Question 42: Centre of Mass and Rotational Motion

Question

An object of mass $$1 \mathrm{~kg}$$ is allowed to hang tangentially from the rim of the wheel of radius R. When released from the rest, the block falls vertically through $$4 \mathrm{~m}$$ height in 2 seconds. The moment of inertia is $$1 \mathrm{~kg} \mathrm{~m}^2$$. The radius of the wheel $$\mathrm{R}$$ 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

0.025 m

Answer

1 m

Answer

0.25 m

Answer

0.5 m

Centre of Mass and Rotational Motion

An object of mass $1 \mathrm{~kg} is allowed to hang tangentially from the rim of the wheel of radius R. When released from the rest, the block falls vertically through 4 \mathrm{~m} height in 2 seconds. The moment of inertia is 1 \mathrm{~kg} \mathrm{~m}^2. The radius of the wheel \mathrm{R}$ is

Choose the correct answer: