$$\omega_0 = 45 \times \frac{2\pi}{60} = \frac{3\pi}{2} \text{ rad/s}$$

The record player decelerates at a constant rate. Since it "slows down," the angular acceleration is negative:

$$\alpha = -0.8 \text{ rad/s}^2$$

We use the kinematic equation for rotational motion:

$$\omega_f^2 = \omega_0^2 + 2\alpha \theta$$

When the record player stops, $$\omega_f = 0$$. Thus,

$$0 = \omega_0^2 + 2\alpha \theta$$

Solving for the angular displacement ($\theta$):

$$\theta = -\frac{\omega_0^2}{2\alpha}$$

Substituting the values:

$$\theta = -\frac{\left(\frac{3\pi}{2}\right)^2}{2(-0.8)} = \frac{\left(\frac{3\pi}{2}\right)^2}{1.6}$$

Simplify the expression:

First, compute the square:

$$\left(\frac{3\pi}{2}\right)^2 = \frac{9\pi^2}{4}$$

Then, substitute back in:

$$\theta = \frac{\frac{9\pi^2}{4}}{1.6} = \frac{9\pi^2}{4 \times 1.6} = \frac{9\pi^2}{6.4}$$

For a numerical approximation:

Using $$\pi^2 \approx 9.87$$,

$$\theta \approx \frac{9 \times 9.87}{6.4} \approx \frac{88.83}{6.4} \approx 13.88 \text{ rad}$$

Therefore, the angular displacement is approximately $$13.88 \text{ rad}$$.

The correct option is:

Option A: 13.88 rad

Rotational Motion

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