Rotational Kinematics — Worked Solutions

These solutions belong to Rotational Kinematics. Each step states why it is useful. Try the questions first, then use this page to check your reasoning.

Question 1

A wheel has ω = −6 rad/s and α = +2 rad/s². What is happening at this instant?

1. Use the stated counterclockwise-positive convention: negative ω is clockwise. Why: The sign of angular velocity gives turning sense.

2. Positive α moves −6 rad/s toward zero, so the wheel is slowing at this instant. Why: Opposite signs of ω and α reduce angular speed until the wheel reaches rest.

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Question 2

What does signed area under an angular-velocity versus time graph give?

1. Multiply the graph units: (rad/s) × s = rad. Why: This identifies angular displacement rather than angular acceleration.

2. Add areas with their signs to obtain Δθ. Why: Clockwise portions below zero offset counterclockwise portions above zero.

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Question 3

A wheel rotates through one complete counterclockwise turn. What is its unwrapped angular displacement?

1. One complete turn sweeps 360°, or 2π rad. Why: Radians measure the angle traveled, not just the final orientation.

2. Counterclockwise is positive, so the unwrapped Δθ is +2π rad. Why: Matching initial and final orientation does not erase the complete turn.

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Question 4

A wheel initially turns at +12 rad/s and comes to rest with constant α = −3.0 rad/s². Find the stopping time and angular displacement.

1. Use ω_f = ω_i + αt and set ω_f = 0. Why: The question supplies constant angular acceleration until the stop.

2. 0 = 12 − 3.0t gives t = 4.0 s. Why: The negative acceleration reduces the initial positive angular velocity.

3. Δθ = ½(12 + 0)(4.0) = +24 rad. Why: The positive triangular area under ω versus t is angular displacement before the stop.

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Question 5

A mark on a wheel has camera orientations 350°, 30°, 90°, and 170° at t = 0, 1, 2, and 3 s. It turns continuously counterclockwise and completes less than one turn between frames. Construct its unwrapped angular displacement from the first frame in radians. Use the data to decide whether constant angular acceleration is a reasonable model, and state one limitation of drawing that conclusion from only these four frames.

1. Unwrap the positions as 350°, 390°, 450°, and 530°. Relative to the first frame, Δθ values are 0°, 40°, 100°, and 180°, or 0, 2π/9, 5π/9, and π rad. Why: The stated turning sense and less-than-one-turn interval resolve the wrap at 360°. Subtracting the first position sets a convenient angular origin.

2. The average angular velocities over the three one-second intervals are 2π/9, π/3, and 4π/9 rad/s. Plot them at the interval midpoints 0.5, 1.5, and 2.5 s; their increase is π/9 rad/s each second. Why: For constant angular acceleration, midpoint interval averages follow a straight ω–t trend. These data are consistent with α = π/9 rad/s², about 0.349 rad/s².

3. Call constant α a reasonable model over the measured interval, rather than a proven instantaneous behavior. Obtain more frames or repeat the run to test it. Why: Four position samples can hide changing acceleration between frames, and angle-reading uncertainty can affect the inferred slopes.

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