Newton's Second Law in Rotational Form Practice

Sample questions

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1. Sample 1difficulty 1/5

   A disk is in <strong>rotational equilibrium</strong>. Which is true?

   • A

     Its angular acceleration is zero.

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   • B

     Its angular velocity is zero.

   • C

     It must be at rest.

   • D

     Both A and C.

   Why

   Rotational equilibrium = zero net torque = zero angular acceleration. The disk may rotate at any constant angular velocity.

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2. Sample 2difficulty 2/5

   A flywheel ($I = 0.50~\text{kg·m}^2$) is acted on by a net torque of $4.0~\text{N·m}$. Its angular acceleration is

   • A

     $0.13~\text{rad/s}^2$

   • B

     $20~\text{rad/s}^2$

   • C

     $8.0~\text{rad/s}^2$

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   • D

     $2.0~\text{rad/s}^2$

   Why

   $\alpha = \tau/I = 4.0/0.50 = 8.0~\text{rad/s}^2$.

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3. Sample 3difficulty 2/5

   A wheel ($I = 0.20~\text{kg·m}^2$) experiences a constant net torque of $0.80~\text{N·m}$. After $5.0~\text{s}$, starting from rest, its angular speed is

   • A

     $10~\text{rad/s}$

   • B

     $20~\text{rad/s}$

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   • C

     $2~\text{rad/s}$

   • D

     $4~\text{rad/s}$

   Why

   $\alpha = 0.80/0.20 = 4.0~\text{rad/s}^2$. $\omega = \alpha t = 20~\text{rad/s}$.

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4. Sample 4difficulty 2/5

   A net torque of $12~\text{N·m}$ acts on a wheel with moment of inertia $I = 0.30~\text{kg·m}^2$. What is the angular acceleration?

   • A

     $2.5~\text{rad/s}^2$

   • B

     $40~\text{rad/s}^2$

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   • C

     $4.0~\text{rad/s}^2$

   • D

     $12~\text{rad/s}^2$

   Why

   $\alpha = \tau/I = 12/0.30 = 40~\text{rad/s}^2$.

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5. Sample 5difficulty 3/5

   A solid disk pulley of mass $M = 2.0~\text{kg}$ and radius $R = 0.10~\text{m}$ is free to rotate on its axle. A string wound around it holds a $1.0~\text{kg}$ mass. Using $g \approx 10~\text{m/s}^2$, what is the hanging mass's downward acceleration? (Disk: $I = \tfrac{1}{2}MR^2$.)

   • A

     $10~\text{m/s}^2$

   • B

     $5.0~\text{m/s}^2$

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   • C

     $2.5~\text{m/s}^2$

   • D

     $6.7~\text{m/s}^2$

   Why

   Equations: $mg - T = ma$ and $TR = I\alpha = \tfrac{1}{2}MR^2 (a/R)$, so $T = \tfrac{1}{2}M a$. Substituting: $mg = a(m + M/2) \Rightarrow a = g\,m/(m + M/2) = 10(1)/(1 + 1) = 5.0~\text{m/s}^2$.

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