Question 66: Centre of Mass and Rotational Motion

Question

A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 1200 rpm . It is brought to rest in 10 s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are $\_\_\_\_$ and $\_\_\_\_$ respectively.

Accepted Answer

The mass of the solid sphere is $M=5 \mathrm{~kg}$ and its radius is $R=4 \mathrm{~cm}=0.04 \mathrm{~m}$
For a solid sphere rotating about its centre, the moment of inertia is $I=\frac{2}{5} M R^2$
$$ I=\frac{2}{5} 	imes 5 	imes(0.04)^2 $$
$$\Rightarrow $$ $$ \mathrm{I}=2 	imes 0.0016=0.0032 \mathrm{~kg} \cdot \mathrm{~m}^2 $$
The initial angular velocity is $\omega_0=1200 \mathrm{rpm}$ (rotations per minute).
The rate of rotation, usually measured in radians per second $\left(\frac{\mathrm{rad}}{\mathrm{s}}ight)$.
$$ \omega_0=2 \pi 	imes \frac{1200}{60}=40 \pi \mathrm{rad} / \mathrm{s} $$
The final angular velocity is $\omega=0$
The time taken before it comes to rest is $\mathrm{t}=10 \mathrm{~s}$.
Using the rotational kinematic equation,
$$ \omega=\omega_0+\alpha t $$
$$\Rightarrow $$ $0=40 \pi+\alpha(10)$
$$\Rightarrow $$ $$ \alpha=-\frac{40 \pi}{10}=-4 \pi \mathrm{rad} / \mathrm{s}^2 $$
The negative sign shows the solid sphere is slowing down.
Using the law of rotational motion, $	au=\mathrm{I} \alpha$, where $	au$ is the torque acting and $\alpha$ is the resultant angular acceleration.
$$ 	au=I 	imes|\alpha| $$
$$\Rightarrow $$ $$ 	au=0.0032 	imes 4 \pi=0.0128 \pi \mathrm{Nm} $$
Using kinematic equation for rotational motion $	heta=\omega_0 \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^2$, the total angle covered is,
$	heta=40 \pi 	imes 10+\frac{1}{2} 	imes(-4 \pi) 	imes 10^2$
$$\Rightarrow $$ $$ 	heta=400 \pi-2 \pi 	imes 100=200 \pi 	ext { radians } $$
Since one full rotation is $2 \pi$ radians, the number of rotations is:
$$ \mathrm{N}=\frac{	heta}{2 \pi}=\frac{200 \pi}{2 \pi}=100 	ext { rotations } $$
Therefore, the magnitude of applied torque is $0.0128 \pi \mathrm{Nm}$ and the number of rotations is 100 . Thus, the correct option is (D).

Answer

$0.128 \pi \mathrm{Nm}, 100$

Answer

$0.0128 \pi \mathrm{Nm}, 50$

Answer

$0.128 \pi \mathrm{Nm}, 50$

Answer

$0.0128 \pi \mathrm{Nm}, 100$

Centre of Mass and Rotational Motion

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