Rotational Motion Quiz Comprehensive Review

Rotational Motion Quiz Review

Handouts to Review

• Rotational Motion Part 1

• Rotational Motion Part 2

Definitions

1. Rotational Motion - the motion of an object around a circular path about an axis

2. Angular Displacement – the angle through which a body moves in a circular path

3. Angular Velocity – the rate of angular displacement of an object that moves about a circle

4. Tangential Velocity - the linear velocity of an object experiencing rotational motion

5. Rotational Inertia – the property of an object to resist changes to its rotational motion

6. Torque - a force acting at a distance from an axis of rotation; sometimes called a "turning force"

7. Center of Mass - the average position of all the mass that makes up an object

8. Centripetal Force - any force that acts toward a fixed center which causes an object to move in a circular path

Review Questions

Question 1

With regard to angular distances, how do radians relate to revolutions relate to degrees?

1 \text{ rev} = 2\pi \text{ radians} = 360^\circ

Question 2

The object shown in the picture below travels an angular distance of 10^\circ in 0.035 s. What is the angular distance traveled in radians? in revolutions? What is the angular velocity of the object.

• Angular displacement in revolutions:
  10^\circ \cdot \frac{1 \text{ rev}}{360^\circ} = 0.028 \text{ rev}

• Angular displacement in radians:
  10^\circ \cdot \frac{\pi \text{ rad}}{180^\circ} = 0.17 \text{ rad}

• Angular velocity:
  \omega = \frac{\Delta \theta}{\Delta t} = \frac{0.17 \text{ rad}}{0.035 \text{ s}} = 4.9 \text{ rad/s}

Question 3

What is the angular displacement of the Earth between Autumn and Summer? Give your answer in revolutions, radians and degrees.

• 0.75 rev counterclockwise (ccw)

• 270^\circ ccw

• 4.71 rad ccw

Question 4

A hamster is running on a suspended wheel. A) If the wheel goes around once in 0.73 s, what is its angular velocity in radians per second (rad/s)? B) If the wheel has a radius of 35 cm, what is the hamster's tangential velocity in meters per second (m/s)?

A)
\omega = \frac{\Delta \theta}{\Delta t} = \frac{2\pi \text{ rad}}{0.73 \text{ s}} = 8.6 \text{ rad/s} ccw

B)
Given: r = 35 \text{ cm} = 0.35 \text{ m}
v = r\omega = 0.35 \text{ m} \cdot 8.6 \text{ rad/s} = 3.0 \text{ m/s}

Question 5

What is the rotational inertia of a kickball? An officially sized kickball has a diameter or 22 cm and a mass of 0.549 kg.

Given:
radius = 11 \text{ cm} = 0.11 \text{ m}
I = \frac{2}{5}mr^2 = \frac{2}{5} \cdot 0.549 \text{ kg} \cdot (0.11 \text{ m})^2 = 0.0027 \text{ kg} \cdot \text{m}^2

Question 6

A bowling ball is found to have a rotational inertia of 0.25 kg.m². Will it be harder to rotate than the kickball in the last problem? How do you know?

Since the rotational inertia of the bowling ball is greater than the rotational inertia of the kickball, it will be harder to rotate the bowling ball.

0.25 > 0.0027

Question 7

A 4.0 N force F is applied to a bar that can pivot around its center as shown below. The force is perpendicular to the bar and is r = 0.25 m away from the center. What is the torque on the bar?

\tau = rF = 0.25 \text{ m} \cdot 4.0 \text{ N} = 1.0 \text{ N} \cdot \text{m} ccw

Question 8

What are two ways the torque in the previous problem could be increased in order to better rotate the bar?

1. Increase the force.

2. Increase the lever arm.
   \tau \propto rF

Question 9

Is the seesaw shown below balanced? If it is not balanced, which way will it rotate?

Given Torques:
\tau1 = r1F1 = 0.85 \text{ m} \cdot 375 \text{ N} = 320 \text{ N} \cdot \text{m} \tau2 = r2F2 = 0.60 \text{ m} \cdot 405 \text{ N} = 240 \text{ N} \cdot \text{m}

Net Torque:
\Sigma \tau = 320 + (-240) = 80 \text{ N} \cdot \text{m}

The seesaw is not balanced; it will rotate ccw.

Question 10

Describe the path of a broom thrown through the air.

1. The center of mass follows a parabolic path.

2. The broom rotates around the center of mass.

Question 11

A 25 kg girl is riding a carousel with a radius of 5.0 m. What is the centripetal force on the girl if her velocity is 6.0 m/s?

F_c = m \frac{v^2}{r} = 25 \text{ kg} \cdot \frac{(6.0 \text{ m/s})^2}{5.0 \text{ m}} = 180 \text{ N}

Question 12

Referring to the carousel in the last problem, how would the girl's tangential velocity be affected if she sat on a horse close to the center versus a horse on the outside edge?

v = r\omega
v \propto r

The closer to the middle, the smaller the tangential velocity.