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(C) The kinematic equation for rotational motion starting from rest $(\omega_0 = 0)$ is given by $	heta(t) = \frac{1}{2} \alpha t^2$.The angle covere

Vedclass · Accepted Answer

$3:5$

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(C) The kinematic equation for rotational motion starting from rest $(\omega_0 = 0)$ is given by $	heta(t) = \frac{1}{2} \alpha t^2$.The angle covered in the $n^{	ext{th}}$ second is given by $	heta_n = 	heta(n) - 	heta(n-1) = \frac{1}{2} \alpha [n^2 - (n-1)^2] = \frac{1}{2} \alpha (2n - 1)$.For the $2^{	ext{nd}}$ second $(n=2)$:$	heta = \frac{1}{2} \alpha (2(2) - 1) = \frac{1}{2} \alpha (3) = 1.5 \alpha$.For the $3^{	ext{rd}}$ second $(n=3)$:$	heta^{\prime} = \frac{1}{2} \alpha (2(3) - 1) = \frac{1}{2} \alpha (5) = 2.5 \alpha$.The ratio is $\frac{	heta}{	heta^{\prime}} = \frac{1.5 \alpha}{2.5 \alpha} = \frac{1.5}{2.5} = \frac{3}{5}$.

$A$ disc at rest is subjected to a uniform angular acceleration about its axis. Let $\theta$ and $\theta^{\prime}$ be the angle made by the disc in the $2^{\text{nd}}$ and $3^{\text{rd}}$ second of its motion,respectively. The ratio $\frac{\theta}{\theta^{\prime}}$ is

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