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Question

(B) The moment of inertia of a solid sphere is given by $I = \frac{2}{5}MR^2$. Since the density $\rho$ is uniform,$M = \rho \cdot V = \rho \cdot \fra

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) The moment of inertia of a solid sphere is given by $I = \frac{2}{5}MR^2$. Since the density $\rho$ is uniform,$M = \rho \cdot V = \rho \cdot \frac{4}{3}\pi R^3$. Thus,$M \propto R^3$. When the radius changes from $R$ to $R/2$,the mass changes from $M$ to $M' = M \cdot (1/2)^3 = M/8$. Since there are no external torques acting on the sphere,the angular momentum is conserved: $L_1 = L_2$. $I_1 \omega_1 = I_2 \omega_2$. Substituting the values: $(\frac{2}{5}MR^2)\omega_1 = (\frac{2}{5}M'(R/2)^2)\omega_2$. $(\frac{2}{5}MR^2)\omega_1 = (\frac{2}{5} \cdot \frac{M}{8} \cdot \frac{R^2}{4})\omega_2$. $MR^2 \omega_1 = (\frac{M R^2}{32}) \omega_2$. $\omega_2 = 32 \omega_1$. Comparing this with $x\omega_1$,we get $x = 32$.

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