Question 68: Centre of Mass and Rotational Motion

Question

A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first 2 s it rotates through an angle $	heta_1$ and in the next 2 s it rotates through an angle $	heta_2$. The ratio $\frac{	heta_2}{	heta_1}$ is $\_\_\_\_$ .

Accepted Answer

For constant angular acceleration $(\alpha)$, the angular displacement $(	heta)$ after time t starting from rest $\left(\omega_0=0ight)$ is given by:
$$ 	heta=\omega_0 \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^2 $$
$$\Rightarrow $$ $$ 	heta=\frac{1}{2} \alpha t^2 $$
Att $=2 \mathrm{~s}$ :
$$ 	heta_1=\frac{1}{2} \alpha(2)^2=2 \alpha $$
To find $	heta_2$, we first find the total angle $	heta_{	ext {total }}$ rotated in $4 s$ (the first $2 s+$ the next $2 s$ ), i.e., at $t=4 s$ :
$$ 	heta_{	ext {total }}=\frac{1}{2} \alpha(4)^2=\frac{1}{2} \alpha(16)=8 \alpha $$
The angle $	heta_2$ is the displacement between $\mathrm{t}=2 \mathrm{~s}$ and $\mathrm{t}=4 \mathrm{~s}$ :
$$ 	heta_2=	heta_{	ext {total }}-	heta_1 $$
$$\Rightarrow $$ $$ 	heta_2=8 \alpha-2 \alpha=6 \alpha $$
So, the ratio of angular displacement is,
$$ \frac{	heta_2}{	heta_1}=\frac{6 \alpha}{2 \alpha}=3 $$
Therefore, the required ratio is 3 .
Hence, the correct option is (B).

Answer

6

Answer

3

Answer

4

Answer

${\frac{1}{3}}$

Centre of Mass and Rotational Motion

A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first 2 s it rotates through an angle \theta_1 and in the next 2 s it rotates through an angle \theta_2. The ratio \frac{\theta_2}{\theta_1} is \_\_\_\_ .

Choose the correct answer: