Rotational Kinematics - AP Physics C Mechanics

1. The angular velocity of a motor is given by $\omega(t) = At^3 + B$, where $A=1.0 \, \text{rad/s}^4$ and $B=2.0 \, \text{rad/s}$. What is the average angular acceleration over the interval from $t=0 \, \text{s}$ to $t=2.0 \, \text{s}$?
   Answer: $4.0 \, \text{rad/s}^2$ Average angular acceleration is the change in angular velocity divided by the time interval, $\alpha_{avg} = \frac{\Delta\omega}{\Delta t}$. First, find the angular velocities: $\omega(2) = (1.0)(2.0)^3 + 2.0 = 8.0 + 2.0 = 10.0 \, \text{rad/s}$. $\omega(0) = (1.0)(0)^3 + 2.0 = 2.0 \, \text{rad/s}$. The change in angular velocity is $\Delta\omega = 10.0 - 2.0 = 8.0 \, \text{rad/s}$. The time interval is $\Delta t = 2.0 \, \text{s}$. So, $\alpha_{avg} = \frac{8.0 \, \text{rad/s}}{2.0 \, \text{s}} = 4.0 \, \text{rad/s}^2$.
2. The graph of angular velocity $\omega$ versus time $t$ for a rotating rigid body is a straight line with a positive slope, passing through the origin. What does this indicate about the body's motion?
   Answer: The body is undergoing a constant positive angular acceleration, starting from rest. The angular acceleration $\alpha$ is the slope of the $\omega$ versus $t$ graph. A straight line with a positive slope indicates a constant positive angular acceleration. Since the line passes through the origin, the initial angular velocity at $t=0$ is zero, meaning the body started from rest.
3. A grinding wheel, initially at rest, accelerates with a uniform angular acceleration of $4.0 \, \text{rad/s}^2$ for $5.0 \, \text{s}$. What is its final angular velocity?
   Answer: $20 \, \text{rad/s}$ For constant angular acceleration, the final angular velocity is given by the kinematic equation $\omega = \omega_0 + \alpha t$. Since the wheel starts from rest, $\omega_0 = 0$. Plugging in the values: $\omega = 0 + (4.0 \, \text{rad/s}^2)(5.0 \, \text{s}) = 20 \, \text{rad/s}$.
4. The angular position of a spinning wheel is given by $\theta(t) = \frac{1}{3}t^3 - \frac{5}{2}t^2 + 6t$, where $t$ is in seconds and $\theta$ is in radians. For $t>0$, at what time does the wheel first momentarily reverse its direction of rotation?
   Answer: $2.0 \, \text{s}$ The wheel reverses its direction of rotation when its angular velocity $\omega$ changes sign, which occurs when $\omega(t)=0$. To find $\omega(t)$, we differentiate $\theta(t)$ with respect to time: $\omega(t) = \frac{d\theta}{dt} = t^2 - 5t + 6$. Setting $\omega(t) = 0$ gives $t^2 - 5t + 6 = 0$, which factors to $(t-2)(t-3) = 0$. The solutions are $t=2 \, \text{s}$ and $t=3 \, \text{s}$. The first time this occurs for $t>0$ is at $t=2.0 \, \text{s}$.
5. A disk rotates according to $\omega(t) = 6 - 2t$ rad/s. If the disk starts at $\theta_0 = 0$ at t = 0, what is its angular position when it momentarily stops?
   Answer: 9 rad The disk stops when $\omega = 0$: $6 - 2t = 0$, so $t = 3$ s. To find angular position, integrate the angular velocity: $\theta(t) = \int_0^t \omega(t')dt' = \int_0^t (6 - 2t')dt' = 6t - t^2 + C$. Since $\theta(0) = 0$, we have $C = 0$, so $\theta(t) = 6t - t^2$. At $t = 3$ s: $\theta(3) = 6(3) - 3^2 = 18 - 9 = 9$ rad. Choice A (6 rad) might result from using only $6t$ and ignoring the $-t^2$ term. Choice C (12 rad) might come from errors in integration. Choice D (18 rad) might result from forgetting to subtract the $t^2$ term.
6. A wheel starts from rest and rotates with constant angular acceleration $\alpha$. After rotating through an angle $\theta_1$, its angular velocity is $\omega_1$. After rotating through an additional angle $\theta_2$, its angular velocity becomes $\omega_2$. Which expression correctly relates these quantities?
   Answer: $\omega_2^2 = \omega_1^2 + 2\alpha\theta_2$ This problem requires careful application of rotational kinematic equations. The wheel starts from rest and after angle $\theta_1$, has angular velocity $\omega_1$. Then after an additional angle $\theta_2$, it has angular velocity $\omega_2$. For the second phase of motion (from $\omega_1$ to $\omega_2$ through angle $\theta_2$), we use $\omega_f^2 = \omega_i^2 + 2\alpha\theta$, which gives $\omega_2^2 = \omega_1^2 + 2\alpha\theta_2$. Choice B incorrectly uses the total angle from the start. Choice C incorrectly takes the square root of the angle term. Choice D ignores the initial angular velocity $\omega_1$ for the second phase.
7. The angular position of a rotating object is given by $\theta(t) = 2t^3$, where $\theta$ is in radians and $t$ is in seconds. What is the average angular velocity of the object over the time interval from $t=1 \, \text{s}$ to $t=3 \, \text{s}$?
   Answer: $26 \, \text{rad/s}$ Average angular velocity is defined as the total angular displacement divided by the time interval, $\omega_{avg} = \frac{\Delta\theta}{\Delta t}$. First, find the angular positions at the endpoints of the interval: $\theta(3) = 2(3)^3 = 54 \, \text{rad}$ and $\theta(1) = 2(1)^3 = 2 \, \text{rad}$. The angular displacement is $\Delta\theta = \theta(3) - \theta(1) = 54 - 2 = 52 \, \text{rad}$. The time interval is $\Delta t = 3 - 1 = 2 \, \text{s}$. Therefore, $\omega_{avg} = \frac{52 \, \text{rad}}{2 \, \text{s}} = 26 \, \text{rad/s}$.
8. A ceiling fan blade rotating at $10 \, \text{rad/s}$ is switched off and slows down with a constant angular acceleration of magnitude $2.0 \, \text{rad/s}^2$. Through what angle does the blade rotate before coming to rest?
   Answer: $25 \, \text{rad}$ Use the rotational kinematic equation $\omega^2 = \omega_0^2 + 2\alpha\Delta\theta$. The final angular velocity is $\omega = 0$, the initial angular velocity is $\omega_0 = 10 \, \text{rad/s}$, and the angular acceleration is $\alpha = -2.0 \, \text{rad/s}^2$ (negative because it's slowing down). Solving for $\Delta\theta$: $0^2 = (10)^2 + 2(-2.0)\Delta\theta$. This gives $0 = 100 - 4.0\Delta\theta$, so $4.0\Delta\theta = 100$, and $\Delta\theta = 25 \, \text{rad}$.
9. A spinning top has an angular velocity given by $\omega(t) = \omega_0 e^{-kt}$, where $\omega_0$ and $k$ are positive constants. If the top's angular position at $t=0$ is $\theta=0$, what is its angular position $\theta(t)$ for $t>0$?
   Answer: $\theta(t) = \frac{\omega_0}{k}(1-e^{-kt})$ Angular position $\theta$ is the integral of angular velocity $\omega$ with respect to time. $\theta(t) = \int_0^t \omega(t') dt' + \theta(0)$. Given $\theta(0)=0$, we have $\theta(t) = \int_0^t \omega_0 e^{-kt'} dt' = \omega_0 \left[ -\frac{1}{k}e^{-kt'} \right]_0^t = -\frac{\omega_0}{k}(e^{-kt} - e^0) = -\frac{\omega_0}{k}(e^{-kt} - 1) = \frac{\omega_0}{k}(1-e^{-kt})$.
10. A disk with moment of inertia $I=0.30\ \text{kg}\cdot\text{m}^2$ is acted on by a constant torque $\tau=1.5\ \text{N}\cdot\text{m}$ for $t=2.0\ \text{s}$, starting from rest; friction is negligible. Key equations: $\tau=I\alpha$ and $\theta=\tfrac12\alpha t^2$. Using the given conditions, calculate the total angular displacement during the 2.0 s interval.
    Answer: $\theta=10\ \text{rad}$ This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically calculating angular displacement under constant angular acceleration. This parallels the translational kinematic equation for displacement under constant acceleration. For a disk with I = 0.30 kg·m² experiencing torque τ = 1.5 N·m for t = 2.0 s starting from rest, we need to find total angular displacement. Choice A is correct because first we find α = τ/I = 1.5/0.30 = 5.0 rad/s², then use θ = ½αt² = ½(5.0)(2.0)² = ½(5.0)(4.0) = 10 rad. Choice C at 20 rad might result from forgetting the factor of ½ in the kinematic equation. To help students: draw parallels between rotational and translational kinematic equations; emphasize that the ½ factor comes from integration just as in linear motion; and practice problems starting from rest to simplify calculations.
11. A motor propeller completes 1200 revolutions in one minute. Assuming the rotation is uniform, what is its angular velocity in radians per second?
    Answer: $40\pi \, \text{rad/s}$ First, convert revolutions per minute (rpm) to radians per second. There are $2\pi$ radians in one revolution and 60 seconds in one minute. Angular velocity $\omega = (1200 \, \frac{\text{rev}}{\text{min}}) \times (\frac{1 \, \text{min}}{60 \, \text{s}}) \times (\frac{2\pi \, \text{rad}}{1 \, \text{rev}}) = (20 \, \frac{\text{rev}}{\text{s}}) \times (2\pi \, \frac{\text{rad}}{\text{rev}}) = 40\pi \, \text{rad/s}$.
12. The angular acceleration of a wheel is given by $\alpha(t) = 6t - 12$, in units of rad/s$^2$. The wheel starts from rest at $t=0$. At what time $t>0$ does the wheel's direction of rotation reverse?
    Answer: $t=4 \, \text{s}$ To find when the direction reverses, we must find when the angular velocity $\omega$ is zero (and changing sign). First, find $\omega(t)$ by integrating $\alpha(t)$: $\omega(t) = \int (6t-12) dt = 3t^2 - 12t + C$. Since the wheel starts from rest, $\omega(0)=0$, which implies $C=0$. So, $\omega(t) = 3t^2 - 12t = 3t(t-4)$. The angular velocity is zero at $t=0$ and $t=4 \, \text{s}$. For $0 < t < 4$, $\omega(t)$ is negative. For $t > 4$, $\omega(t)$ is positive. Therefore, the direction of rotation reverses at $t=4 \, \text{s}$.
13. A rigid body rotates with a constant positive angular velocity $\omega$ about a fixed axis. Which of the following statements correctly describes its angular acceleration $\alpha$ and its angular position $\theta$, assuming $\theta(0)=0$?
    Answer: $\alpha=0$ and $\theta$ increases linearly with time. If the angular velocity $\omega$ is constant, the angular acceleration $\alpha = \frac{d\omega}{dt}$ must be zero. The angular position is the integral of the constant angular velocity, $\theta(t) = \int \omega dt = \omega t + C$. Since $\theta(0)=0$, the constant of integration is zero, so $\theta(t) = \omega t$. This shows that the angular position increases linearly with time.
14. A wheel with moment of inertia $I=1.2\ \text{kg}\cdot\text{m}^2$ must speed up from $\omega_0=5.0\ \text{rad/s}$ to $\omega_f=17\ \text{rad/s}$ in $t=4.0\ \text{s}$ under a constant net torque (friction included). Key equations: $\alpha=(\omega_f-\omega_0)/t$ and $\tau=I\alpha$. Using the given conditions, what is the torque required to achieve this change?
    Answer: $\tau=3.6\ \text{N}\cdot\text{m}$ This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically calculating required torque for a desired angular acceleration. The problem requires working backwards from kinematic information to find the necessary torque. For a wheel with I = 1.2 kg·m² accelerating from ω0 = 5.0 rad/s to ωf = 17 rad/s in t = 4.0 s, we first find the angular acceleration. Choice B is correct because α = (ωf - ω0)/t = (17 - 5.0)/4.0 = 12/4.0 = 3.0 rad/s², and then τ = Iα = 1.2 × 3.0 = 3.6 N·m. Choice C at 14 N·m might result from calculation errors or using wrong values. To help students: emphasize the two-step process of finding acceleration first, then torque; practice problems that work backwards from desired motion to required forces/torques; and always verify that calculated values make physical sense.
15. A rotating object starts from rest at $t=0$ and is subjected to a time-dependent angular acceleration given by $\alpha(t) = 6t \, \text{rad/s}^2$. What is its total angular displacement after the first $2.0 \, \text{s}$?
    Answer: $8.0 \, \text{rad}$ To find the angular displacement, we must integrate the angular acceleration twice. First, find the angular velocity $\omega(t)$ by integrating $\alpha(t)$: $\omega(t) = \int \alpha(t) dt = \int 6t dt = 3t^2 + C$. Since the object starts from rest, $\omega(0) = 0$, so $C=0$. Thus, $\omega(t) = 3t^2$. Next, find the angular displacement $\Delta\theta$ by integrating $\omega(t)$ from $t=0$ to $t=2.0$ s: $\Delta\theta = \int_0^2 3t^2 dt = \left[ t^3 \right]_0^2 = 2^3 - 0^3 = 8.0 \, \text{rad}$.
16. A merry-go-round is rotating counter-clockwise and its rate of rotation is decreasing. If the counter-clockwise direction is defined as positive, which of the following correctly describes the signs of its angular velocity $\omega$ and angular acceleration $\alpha$?
    Answer: $\omega$ is positive, $\alpha$ is negative. Since the merry-go-round is rotating counter-clockwise and this direction is positive, its angular velocity $\omega$ is positive. Since it is slowing down, its angular acceleration $\alpha$ must be in the opposite direction to its angular velocity. Therefore, the angular acceleration $\alpha$ is negative.
17. A gyroscope has spin angular momentum magnitude $L=0.40\ \text{kg}\cdot\text{m}^2/\text{s}$. Gravity exerts a torque of magnitude $\tau=0.080\ \text{N}\cdot\text{m}$ about the pivot, producing steady precession; assume the spin magnitude stays constant. Key equation: $\Omega=\tau/L$. Using the given conditions, determine the precession rate $\Omega$.
    Answer: $\Omega=0.20\ \text{rad/s}$ This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically gyroscopic precession. Precession occurs when a torque acts perpendicular to a spinning object's angular momentum, causing the spin axis to rotate. For a gyroscope with angular momentum L = 0.40 kg·m²/s experiencing torque τ = 0.080 N·m, the precession rate is found using the gyroscopic equation. Choice A is correct because Ω = τ/L = 0.080/0.40 = 0.20 rad/s. Choice B at 5.0 rad/s would result from inverting the fraction or other calculation errors. To help students: explain that precession is different from ordinary rotation; emphasize that the precession equation Ω = τ/L applies when spin angular momentum is much larger than precession angular momentum; and use demonstrations or videos to visualize gyroscopic motion.
18. A carousel starts from rest and accelerates at a constant rate $\alpha_1$ for a time $t_1$. It then rotates at a constant angular velocity for an additional time $t_2$. What is the total angular displacement $\Delta\theta$ of the carousel during the entire time interval $t_1 + t_2$?
    Answer: $\frac{1}{2}\alpha_1 t_1^2 + \alpha_1 t_1 t_2$ The motion has two parts. Part 1 (acceleration): $\Delta\theta_1 = \omega_0 t_1 + \frac{1}{2}\alpha_1 t_1^2 = 0 + \frac{1}{2}\alpha_1 t_1^2$. The angular velocity at the end of this part is $\omega_1 = \omega_0 + \alpha_1 t_1 = \alpha_1 t_1$. Part 2 (constant velocity): $\Delta\theta_2 = \omega_1 t_2 = (\alpha_1 t_1) t_2$. The total displacement is the sum: $\Delta\theta = \Delta\theta_1 + \Delta\theta_2 = \frac{1}{2}\alpha_1 t_1^2 + \alpha_1 t_1 t_2$.
19. A flywheel's angular velocity is described by the function $\omega(t) = C - Dt^2$, where $C$ and $D$ are positive constants and $t$ is time. What expression represents the angular acceleration $\alpha(t)$ of the flywheel?
    Answer: $\alpha(t) = -2Dt$ Angular acceleration $\alpha$ is the time derivative of angular velocity $\omega$. Taking the derivative of the given function $\omega(t) = C - Dt^2$ with respect to time $t$ gives $\alpha(t) = \frac{d\omega}{dt} = \frac{d}{dt}(C - Dt^2) = -2Dt$.
20. A rigid body, initially rotating with angular velocity $\omega_0$, is subjected to a time-varying angular acceleration $\alpha(t)$. The change in the body's angular velocity from $t=0$ to a final time $t=t_f$ is equal to which of the following?
    Answer: The area under the $\alpha$ versus $t$ graph from $t=0$ to $t=t_f$ By definition, $\alpha = \frac{d\omega}{dt}$. Rearranging gives $d\omega = \alpha dt$. To find the total change in angular velocity, $\Delta\omega$, we integrate this expression over the time interval: $\Delta\omega = \int_{0}^{t_f} \alpha(t) dt$. The definite integral of a function represents the area under the curve of that function. Thus, the change in angular velocity is the area under the $\alpha$ versus $t$ graph.

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