Question 44: Centre of Mass and Rotational Motion

Question

A record player is spinning at an angular velocity of $$45 \mathrm{~rpm}$$ just before it is turned off. It then decelerates at a constant rate of $$0.8 \mathrm{~rad} \mathrm{~s}^{-1}$$. The angular displacement is

Accepted Answer

Let's break down the solution step by step:
Convert the initial angular velocity from revolutions per minute (rpm) to radians per second (rad/s):
Since $$1 	ext{ revolution} = 2\pi 	ext{ rad}$$ and $$1 	ext{ minute} = 60 	ext{ s}$$,
$$\omega_0 = 45 	imes \frac{2\pi}{60} = \frac{3\pi}{2} 	ext{ rad/s}$$
The record player decelerates at a constant rate. Since it "slows down," the angular acceleration is negative:
$$\alpha = -0.8 	ext{ rad/s}^2$$
We use the kinematic equation for rotational motion:
$$\omega_f^2 = \omega_0^2 + 2\alpha 	heta$$
When the record player stops, $$\omega_f = 0$$. Thus,
$$0 = \omega_0^2 + 2\alpha 	heta$$
Solving for the angular displacement ($	heta$):
$$	heta = -\frac{\omega_0^2}{2\alpha}$$
Substituting the values:
$$	heta = -\frac{\left(\frac{3\pi}{2}ight)^2}{2(-0.8)} = \frac{\left(\frac{3\pi}{2}ight)^2}{1.6}$$
Simplify the expression:
First, compute the square:
$$\left(\frac{3\pi}{2}ight)^2 = \frac{9\pi^2}{4}$$
Then, substitute back in:
$$	heta = \frac{\frac{9\pi^2}{4}}{1.6} = \frac{9\pi^2}{4 	imes 1.6} = \frac{9\pi^2}{6.4}$$
For a numerical approximation:
Using $$\pi^2 \approx 9.87$$,
$$	heta \approx \frac{9 	imes 9.87}{6.4} \approx \frac{88.83}{6.4} \approx 13.88 	ext{ rad}$$
Therefore, the angular displacement is approximately $$13.88 	ext{ rad}$$.
The correct option is:
Option A: 13.88 rad

Answer

13.88 rad

Answer

1.66 rad

Answer

55.44 rad

Answer

56.25 rad

Centre of Mass and Rotational Motion

A record player is spinning at an angular velocity of $45 \mathrm{~rpm} just before it is turned off. It then decelerates at a constant rate of 0.8 \mathrm{~rad} \mathrm{~s}^{-1}$. The angular displacement is

Choose the correct answer: