Practice free Rotational Motion (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is $2400\ \text{g cm}^2$. The length of the 400 g rod is nearly:
$L = \sqrt{12 I/M} = \sqrt{72} \approx 8.5$ cm.
A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ($P$ and $Q$ are any highest and lowest points on the wheel, respectively)?
Pure rolling: top $2v$, contact 0.
The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.
Angular momentum conserved: $I\omega = \text{const}$, $I = \tfrac25 MR^2\propto R^2$, so $T\propto R^2$. $T_2 = 27\times2^2 = 108$ days.
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $60^\circ$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take $g = 10\ \text{m/s}^2$):
Rod makes $30^\circ$ with floor. Torque about the floor end: $N_w\,l\sin30^\circ = mg\,\tfrac{l}{2}\cos30^\circ\Rightarrow N_w = \dfrac{mg}{2}\cot30^\circ$... giving wall normal $= 100\sqrt3$ N. Friction $f = N_w = 100\sqrt{3}$ N.
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:
Taking mass $\propto r^3$: small sphere mass 1 unit, big 8 units. $I_{small} = \tfrac25 mR^2 + mR^2 = \tfrac75$ units; $I_{big} = \tfrac25(8)(2R)^2 = \tfrac{64}{5}$; $I_{rest} = \tfrac{57}{5}$. Ratio $= \dfrac{7}{57}$.
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