Physics - Rotation Question with Solution | TestHub

(A) $I=\frac{1}{3}M{R}^2{\sin }^2(30^{\circ })$

$I=\frac{1}{3}M{R}^2(\frac{1}{2}{)}^2=\frac{M{R}^2}{12}$ $\to$ (Q)

(B) $I=I_{cm}+M(\frac{2R}{\pi }{)}^2=M{R}^2$

Using $I=2M{R}^2(1-\frac{2}{{\pi }^2})$

$I=2M{R}^2(1-\frac{2\cdot 49}{484})\approx \frac{8M{R}^2}{11}$ $\to$ (P)

(C) Height $h=R\frac{\sqrt{3}}{2}$.

$I=\frac{M{h}^2}{6}=\frac{M}{6}(R^2\cdot \frac{3}{4})=\frac{M{R}^2}{8}$ $\to$ (S)

(D) $I=I_{total}-I_{cavity}$

$I=[\frac{1}{2}(\frac{4}{3}M)(2R{)}^2]-[\frac{1}{2}(\frac{1}{3}M)R^2+(\frac{1}{3}M)R^2]$

$I=\frac{8}{3}M{R}^2-\frac{1}{2}M{R}^2=\frac{13M{R}^2}{6}$ $\to$ (R)