JEE MAIN - Physics (2023 - 13th April Morning Shift - No. 27)
A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is $$\pi: 22$$ then, the value of its angular speed will be ____________ $$\mathrm{rad} / \mathrm{s}$$.
Answer
4
Explanation
Given that the solid sphere is rolling without slipping, we have:
Angular momentum $$L = \left(I_{\text{com}}\right)(\omega)$$
Kinetic energy $$K = \frac{1}{2}(I_{\text{com}})(\omega^2) + \frac{1}{2}MV_{\text{com}}^2$$
For a solid sphere, the moment of inertia is $$I_{\text{com}} = \frac{2}{5}MR^2$$, and the relationship between linear and angular velocity is $$V_{\text{com}} = R\omega$$.
Substituting these values into the expressions for $$L$$ and $$K$$:
$$L = \frac{2}{5}MR^2 \frac{V_{\text{com}}}{R} = \frac{2MRV_{\text{com}}}{5}$$
$$K = \frac{1}{2}\left(\frac{2}{5}MR^2\right) \frac{V_{\text{com}}^2}{R^2} + \frac{1}{2}MV_{\text{com}}^2 = \frac{7}{10}MV_{\text{com}}^2$$
Now, the given ratio of $$\frac{L}{K}$$ is $$\frac{\pi}{22}$$:
$$\frac{L}{K} = \frac{4}{7} \frac{R}{V_{\text{com}}} = \frac{\pi}{22}$$
Since $$V_{\text{com}} = R\omega$$, we can substitute this relationship into the equation and solve for $$\omega$$:
$$\frac{4}{7} \frac{R}{R\omega} = \frac{\pi}{22}$$
$$\frac{4}{7\omega} = \frac{\pi}{22}$$
$$\omega = \frac{4}{7} \times \frac{22}{\pi} \times 7 = 4$$
Thus, the value of the angular speed is $$4 \ \text{rad/s}$$.
Angular momentum $$L = \left(I_{\text{com}}\right)(\omega)$$
Kinetic energy $$K = \frac{1}{2}(I_{\text{com}})(\omega^2) + \frac{1}{2}MV_{\text{com}}^2$$
For a solid sphere, the moment of inertia is $$I_{\text{com}} = \frac{2}{5}MR^2$$, and the relationship between linear and angular velocity is $$V_{\text{com}} = R\omega$$.
Substituting these values into the expressions for $$L$$ and $$K$$:
$$L = \frac{2}{5}MR^2 \frac{V_{\text{com}}}{R} = \frac{2MRV_{\text{com}}}{5}$$
$$K = \frac{1}{2}\left(\frac{2}{5}MR^2\right) \frac{V_{\text{com}}^2}{R^2} + \frac{1}{2}MV_{\text{com}}^2 = \frac{7}{10}MV_{\text{com}}^2$$
Now, the given ratio of $$\frac{L}{K}$$ is $$\frac{\pi}{22}$$:
$$\frac{L}{K} = \frac{4}{7} \frac{R}{V_{\text{com}}} = \frac{\pi}{22}$$
Since $$V_{\text{com}} = R\omega$$, we can substitute this relationship into the equation and solve for $$\omega$$:
$$\frac{4}{7} \frac{R}{R\omega} = \frac{\pi}{22}$$
$$\frac{4}{7\omega} = \frac{\pi}{22}$$
$$\omega = \frac{4}{7} \times \frac{22}{\pi} \times 7 = 4$$
Thus, the value of the angular speed is $$4 \ \text{rad/s}$$.
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