AP Physics C: Mechanics Rotation — Worked Answer Explanations

Unit 5 · 12 questions explained

1. Question 1 · Easy

   A wheel starts from rest and reaches an angular velocity of $10$ rad/s in $5$ s under constant angular acceleration. What is the angular acceleration?

   • A
     $0.5$ rad/s²

     Why not A: Divided by the square of time rather than time itself.
   • B
     $2$ rad/s²Correct
   • C
     $50$ rad/s²

     Why not C: Multiplied $\omega$ by $t$ rather than dividing.
   • D
     $5$ rad/s²

     Why not D: Divided $t$ by $\omega$ instead of $\omega$ by $t$.
   Explanation

   $\alpha =\frac{\Delta \omega }{\Delta t}=\frac{10-0}{5}=2$ rad/s².

   Key takeaway

   Angular acceleration $\alpha = \Delta\omega/\Delta t$ is the rotational analog of linear acceleration $a = \Delta v/\Delta t$.

2. Question 2 · Easy

   A uniform solid disk of mass $M$ and radius $R$ has moment of inertia about its central axis $I=\frac{1}{2}M{R}^2$. A net torque $\tau =4$ N·m is applied to a disk with $M=2$ kg and $R=1$ m. What is the angular acceleration?

   • A
     $2$ rad/s²

     Why not A: Used $I=M{R}^2$ (full, not half) for the disk.
   • B
     $4$ rad/s²Correct
   • C
     $8$ rad/s²

     Why not C: Used $I=\frac{1}{4}M{R}^2$ instead of $\frac{1}{2}M{R}^2$.
   • D
     $1$ rad/s²

     Why not D: Forgot to account for $R^2$ in $I$, using $I=M$ instead.
   Explanation

   $I=\frac{1}{2}M{R}^2=\frac{1}{2}(2)(1{)}^2=1$ kg·m². $\alpha =\frac{\tau }{I}=\frac{4}{1}=4$ rad/s².

   Key takeaway

   Newton's second law for rotation: $\tau_{net} = I\alpha$. Know the moments of inertia of common shapes.

3. Question 3 · Easy

   A particle of mass $m=0.5$ kg moves in a circle of radius $r=2$ m at speed $v=4$ m/s. What is its angular momentum about the center?

   • A
     $2$ kg·m²/s

     Why not A: Used $L=m{r}^2$ (omitted $v$ or used $r$ twice).
   • B
     $4$ kg·m²/sCorrect
   • C
     $8$ kg·m²/s

     Why not C: Used $L=mv{r}^2$ (extra factor of $r$).
   • D
     $0.5$ kg·m²/s

     Why not D: Divided $mv$ by $r$ instead of multiplying.
   Explanation

   $L=mvr=0.5\times 4\times 2=4$ kg·m²/s. Equivalently, $L=I\omega$ where $I=m{r}^2=2$ kg·m² and $\omega =v/r=2$ rad/s, giving $L=4$ kg·m²/s.

   Key takeaway

   For a particle in circular motion, $L = mvr = I\omega$. Both formulas give the same result.

4. Question 4 · Easy

   A solid cylinder ($I=\frac{1}{2}M{R}^2$) rolls without slipping down a frictionless incline of height $h$. Using energy conservation, find the speed at the bottom.

   • A
     $\sqrt{2gh}$

     Why not A: Used energy for a sliding (non-rolling) object — ignored rotational KE.
   • B
     $\sqrt{\frac{4gh}{3}}$Correct
   • C
     $\sqrt{gh}$

     Why not C: Used $\frac{1}{2}mgh=\frac{1}{2}m{v}^2$ (factor of $\frac{1}{2}$ error).
   • D
     $\sqrt{\frac{2gh}{3}}$

     Why not D: Wrote $mgh=\frac{1}{2}I{v}^2$ instead of including both translational and rotational KE.
   Explanation

   Energy conservation: $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$. For rolling without slipping, $\omega =v/R$ and $I=\frac{1}{2}m{R}^2$:
   $$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\cdot \frac{1}{2}m{R}^2\cdot \frac{v^2}{R^2}=\frac{1}{2}m{v}^2+\frac{1}{4}m{v}^2=\frac{3}{4}m{v}^2.$$
   $$v=\sqrt{\frac{4gh}{3}}.$$

   Key takeaway

   Rolling objects have both translational ($\frac{1}{2}mv^2$) and rotational ($\frac{1}{2}I\omega^2$) kinetic energy. Use $\omega = v/R$ to combine them.

5. Question 5 · Medium

   Using the integral definition $I=\int r^{2}dm$, find the moment of inertia of a thin uniform rod of mass $M$ and length $L$ about one end.

   • A
     $\frac{1}{12}M{L}^2$

     Why not A: Moment about the center of mass, not the end.
   • B
     $\frac{1}{2}M{L}^2$

     Why not B: Used ${\int }_0^{L}rdr$ instead of ${\int }_0^L{r}^{2}dr$.
   • C
     $\frac{1}{3}M{L}^2$Correct
   • D
     $M{L}^2$

     Why not D: Treated all mass as concentrated at the far end ($I=M{L}^2$).
   Explanation

   Linear mass density $\lambda =M/L$. $dm=\lambda dr$.
   $$I={\int }_0^L{r}^2\lambda dr=\frac{M}{L}\cdot \frac{L^3}{3}=\frac{1}{3}M{L}^2.$$
   Note: about the center, $I_{cm}=\frac{1}{12}M{L}^2$. The parallel-axis theorem gives $I_{end}=I_{cm}+M(L/2{)}^2=\frac{1}{12}M{L}^2+\frac{1}{4}M{L}^2=\frac{1}{3}M{L}^2$, confirming the integral.

   Key takeaway

   $I = \int r^2\,dm$ with $dm = \lambda\,dr$ for a rod. Integration from $0$ to $L$ gives $\frac{1}{3}ML^2$ about one end.

6. Question 6 · Medium

   A torque $\tau (t)=3t$ N·m acts on a flywheel with $I=6$ kg·m² starting from rest. Find $\omega$ at $t=4$ s.

   • A
     $2$ rad/s

     Why not A: Evaluated $\tau (4)/I$ without integrating over time.
   • B
     $4$ rad/sCorrect
   • C
     $8$ rad/s

     Why not C: Computed $\int \tau dt=24$ and forgot to divide by $I=6$.
   • D
     $24$ rad/s

     Why not D: Computed $\int \tau dt$ and used it directly as $\omega$ without dividing by $I$.
   Explanation

   $I\alpha =\tau (t)\Rightarrow \alpha (t)=\tau /I=3t/6=t/2$. Integrate: $\omega (t)={\int }_0^t\frac{t^{\prime }}{2}d{t}^{\prime }=\frac{t^2}{4}$. At $t=4$: $\omega =16/4=4$ rad/s.

   Key takeaway

   For variable torque, $\alpha(t) = \tau(t)/I$, then integrate $\alpha$ to find $\omega(t)$.

7. Question 7 · Medium

   A figure skater with $I_1=4$ kg·m² spins at $2$ rad/s and pulls in arms to reach $I_2=1$ kg·m². What is the new angular velocity?

   • A
     $2$ rad/s

     Why not A: Assumed angular velocity is unchanged when arms are pulled in.
   • B
     $4$ rad/s

     Why not B: Divided moment of inertia change ($I_1/I_2=4$) but applied it incorrectly, getting $4\times 1=4$.
   • C
     $8$ rad/sCorrect
   • D
     $0.5$ rad/s

     Why not D: Divided rather than multiplied (inverted the ratio).
   Explanation

   Conservation of angular momentum (no external torque): $I_1{\omega }_1=I_2{\omega }_2$. ${\omega }_2=\frac{I_1{\omega }_1}{I_2}=\frac{4\times 2}{1}=8$ rad/s.

   Key takeaway

   When no external torque acts, $L = I\omega$ is conserved. Decreasing $I$ increases $\omega$ proportionally.

8. Question 8 · Medium

   A uniform disk ($I_{cm}=\frac{1}{2}M{R}^2$) has a small bolt of mass $m$ attached at its rim. Using the parallel-axis theorem, what is the total moment of inertia about the disk's center?

   • A
     $\frac{1}{2}M{R}^2$

     Why not A: Ignored the bolt's contribution.
   • B
     $\frac{1}{2}(M+m)R^2$

     Why not B: Treated both mass and bolt as if both had $\frac{1}{2}{R}^2$ moment arms.
   • C
     $\frac{1}{2}M{R}^2+m{R}^2$Correct
   • D
     $\frac{1}{2}M{R}^2+\frac{1}{2}m{R}^2$

     Why not D: Applied disk formula to the bolt, but the bolt is a point mass at radius $R$, so its $I=m{R}^2$.
   Explanation

   The disk contributes $I_{disk}=\frac{1}{2}M{R}^2$. The bolt is a point mass at radius $R$, so $I_{bolt}=m{R}^2$. Total: $I=\frac{1}{2}M{R}^2+m{R}^2$. No parallel-axis shift is needed for the bolt since it is already at radius $R$ from the axis.

   Key takeaway

   The parallel-axis theorem $I = I_{cm} + Md^2$ shifts an object's moment of inertia to a parallel axis. For a point mass at distance $R$, $I = mR^2$ directly.

9. Question 9 · Hard

   A net torque ${\tau }_{net}=10$ N·m acts on a solid sphere ($I=\frac{2}{5}M{R}^2$, $M=5$ kg, $R=0.2$ m). Find $\alpha$.

   • A
     $50$ rad/s²

     Why not A: Used $I=M{R}^2$ (no $2/5$ factor).
   • B
     $125$ rad/s²Correct
   • C
     $250$ rad/s²

     Why not C: Used $I=\frac{1}{5}M{R}^2$ instead of $\frac{2}{5}M{R}^2$.
   • D
     $10$ rad/s²

     Why not D: Set $\alpha =\tau /M$ (used mass instead of moment of inertia).
   Explanation

   $I=\frac{2}{5}M{R}^2=\frac{2}{5}(5)(0.04)=0.08$ kg·m². $\alpha =\frac{\tau }{I}=\frac{10}{0.08}=125$ rad/s².

   Key takeaway

   Know the moments of inertia: sphere $\frac{2}{5}MR^2$, disk $\frac{1}{2}MR^2$, rod (center) $\frac{1}{12}ML^2$, rod (end) $\frac{1}{3}ML^2$.

10. Question 10 · Hard

    A student derives the moment of inertia of a thin spherical shell of mass $M$ and radius $R$ about a diameter. Which result is correct?

    • A
      $\frac{1}{2}M{R}^2$

      Why not A: Moment of inertia for a solid disk, not a spherical shell.
    • B
      $\frac{2}{5}M{R}^2$

      Why not B: Moment of a solid sphere; a hollow shell has larger $I$ for the same $M$, $R$.
    • C
      $\frac{2}{3}M{R}^2$Correct
    • D
      $M{R}^2$

      Why not D: Corresponds to all mass at radius $R$ with no angular factor (hoop with all mass at the equator only).
    Explanation

    Integrating $dI=r_{\perp }^{2}dm$ over the shell surface (with $r_{\perp }=R\sin \varphi$ and $dm=\frac{M}{4\pi R^2}\cdot 2\pi R^2\sin \varphi d\varphi$):
    $$I={\int }_0^{\pi }{R}^2{\sin }^2\varphi \cdot \frac{M}{2}\sin \varphi d\varphi =\frac{M{R}^2}{2}{\int }_0^{\pi }{\sin }^3\varphi d\varphi =\frac{M{R}^2}{2}\cdot \frac{4}{3}=\frac{2}{3}M{R}^2.$$

    Key takeaway

    Hollow spherical shell: $I = \frac{2}{3}MR^2$. Greater than solid sphere ($\frac{2}{5}MR^2$) because mass is farther from the axis on average.

11. Question 11 · Hard

    A uniform rod ($M$, $L$) is held horizontal, pivoted at one end, and released. Using $\tau =I\alpha$ and torque from gravity, find the initial angular acceleration and the linear acceleration of the free end.

    • A
      $\alpha =\frac{g}{L}$; $a_{tip}=g$

      Why not A: Used $\tau =MgL$ and $I=M{L}^2$ (forgot factor of $\frac{1}{3}$).
    • B
      $\alpha =\frac{3g}{2L}$; $a_{tip}=\frac{3g}{2}$Correct
    • C
      $\alpha =\frac{3g}{L}$; $a_{tip}=3g$

      Why not C: Used torque $=Mg(L/2)$ but moment $I=\frac{1}{2}M{L}^2$ (disk instead of rod).
    • D
      $\alpha =\frac{g}{2L}$; $a_{tip}=\frac{g}{2}$

      Why not D: Used $I=\frac{1}{2}M{L}^2$ (disk formula) with correct torque.
    Explanation

    Torque about pivot from gravity (acts at center $L/2$): $\tau =Mg\frac{L}{2}$. Moment of inertia of rod about end: $I=\frac{1}{3}M{L}^2$. Angular acceleration:
    $$\alpha =\frac{\tau }{I}=\frac{Mg(L/2)}{\frac{1}{3}M{L}^2}=\frac{3g}{2L}.$$
    Linear acceleration of the tip: $a_{tip}=\alpha L=\frac{3g}{2}$. Note $a_{tip}>g$ — the tip falls faster than free-fall because the pivot constrains the base.

    Key takeaway

    For a pivoted rod, $\tau = Mg(L/2)$ (gravity at center) and $I_{end} = \frac{1}{3}ML^2$, giving $\alpha = 3g/(2L)$. The tip's linear acceleration $3g/2$ exceeds $g$.

12. Question 12 · Hard

    A disk ($I_1=2$ kg·m², ${\omega }_1=6$ rad/s) drops onto a stationary disk ($I_2=3$ kg·m²) on the same frictionless axle. They reach a common angular velocity ${\omega }_f$. How much kinetic energy is lost?

    • A
      $0$ J

      Why not A: Confused angular momentum conservation with kinetic energy conservation — KE is not conserved in this inelastic coupling.
    • B
      $14.4$ J

      Why not B: Computed the final KE ($14.4$ J) rather than the energy lost ($\Delta KE=K{E}_i-K{E}_f$).
    • C
      $21.6$ JCorrect
    • D
      $36$ J

      Why not D: Reported the initial kinetic energy as the loss.
    Explanation

    Angular momentum conservation: $I_1{\omega }_1=(I_1+I_2){\omega }_f\Rightarrow {\omega }_f=\frac{2(6)}{5}=2.4$ rad/s.

    $K{E}_i=\frac{1}{2}(2)(6^2)=36$ J.

    $K{E}_f=\frac{1}{2}(5)(2.4^2)=\frac{1}{2}(5)(5.76)=14.4$ J.

    $\Delta KE=36-14.4=21.6$ J lost.

    General formula: $\Delta KE=\frac{I_1{I}_2{\omega }_1^2}{2(I_1+I_2)}=\frac{(2)(3)(36)}{2(5)}=\frac{216}{10}=21.6$ J. ✓

    Key takeaway

    Disk-coupling is a rotational perfectly-inelastic collision: $L$ is conserved, $KE$ is not. $\Delta KE = \frac{I_1 I_2}{2(I_1+I_2)}\omega_1^2$.