Unit 5: Rotational Motion (Part 1)

Kinematics, Connections to Linear Motion, and Torque

Section 5.1

Rotational Kinematics

Rotational kinematics describes motion using angular variables. These variables are analogous to linear kinematics. For variable acceleration, calculus must be used: \( \omega = \displaystyle\frac{d\theta}{dt} \) and \( \alpha = \displaystyle\frac{d\omega}{dt} \).

Q1: A wheel’s angular velocity is \( \omega(t) = 4t^3 – 2t \). What is the angular acceleration at \( t = 2 \text{ s} \)?

(A) \( 14 \text{ rad/s}^2 \)
(B) \( 28 \text{ rad/s}^2 \)
(C) \( 46 \text{ rad/s}^2 \)
(D) \( 50 \text{ rad/s}^2 \)

Differentiate the angular velocity to find angular acceleration:
\( \alpha(t) = \displaystyle\frac{d\omega}{dt} = 12t^2 – 2 \).
Evaluate at \( t = 2 \): \( \alpha(2) = 12(2^2) – 2 = 48 – 2 = 46 \text{ rad/s}^2 \).

Correct Answer: C

Q2: A disk starting from rest has constant \( \alpha = 4 \text{ rad/s}^2 \). How many revolutions are made in the first \( 10 \text{ s} \)?

(A) \( \displaystyle\frac{100}{\pi} \)
(B) \( \displaystyle\frac{200}{\pi} \)
(C) \( 200 \)
(D) \( 400 \)

Find total angular displacement: \( \Delta\theta = \omega_0t + \displaystyle\frac{1}{2}\alpha t^2 = 0 + \displaystyle\frac{1}{2}(4)(10^2) = 200 \text{ rad} \).
Convert to revolutions: \( \text{Rev} = \displaystyle\frac{200}{2\pi} = \displaystyle\frac{100}{\pi} \).

Correct Answer: A

Section 5.2

Linear and Rotational Connections

Points on a rigid body share the same angular motion but have different linear properties based on their distance \( r \) from the axis. Use \( v_t = r\omega \) and \( a_t = r\alpha \).

Q1: A turntable (\( r = 0.2 \text{ m} \)) rotates at \( 33 \text{ rpm} \). What is the tangential speed on the rim?

(A) \( 0.11\pi \text{ m/s} \)
(B) \( 0.22\pi \text{ m/s} \)
(C) \( 0.66\pi \text{ m/s} \)
(D) \( 1.32\pi \text{ m/s} \)

Convert rpm to rad/s: \( \omega = 33 \displaystyle\frac{\text{rev}}{\text{min}} \times \displaystyle\frac{2\pi \text{ rad}}{60 \text{ s}} = 1.1\pi \text{ rad/s} \).
Find speed: \( v_t = r\omega = (0.2)(1.1\pi) = 0.22\pi \text{ m/s} \).

Correct Answer: B

Q2: A wheel rolls without slipping. If the center speed is \( V \), what is the speed at the very top relative to the ground?

(A) \( 0 \)
(B) \( V \)
(C) \( 2V \)
(D) \( \sqrt{2}V \)

The top point has translational velocity \( V \) and rotational velocity \( r\omega = V \).
Total speed relative to the ground is the vector sum: \( V + V = 2V \).

Correct Answer: C

Section 5.3

Torque

Torque is the cross product of position and force: \( \vec{\tau} = \vec{r} \times \vec{F} \). The magnitude is \( \tau = rF \sin\theta \). It measures the tendency of a force to rotate an object about an axis.

Q1: A force \( \vec{F} = 3\hat{j} \text{ N} \) is applied at \( \vec{r} = 2\hat{i} \text{ m} \). What is the torque vector?

(A) \( 6\hat{k} \text{ Nm} \)
(B) \( -6\hat{k} \text{ Nm} \)
(C) \( 6\hat{i} \text{ Nm} \)
(D) \( 0 \text{ Nm} \)

\( \vec{\tau} = \vec{r} \times \vec{F} = (2\hat{i}) \times (3\hat{j}) = 6(\hat{i} \times \hat{j}) = 6\hat{k} \text{ Nm} \).

Correct Answer: A

Q2: An object is in equilibrium. Which statement must be true?

(A) Only the net force is zero
(B) Only the net torque is zero
(C) Both net force and net torque are zero
(D) The object must be at rest

For static or dynamic equilibrium, translational equilibrium (\( \sum F = 0 \)) and rotational equilibrium (\( \sum \tau = 0 \)) must both be satisfied.

Correct Answer: C

Rotational Formulas Recap

Angular Velocity\( \omega = \displaystyle\frac{d\theta}{dt} \)

Angular Accel\( \alpha = \displaystyle\frac{d\omega}{dt} \)

Tangential Speed\( v_t = r\omega \)

Centripetal Accel\( a_c = r\omega^2 \)

Torque Vector\( \vec{\tau} = \vec{r} \times \vec{F} \)

Torque Magnitude\( \tau = rF \sin\theta \)

Unit 5: Rotational Motion (Part 2)

Inertia, Rotational Equilibrium, and Newton’s Second Law

Section 5.4

Rotational Inertia (Moment of Inertia)

Rotational inertia (\( I \)) represents an object’s resistance to angular acceleration. For point masses, \( I = \sum m_i r_i^2 \). For continuous bodies, use calculus: \( I = \int r^2 dm \). The Parallel Axis Theorem (\( I = I_{cm} + Md^2 \)) allows you to find the inertia about any axis parallel to one through the center of mass.

Q1: A system consists of two point masses, \( m \) and \( 2m \), separated by a distance \( L \). What is the rotational inertia of the system about an axis passing through the midpoint of the two masses?

(A) \( \displaystyle\frac{3}{4}mL^2 \)
(B) \( \displaystyle\frac{3}{2}mL^2 \)
(C) \( 3mL^2 \)
(D) \( \displaystyle\frac{1}{2}mL^2 \)

The midpoint is at a distance of \( L/2 \) from each mass.
\( I = \sum m_i r_i^2 = m(L/2)^2 + 2m(L/2)^2 \)
\( I = m(L^2/4) + 2m(L^2/4) = \displaystyle\frac{3}{4}mL^2 \).

Correct Answer: A

Q2: A thin uniform rod of mass \( M \) and length \( L \) has a rotational inertia \( I_{center} = \displaystyle\frac{1}{12}ML^2 \). What is its rotational inertia about an axis through one of its ends?

(A) \( \displaystyle\frac{1}{6}ML^2 \)
(B) \( \displaystyle\frac{1}{4}ML^2 \)
(C) \( \displaystyle\frac{1}{3}ML^2 \)
(D) \( \displaystyle\frac{7}{12}ML^2 \)

Use the Parallel Axis Theorem: \( I = I_{cm} + Md^2 \).
The distance from the center to the end is \( d = L/2 \).
\( I = \displaystyle\frac{1}{12}ML^2 + M(L/2)^2 = \displaystyle\frac{1}{12}ML^2 + \displaystyle\frac{1}{4}ML^2 \)
\( I = \displaystyle\frac{1+3}{12}ML^2 = \displaystyle\frac{4}{12}ML^2 = \displaystyle\frac{1}{3}ML^2 \).

Correct Answer: C

Q3: A solid sphere, a solid cylinder, and a thin hoop all have the same mass and radius. Which one has the smallest rotational inertia?

(A) The hoop
(B) The solid cylinder
(C) The solid sphere
(D) They all have the same inertia

Mass closer to the axis results in lower inertia.
– Hoop: \( I = MR^2 \) (constant 1.0)
– Cylinder: \( I = 0.5MR^2 \)
– Sphere: \( I = 0.4MR^2 \)
Since \( 0.4 < 0.5 < 1.0 \), the sphere has the smallest inertia.

Correct Answer: C

Q4: A non-uniform rod of length \( L \) has a linear mass density \( \lambda(x) = kx \). What is the rotational inertia about an axis through the left end (\( x = 0 \))?

(A) \( \displaystyle\frac{1}{3}kL^3 \)
(B) \( \displaystyle\frac{1}{4}kL^4 \)
(C) \( \displaystyle\frac{1}{2}kL^2 \)
(D) \( \displaystyle\frac{1}{5}kL^5 \)

Use \( I = \int x^2 dm \). Here \( dm = (kx) dx \).
\( I = \int_{0}^{L} kx^3 dx = \left[ \displaystyle\frac{1}{4}kx^4 \right]_0^L = \displaystyle\frac{1}{4}kL^4 \).

Correct Answer: B

Q5: If an object’s rotational inertia is doubled while the net torque remains constant, the angular acceleration will:

(A) Double
(B) Quadruple
(C) Remain the same
(D) Be halved

From \( \alpha = \tau/I \), if \( I \) doubles and \( \tau \) is constant, \( \alpha \) becomes half of its original value.

Correct Answer: D

Section 5.5

Rotational Equilibrium and Newton’s First Law

Rotational Equilibrium requires \( \sum \tau = 0 \). For total static equilibrium, both \( \sum \vec{F} = 0 \) and \( \sum \vec{\tau} = 0 \) must be true.

Q1: A uniform beam of length \( 6 \text{ m} \) is supported at its center. A \( 40 \text{ kg} \) boy sits \( 2 \text{ m} \) to the left of the center. Where should a \( 20 \text{ kg} \) girl sit to keep the beam balanced?

(A) \( 2 \text{ m} \) to the right of the center
(B) \( 3 \text{ m} \) to the right of the center
(C) \( 4 \text{ m} \) to the right of the center
(D) The beam cannot be balanced.

Torque balance: \( (40)(2) = (20)(x) \Rightarrow x = 4 \text{ m} \).
Since the beam only extends 3 m on each side of the center, the girl would need to sit off the beam.

Correct Answer: D

Q2: An object rotates at a constant angular velocity. What can be concluded about the net torque?

(A) It is constant and positive
(B) It is zero
(C) It is increasing linearly
(D) It is equal to the rotational inertia

Constant angular velocity means \( \alpha = 0 \). Since \( \sum \tau = I \alpha \), the net torque must be zero.

Correct Answer: B

Q3: A ladder rests against a smooth vertical wall and a rough horizontal floor. Which force prevents the ladder from falling?

(A) The normal force from the wall
(B) The weight of the ladder
(C) The static friction force from the floor
(D) The normal force from the floor

Static friction at the base pushes the ladder toward the wall, preventing the bottom from sliding out and maintaining equilibrium.

Correct Answer: C

Q4: If the net force on an object is zero, is it possible for the object to NOT be in rotational equilibrium?

(A) No, net force zero always implies net torque zero.
(B) Yes, but only if the object is at rest.
(C) Yes, if a force couple acts on it.
(D) No, a net torque always produces a net force.

A “force couple” (equal and opposite forces acting along different lines) has a net force of zero but produces a non-zero net torque.

Correct Answer: C

Q5: To find the tension in a cable supporting a bar pivoted at a wall, which pivot point is most convenient for summing torques?

(A) The center of mass of the sign
(B) The center of the bar
(C) The point where the cable is attached
(D) The pivot at the wall

Summing torques about the wall pivot eliminates the unknown hinge forces from the equation, allowing you to solve for tension directly.

Correct Answer: D

Section 5.6

Newton’s Second Law in Rotational Form

The rotational equivalent of \( F = ma \) is \( \sum \tau = I \alpha \). For systems with both translation and rotation, link them using \( a = R \alpha \) for no-slip conditions.

Q1: A constant torque of \( 20 \text{ Nm} \) is applied to a wheel with \( I = 5 \text{ kgm}^2 \). What is the angular acceleration?

(A) \( 4 \text{ rad/s}^2 \)
(B) \( 100 \text{ rad/s}^2 \)
(C) \( 0.25 \text{ rad/s}^2 \)
(D) \( 15 \text{ rad/s}^2 \)

\( \alpha = \tau/I = 20/5 = 4 \text{ rad/s}^2 \).

Correct Answer: A

Q2: A mass \( m \) hangs from a string around a solid disk pulley of mass \( M \). What is the block’s acceleration? (\( I_{disk} = 0.5MR^2 \))

(A) \( g \)
(B) \( \displaystyle\frac{mg}{m + M} \)
(C) \( \displaystyle\frac{mg}{m + M/2} \)
(D) \( \displaystyle\frac{Mg}{m + M} \)

Block: \( mg – T = ma \). Pulley: \( TR = (0.5MR^2)(a/R) \Rightarrow T = 0.5Ma \).
Substitute \( T \): \( mg – 0.5Ma = ma \Rightarrow a = \displaystyle\frac{mg}{m + M/2} \).

Correct Answer: C

Q3: A solid cylinder rolls without slipping down an incline. What is the friction force? (\( I = 0.5MR^2 \))

(A) \( \displaystyle\frac{1}{2}Mg \sin\theta \)
(B) \( \displaystyle\frac{1}{3}Mg \sin\theta \)
(C) \( \mu Mg \cos\theta \)
(D) \( \displaystyle\frac{2}{3}Mg \sin\theta \)

Linear: \( Mg \sin\theta – f = Ma \). Rotational: \( fR = 0.5MR^2(a/R) \Rightarrow a = 2f/M \).
\( Mg \sin\theta – f = 2f \Rightarrow f = \displaystyle\frac{1}{3}Mg \sin\theta \).

Correct Answer: B

Q4: A thin hoop rolls down an incline. Compared to a solid cylinder of the same mass and radius, the hoop’s acceleration is:

(A) Greater
(B) Smaller
(C) The same
(D) Zero

The hoop has a larger rotational inertia (\( I = MR^2 \)) than the cylinder (\( I = 0.5MR^2 \)), leading to a smaller translational acceleration.

Correct Answer: B

Q5: If initial angular velocity is doubled and a constant frictional torque is applied, the time to stop will:

(A) Remain the same
(B) Double
(C) Quadruple
(D) Be halved

From \( \omega_f = \omega_0 + \alpha t \), where \( \omega_f = 0 \), \( t \) is directly proportional to \( \omega_0 \). Doubling initial velocity doubles the time.

Correct Answer: B

Dynamics of Rotation Recap

Point Inertia\( I = \sum m_i r_i^2 \)

Body Inertia\( I = \int r^2 dm \)

Parallel Axis\( I = I_{cm} + Md^2 \)

Newton II (Rot)\( \sum \tau = I \alpha \)

No-Slip Link\( a = R \alpha \)

Equilibrium\( \sum \tau = 0 \) & \( \sum F = 0 \)

AP Physics C Mechanics Quick Review: Unit 5 Torque and Rotational Dynamics