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## Q. 7.3

COMPACT DISCS

GOAL Apply the rotational kinematics equations in tandem with tangential acceleration and speed.

PROBLEM A compact disc (CD) rotates from rest up to an angular velocity of −31.4 rad/s in a time of 0.892 s. (a) What is the angular acceleration of the disc, assuming the angular acceleration is uniform? (b) Through what angle does the disc turn while coming up to speed? (c) If the radius of the disc is 4.45 cm, find the tangential velocity of a microbe riding on the rim of the disc when t = 0.892 s. (d) What is the magnitude of the tangential acceleration of the microbe at the given time?

STRATEGY We can solve parts (a) and (b) by applying the kinematic equations for angular velocity and angular displacement (Eqs. 7.7 and 7.8).

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

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

Multiplying the radius by the angular acceleration yields the tangential acceleration at the rim, whereas multiplying the radius by the angular velocity gives the tangential velocity at that point.

## Verified Solution

(a) Find the angular acceleration of the disc.

Apply the angular velocity equation $\omega=\omega_i+\alpha t$, taking $\omega_i=0$ at $t=0$ :

$\alpha=\frac{\omega}{t}=\frac{-31.4 \mathrm{~rad} / \mathrm{s}}{0.892 \mathrm{~s}}=-35.2 \mathrm{~rad} / \mathrm{s}^2$

(b) Through what angle does the disc turn?

Use Equation $7.8$ for angular displacement, with $t=0.892 \mathrm{~s}$ and $\omega_i=0$ :

$\Delta \theta=\omega_i t+\frac{1}{2} \alpha t^2=\frac{1}{2}\left(-35.2 \mathrm{~rad} / \mathrm{s}^2\right)(0.892 \mathrm{~s})^2=-14.0 \mathrm{~rad}$

(c) Find the final tangential velocity of a microbe at $r=4.45 \mathrm{~cm}$

Substitute into Equation 7.10:

$v_t = rω$      [7.10]

$v_t=r \omega=(0.044 5 \mathrm{~m})(-31.4 \mathrm{~rad} / \mathrm{s})=-1.40 \mathrm{~m} / \mathrm{s}$

(d) Find the tangential acceleration of the microbe at $r=4.45 \mathrm{~cm}$

Substitute into Equation 7.11:

$a_t = rα$      [7.11]

$a_t=r \alpha=(0.044 5 \mathrm{~m})\left(-35.2 \mathrm{~rad} / \mathrm{s}^2\right)=-1.57 \mathrm{~m} / \mathrm{s}^2$

REMARKS Because $2 \pi \mathrm{~rad}=1 \mathrm{~rev}$, the angular displacement in part (b) corresponds to $2.23$ revolutions in the clockwise direction. In general, dividing the number of radians by 6 gives a rough approximation to the number of revolutions, because $2 \pi \sim 6$.