Rotational Kinematics - AP Physics 1

1. A turntable starts at $\theta=0$ and rotates with constant angular velocity $\omega=4\ \text{rad/s}$ for $3\ \text{s}$. What is the angular displacement?
   Answer: $\Delta\theta=12\ \text{rad}$ This question assesses the calculation of angular displacement for constant angular velocity in rotational kinematics. Angular displacement represents the total angle swept by an object, similar to distance traveled in linear motion. When angular velocity is constant, the displacement is simply the product of velocity and time, without acceleration affecting the relationship. This mirrors how constant linear velocity leads to displacement as velocity times time. Choice D is a distractor that mistakenly equates constant velocity with zero displacement, ignoring that motion still accumulates angle over time. A transferable strategy is to use the basic kinematic equations, identifying constant quantities to select the appropriate formula.
2. A disk rotates with constant $\omega=-4\,\text{rad/s}$ for $5\,\text{s}$. What is its angular displacement?
   Answer: $-20\,\text{rad}$ This problem involves angular displacement with constant angular velocity in rotational kinematics. For constant angular velocity, the angular displacement is Δθ = ωt. With ω = -4 rad/s (negative indicating clockwise rotation) and t = 5 s, we get Δθ = (-4 rad/s)(5 s) = -20 rad. The negative sign indicates the displacement is in the clockwise direction. Choice A (20 rad) has the correct magnitude but wrong sign, ignoring the direction of rotation. Always maintain the sign of angular velocity to correctly determine the direction of angular displacement.
3. A wheel rotates with constant $\omega=-5\ \text{rad/s}$ for $2\ \text{s}$. Which describes its angular displacement?
   Answer: $-10\ \text{rad}$, because the rotation is in the negative direction This question probes understanding angular displacement with negative angular velocity in rotational kinematics. Negative velocity indicates rotation in the opposite direction, affecting the sign of displacement. Displacement is velocity times time, preserving the direction information. Qualitatively, this shows how direction influences total angle covered, unlike speed which ignores sign. Choice A is a distractor that uses the magnitude but assigns a positive sign, ignoring the negative direction. A transferable strategy is to consistently track signs in rotational variables to determine direction-dependent quantities.
4. A wheel's angular velocity is constant at $-5\,\text{rad/s}$ for $2\,\text{s}$. What is its angular acceleration?
   Answer: $0\,\text{rad/s}^2$ This problem involves understanding constant angular velocity motion. When angular velocity is constant, the angular acceleration must be zero by definition: α = dω/dt = 0. The fact that ω = -5 rad/s (negative, indicating clockwise rotation) doesn't change this - if velocity is constant, acceleration is zero regardless of the velocity's sign or magnitude. Choice C (-5 rad/s²) incorrectly assumes the acceleration equals the velocity value, while choice A (-10 rad) confuses displacement with acceleration. Remember: constant velocity always means zero acceleration.
5. A disk's angular position is $\theta(t)=5+3t$ (radians). What is its angular acceleration?
   Answer: $0\,\text{rad/s}^2$ This question examines rotational kinematics through derivatives of angular position. For $\theta(t) = 5 + 3t$, $\omega = d\theta/dt = 3 \, \text{rad/s}$ constant, so $\alpha = d\omega/dt = 0 \, \text{rad/s}^2$, indicating no change in velocity. Linear $\theta(t)$ means constant $\omega$ and zero $\alpha$, qualitatively like uniform linear motion. The constant term in $\theta$ doesn't affect derivatives. Choice A (3 rad/s) mixes up acceleration with velocity. Differentiate position twice to find acceleration, confirming constancy in similar kinematic functions.
6. A turntable rotates at constant $\omega=5\,\text{rad/s}$. Which statement about its angular acceleration is correct?
   Answer: $\alpha=0$ because $\omega$ is constant This problem examines the concept of angular acceleration in rotational kinematics. Angular acceleration α represents the rate of change of angular velocity. When angular velocity ω is constant (not changing with time), the angular acceleration must be zero by definition. Since the turntable rotates at a constant ω = 5 rad/s, there is no change in angular velocity, so α = 0. Choice A incorrectly assumes that rotation itself implies acceleration, confusing velocity with acceleration. The key principle is that constant velocity (whether linear or angular) always means zero acceleration.
7. A wheel's angular velocity is constant and nonzero from $t=0$ to $t=5\ \text{s}$. Which statement must be true?
   Answer: Its angular acceleration is zero during the interval This problem tests conceptual understanding of constant angular velocity in rotational kinematics. When angular velocity is constant and nonzero, the angular acceleration must be zero by definition (α = dω/dt = 0). The wheel continues to rotate at the same rate, so its angular position changes continuously (not constant), and its angular displacement over the interval is nonzero (Δθ = ωΔt ≠ 0). Choice C incorrectly suggests increasing angular velocity, which would require nonzero acceleration. The key insight is that constant velocity (whether linear or angular) always means zero acceleration, while the position continues to change at a steady rate.
8. A wheel rotates with constant $\omega=-3\ \text{rad/s}$ for $2\ \text{s}$. What is the sign of $\Delta\theta$?
   Answer: Negative, because angular velocity is negative This problem tests understanding of the relationship between angular velocity and angular displacement in rotational kinematics. When angular velocity is constant, angular displacement is Δθ = ωt. With ω = -3 rad/s and t = 2 s, we get Δθ = (-3 rad/s)(2 s) = -6 rad. The negative angular displacement indicates rotation in the negative direction (typically clockwise when viewed from above). Choice A incorrectly focuses on the magnitude (speed) rather than the signed velocity. Remember that in rotational kinematics, negative angular velocity produces negative angular displacement, just as negative linear velocity produces negative displacement.
9. A turntable has constant angular acceleration $\alpha=1.5\,\text{rad/s}^2$ from rest. What is $\omega$ at $t=4\,\text{s}$?
   Answer: $6.0\,\text{rad/s}$ This problem tests applying constant angular acceleration from rest. For rotational motion starting from rest ($\omega_0 = 0$) with constant angular acceleration, the angular velocity at time t is $\omega = \omega_0 + \alpha t = 0 + \alpha t$. With $\alpha = 1.5 \, \text{rad/s}^2$ and $t = 4 \, \text{s}$, we get $\omega = 1.5 \times 4 = 6.0 \, \text{rad/s}$. Choice A ($6.0 \, \text{rad}$) has incorrect units for velocity, while choice D ($24 \, \text{rad/s}^2$) might come from incorrectly multiplying all values together. For constant acceleration from rest, final velocity equals acceleration times time.
10. A disk has $\omega_i=10\ \text{rad/s}$ and $\alpha=-2\ \text{rad/s}^2$ for $3\ \text{s}$. What is $\omega_f$?
    Answer: $4\ \text{rad/s}$ This problem requires applying rotational kinematics with constant negative angular acceleration. The disk starts with ω₀ = 10 rad/s and experiences α = -2 rad/s² for t = 3 s. Using ω = ω₀ + αt, we get ω = 10 + (-2)(3) = 10 - 6 = 4 rad/s. The negative acceleration causes the disk to slow down from its initial angular velocity. Choice A (16 rad/s) incorrectly adds the acceleration term instead of subtracting it. To handle problems with deceleration, pay careful attention to the sign of acceleration and ensure it reduces the velocity magnitude.
11. A wheel starts at $\theta=0$ and rotates with constant angular velocity $\omega=4\ \text{rad/s}$ for $3\ \text{s}$. What is its angular displacement?
    Answer: $\Delta\theta=12\ \text{rad}$ This problem tests understanding of rotational kinematics with constant angular velocity. When angular velocity ω is constant, the angular displacement is given by Δθ = ωt, similar to how linear displacement equals velocity times time for constant velocity motion. Here, ω = 4 rad/s and t = 3 s, so Δθ = (4 rad/s)(3 s) = 12 rad. Choice A (4/3 rad) incorrectly divides ω by t instead of multiplying. To solve rotational kinematics problems, identify which quantities are constant and apply the appropriate kinematic equation, remembering that angular quantities follow the same patterns as linear quantities.
12. A wheel's angular velocity is constant and positive. Which statement about angular acceleration is correct?
    Answer: $\alpha=0$ because $\omega$ is constant This question tests recognizing angular acceleration for constant angular velocity in rotational kinematics. Constant velocity means no change in rotational speed, implying zero acceleration. Acceleration only exists if velocity is varying, either in magnitude or direction. Qualitatively, this distinguishes uniform from accelerated rotational motion, akin to linear cases. Choice A is a distractor linking acceleration sign to velocity sign, but acceleration is independent when velocity is constant. A transferable strategy is to check if velocity changes over time to determine if acceleration is zero or nonzero.
13. A wheel starts at $\theta=0$ and rotates with constant $\omega=6\,\text{rad/s}$ for $4\,\text{s}$. What is its angular displacement?
    Answer: $24\,\text{rad}$ This problem tests understanding of rotational kinematics, specifically the relationship between angular displacement, angular velocity, and time. For constant angular velocity, the angular displacement is given by Δθ = ωt, where ω is the angular velocity and t is the time. With ω = 6 rad/s and t = 4 s, we calculate Δθ = (6 rad/s)(4 s) = 24 rad. Choice A (1.5 rad) incorrectly divides ω by t instead of multiplying. The key strategy is to recognize constant angular velocity problems and apply the direct relationship Δθ = ωt.
14. A wheel starts from rest and reaches $\omega=8\,\text{rad/s}$ after $4\,\text{s}$ with constant $\alpha$. What is $\alpha$?
    Answer: $2\,\text{rad/s}^2$ This problem involves finding angular acceleration from initial conditions in rotational kinematics. Starting from rest means ω₀ = 0, and with constant angular acceleration, we use α = Δω/Δt. The angular velocity changes from 0 to 8 rad/s in 4 s, so α = (8 rad/s - 0)/(4 s) = 2 rad/s². Choice C (0.5 rad/s²) incorrectly divides time by the change in velocity instead of the reverse. When an object starts from rest with constant acceleration, the acceleration equals the final velocity divided by the time.
15. A fan's angular velocity is constant and negative: $\omega=-5\,\text{rad/s}$ for $3\,\text{s}$. What is $\Delta\theta$?
    Answer: $-15\,\text{rad}$ This question assesses rotational kinematics by calculating angular displacement with constant negative velocity. Constant ω = -5 rad/s over 3 s gives Δθ = ωΔt = -15 rad, where the negative sign indicates direction opposite to positive reference. Angular displacement accumulates based on velocity's magnitude and direction, akin to linear displacement depending on velocity sign. With no acceleration mentioned, motion is uniform in the negative direction. Choice A (15 rad) ignores the negative sign, a frequent oversight in directional quantities. Remember to include signs in calculations for vector quantities like displacement and velocity in kinematics problems.
16. A wheel starts at $\theta=0$ with $\omega_0=2\ \text{rad/s}$ and $\alpha=1\ \text{rad/s}^2$. What is $\theta$ at $t=2\ \text{s}$?
    Answer: $\theta=6\ \text{rad}$ This problem requires using the kinematic equation for angular position with constant angular acceleration in rotational kinematics. The equation θ = θ₀ + ω₀t + ½αt² gives angular position as a function of time. With θ₀ = 0, ω₀ = 2 rad/s, α = 1 rad/s², and t = 2 s, we calculate: θ = 0 + (2)(2) + ½(1)(2²) = 4 + 2 = 6 rad. Choice D (4 rad/s) has incorrect units, confusing angular position (rad) with angular velocity (rad/s). For constant acceleration problems, use the complete kinematic equation and carefully track units to avoid confusing position, velocity, and acceleration.
17. A wheel starts from rest and rotates with constant $\alpha=2\ \text{rad/s}^2$ for $5\ \text{s}$. What is $\omega_f$?
    Answer: $10\ \text{rad/s}$ This problem tests finding final angular velocity starting from rest with constant acceleration. The wheel starts from rest (ω₀ = 0 rad/s) with α = 2 rad/s² for t = 5 s. Using ω = ω₀ + αt, we get ω = 0 + 2(5) = 10 rad/s. The final angular velocity is directly proportional to both acceleration and time when starting from rest. Choice A (25 rad) incorrectly calculates angular displacement instead of velocity using ½αt². To find final velocity from rest, multiply acceleration by time; to find displacement from rest, use ½αt².
18. A wheel turns from $\theta=1.0\ \text{rad}$ to $\theta=5.0\ \text{rad}$ in $2.0\ \text{s}$. What is its average angular velocity?
    Answer: $\bar\omega=2.0\ \text{rad/s}$ This problem involves calculating average angular velocity in rotational kinematics. Average angular velocity is defined as ω̄ = Δθ/Δt, the total angular displacement divided by the time interval. The angular displacement is Δθ = 5.0 - 1.0 = 4.0 rad, and the time interval is Δt = 2.0 s. Therefore, ω̄ = (4.0 rad)/(2.0 s) = 2.0 rad/s. Choice B (3.0 rad) has incorrect units, confusing angular displacement with angular velocity. To find average angular velocity, always calculate the change in angular position (final minus initial) and divide by the elapsed time, ensuring the result has units of rad/s.
19. A flywheel slows from $18\ \text{rad/s}$ to rest in $6\ \text{s}$ with constant $\alpha$. What is $\alpha$?
    Answer: $-3\ \text{rad/s}^2$ This problem requires finding the angular acceleration when an object decelerates to rest. The flywheel slows from ω₀ = 18 rad/s to ω = 0 rad/s in t = 6 s. Using ω = ω₀ + αt, we get 0 = 18 + α(6), which gives α = -18/6 = -3 rad/s². The negative sign indicates deceleration opposing the initial rotation direction. Choice B (-108 rad/s²) incorrectly multiplies 18 × 6 instead of dividing. When solving deceleration problems, the final velocity is often zero, and the acceleration must be negative to reduce the initial velocity.
20. A rotor has constant angular acceleration $\alpha=3\ \text{rad/s}^2$ and initial $\omega_0=1\ \text{rad/s}$. What is $\omega$ at $t=4\ \text{s}$?
    Answer: $\omega=13\ \text{rad/s}$ This problem applies the kinematic equation for angular velocity with constant angular acceleration in rotational kinematics. The equation ω = ω₀ + αt relates final angular velocity to initial velocity, acceleration, and time. Given ω₀ = 1 rad/s, α = 3 rad/s², and t = 4 s, we calculate: ω = 1 + (3)(4) = 1 + 12 = 13 rad/s. Choice C (4 rad) has incorrect units (radians instead of rad/s), confusing angular position with angular velocity. To solve problems with constant angular acceleration, identify the known quantities and select the appropriate kinematic equation, ensuring all terms have consistent units.

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