JEE Main 2023 — Rotational Motion Question with Solution

Given that the solid sphere is rolling without slipping, we have:

Angular momentum $$L=\left(I_{\text{com}}\right)(\omega )$$

Kinetic energy $$K=\frac{1}{2}(I_{\text{com}})({\omega }^2)+\frac{1}{2}M{V}_{\text{com}}^2$$

For a solid sphere, the moment of inertia is $$I_{\text{com}}=\frac{2}{5}M{R}^2$$, and the relationship between linear and angular velocity is $$V_{\text{com}}=R\omega$$.

Substituting these values into the expressions for $$L$$ and $$K$$:

$$L=\frac{2}{5}M{R}^2\frac{V_{\text{com}}}{R}=\frac{2MR{V}_{\text{com}}}{5}$$

$$K=\frac{1}{2}\left(\frac{2}{5}M{R}^2\right)\frac{V_{\text{com}}^2}{R^2}+\frac{1}{2}M{V}_{\text{com}}^2=\frac{7}{10}M{V}_{\text{com}}^2$$

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 $$4\text{rad/s}$$.