Rotation Ques 29
- A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of $\theta$, where $\theta$ is the angle by which it has rotated, is given as $k \theta^{2}$. If its moment of inertia is $I$, then the angular acceleration of the disc is
(2019 Main, 9 April I)
(a) $\frac{k}{2 I} \theta$
(b) $\frac{k}{I} \theta$
(c) $\frac{k}{4 I} \theta$
(d) $\frac{2 k}{I} \theta$
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Answer:
Correct Answer: 29.(d)
Solution:
Formula:
- Given, kinetic energy $=k \theta^{2}$
We know that, kinetic energy of a rotating body about its axis $=\frac{1}{2} I \omega^{2}$
where, $I$ is moment of inertia and $\omega$ is angular velocity.
$$ \begin{aligned} & \text { So, } & \frac{1}{2} I \omega^{2} & =k \theta^{2} \text { or } \omega^{2}=\frac{2 k \theta^{2}}{I} \\ \Rightarrow & & \omega & =\sqrt{\frac{2 k}{I}} \theta \end{aligned} $$
Differentiating the above equation w.r.t. time on both sides, we get
$$ \frac{d \omega}{d t}=\sqrt{\frac{2 k}{I}} \cdot \frac{d \theta}{d t}=\sqrt{\frac{2 k}{I}} \cdot \omega \quad \because \omega=\frac{d \theta}{d t} $$
$\therefore$ Angular acceleration,
$$ \begin{aligned} \alpha & =\frac{d \omega}{d t}=\sqrt{\frac{2 k}{I}} \cdot \omega=\sqrt{\frac{2 k}{I}} \cdot \sqrt{\frac{2 k}{I}} \theta \text { [using Eq. (i)] } \\ \text { or } \quad \alpha & =\frac{2 k}{I} \theta \end{aligned} $$
Alternate Solution
$$ \begin{array}{ll} \text { As, } & \omega^{2}=\frac{2 k \theta^{2}}{I} \\ \Rightarrow & 2 \omega \frac{d \omega}{d t}=\frac{2 k}{I} \cdot 2 \theta \frac{d \theta}{d t} \text { or } \omega d \omega=\frac{2 k}{I} \theta d \theta \\ & \omega \frac{d \omega}{d \theta}(=\alpha)=\frac{2 k}{I} \cdot \theta \text { or } \alpha=\frac{2 k}{I} \theta \end{array} $$
BETA
