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Q. 7.2

A ROTATING WHEEL

GOAL Apply the rotational kinematic equations.

PROBLEM A wheel rotates with a constant angular acceleration of 3.50 rad/s². If the angular velocity of the wheel is 2.00 rad/s at t = 0, (a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and in revolutions. (b) What is the angular velocity of the wheel at t = 2.00 s? (c) What angular displacement (in revolutions) results while the angular velocity found in part (b) doubles?

STRATEGY The angular acceleration is constant, so this problem just requires substituting given values into Equations 7.7–7.9.

$\omega = \omega_i + \alpha t$             [7.7]

$\omega^2 = \omega_i^2 + 2\alpha \Delta \theta$    [7.9]

Note that both the angular acceleration and initial angular velocity are positive, meaning the rotation is strictly in the counterclockwise (positive angular) direction.

Verified Solution

(a) Find the angular displacement after $2.00 \mathrm{~s}$, in both radians and revolutions.

Use Equation 7.8,

$\Delta \theta = \omega_i t + \frac{1}{2}\alpha t^2$    [7.8]

setting $\omega_i=2.00 \mathrm{rad} / \mathrm{s}, \alpha=3.5 \mathrm{rad} / \mathrm{s}^2$, and $t=2.00 \mathrm{~s}:$

\begin{aligned}\Delta \theta &=\omega_i t+\frac{1}{2} \alpha t^2 \\&=(2.00 \mathrm{rad} / \mathrm{s})(2.00 \mathrm{~s})+\frac{1}{2}\left(3.50 \mathrm{rad} / \mathrm{s}^2\right)(2.00 \mathrm{~s})^2 \\&=11.0 \mathrm{rad}\end{aligned}

$\Delta \theta=(11.0 \mathrm{rad})(1.00 \mathrm{rev} / 2 \pi \mathrm{rad})=1.75 \mathrm{rev}$

(b) What is the angular velocity of the wheel at $t=2.00 \mathrm{~s}$ ?

Substitute the same values into Equation 7.7:

\begin{aligned}\omega &=\omega_i+\alpha t=2.00 \mathrm{rad} / \mathrm{s}+\left(3.50 \mathrm{rad} / \mathrm{s}^2\right)(2.00 \mathrm{~s}) \\&=9.00 \mathrm{rad} / \mathrm{s}\end{aligned}

(c) What angular displacement (in revolutions) results during the time in which the angular velocity found in part (b) doubles?

Apply the time-independent rotational kinematics equation:

$\omega_f{ }^2-\omega_i{ }^2=2 \alpha \Delta \theta$

Substitute values, noting that $\omega_f=2 \omega_i:$

$(2 \times 9.00 \mathrm{rad} / \mathrm{s})^2-(9.00 \mathrm{rad} / \mathrm{s})^2=2\left(3.50 \mathrm{rad} / \mathrm{s}^2\right) \Delta \theta$

Solve for the angular displacement and convert to revolutions:

$\Delta \theta=(34.7 \mathrm{rad})(1 \mathrm{rev} / 2 \pi \mathrm{rad})=5.52 \mathrm{rev}$

REMARKS The result of part (b) could also be obtained from Equation $7.9$ and the results of part (a).