## Chapter 7

## Q. 7.3

## 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.

## Step-by-Step

## 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.