47\ 	ext{rad/s}}{4.5\ 	ext{s}} = 0.33\ 	ext{rad/s}^2$$ 
 $$T = F r = 13.0\ 	ext{N} 	imes 0.15\ 	ext{m} = 1.95\ 	ext{N-m}$$ 
 $$I = \frac{T}{\alpha} = \frac{1.95}{0.33} = 5.9\ 	ext{kg-m}^2$$

Question 9

A baton consists of two masses (0.30 kg each) at the ends of a 0.66 m rod. What is the moment of inertia about the center?

Accepted Answer

$$I = 2 	imes m r^2$ = 2 	imes 0.30\ 	ext{kg} 	imes (0.33\ 	ext{m})^2$ = 0.066\ 	ext{kg-m}^2$$

Question 10

If you double the distance of two children from the center of a seesaw, by what factor does the moment of inertia increase?

Accepted Answer

It increases by a factor of 4, since $$I = m r^2$ and r$ is doubled.

Question 11

A solid wheel (mass 5.2 kg, diameter 0.55 m) is at rest. You want it to rotate at 12 rev/s in 35 s. What force must you exert on a strap wrapped around the wheel?

Accepted Answer

$$r = 0.275\ 	ext{m}$$ 
 $$I = \frac{1}{2} m r^2$ = \frac{1}{2} 	imes 5.2\ 	ext{kg} 	imes (0.275\ 	ext{m})^2$ = 0.393\ 	ext{kg-m}^2$$ 
 $$\omega = 12\ 	ext{rev/s} 	imes 2\pi = 75.4\ 	ext{rad/s}$$ 
 $$\alpha = \frac{75.4}{35} = 2.15\ 	ext{rad/s}^2$$ 
 $$T = I \alpha = 0.393 	imes 2.15 = 0.845\ 	ext{N-m}$$ 
 $$F = \frac{T}{r} = \frac{0.845}{0.275} = 3.07\ 	ext{N}$$

Question 12

Two people pull on ropes wrapped around the edge of a large wheel (mass 12 kg, diameter 2.4 m). One pulls clockwise with 43 N, the other counterclockwise with 67 N. What is the net torque?

Accepted Answer

$$r = 1.2\ 	ext{m}$$ 
 $$T_{	ext{net}} = (67\ 	ext{N} - 43\ 	ext{N}) 	imes 1.2\ 	ext{m} = 28.8\ 	ext{N-m}$$

Question 13

A chain is wrapped around a pulley and pulled with a force of 16.0 N. The pulley has a radius of 0.20 m. Its rotational speed increases from 0.0 to 17.0 rev/min in 5.00 s. What is the moment of inertia?

Accepted Answer

$$\omega = 17\ 	ext{rev/min} 	imes \frac{2\pi}{60} = 1.78\ 	ext{rad/s}$$ 
 $$\alpha = \frac{1.78}{5.00} = 0.356\ 	ext{rad/s}^2$$ 
 $$T = F r = 16.0\ 	ext{N} 	imes 0.20\ 	ext{m} = 3.2\ 	ext{N-m}$$ 
 $$I = \frac{T}{\alpha} = \frac{3.2}{0.356} = 9.0\ 	ext{kg-m}^2$$

Question 14

A solid wheel accelerates at 3.25 rad/s² when a force of 4.5 N exerts a torque on it. If the wheel is replaced by a wheel with all its mass on the rim ($$I = mr^2$), what force should be exerted to achieve the same angular acceleration?

Accepted Answer

A greater force is needed because the moment of inertia is larger for the rim mass. $$F_{	ext{rim}} = \frac{I_{	ext{rim}} \alpha}{r}$$

Question 15

A bicycle wheel can be accelerated by pulling on the chain (radius 0.14 m) or a string wrapped around the tire (radius 0.38 m). If 15 N is needed on the chain, what force is needed on the string for the same acceleration?

Accepted Answer

$$F_{	ext{string}} = \frac{T}{r_{	ext{string}}} = \frac{15\ 	ext{N} 	imes 0.14\ 	ext{m}}{0.38\ 	ext{m}} = 5.53\ 	ext{N}$$

Question 16

A ball on a low-friction, tilted surface slides downhill without rotating. On a rough surface, it rolls. Why?

Accepted Answer

Friction provides a torque that causes the ball to rotate; without friction, no torque acts to rotate the ball.

Question 17

If you apply a force at an angle less than 90° to the lever arm, how does the torque compare to applying the same force perpendicular?

Accepted Answer

The torque is less, because $$T = Fr \sin 	heta$$ and $$\sin 	heta \u003c 1$$ for $$	heta \u003c 90^\circ$$

Rotational Dynamics: Torque, Moment of Inertia, and Newton's Second Law for Rotation

Guida di studio - Domande di pratica

Study guide - Flashcard