Vardaan Institute - Detailed Solutions: Master Sheet (Rotational Motion)

Section A: Center of Mass Solutions

Find COM of 3 particles ($1\text{kg}, 2\text{kg}, 3\text{kg}$) at vertices of equilateral triangle of side $1\text{ m}$.

Sol: Coordinates: $(0,0)$, $(1,0)$, $(0.5, \frac{\sqrt{3}}{2})$. $X_{com} = \frac{1(0) + 2(1) + 3(0.5)}{6} = \frac{3.5}{6} = \frac{7}{12}\text{ m}$. $Y_{com} = \frac{1(0) + 2(0) + 3(\sqrt{3}/2)}{6} = \frac{\sqrt{3}}{4}\text{ m}$.

Find M.I. of a thin uniform rod about an axis perpendicular through one end.

Sol: By parallel axis theorem: $I_{end} = I_{cm} + M\left(\frac{L}{2}\right)^2 = \frac{1}{12} M L^2 + \frac{1}{4} M L^2 = \frac{1}{3} M L^2$.

Section B: Torque & Angular Momentum Solutions

A wheel of $I = 2\text{ kg}\cdot\text{m}^2$ rotating at $\omega_0 = 50\text{ rad/s}$ stops under torque $\tau = 10\text{ N}\cdot\text{m}$. Find time and revolutions.

Sol: Angular deceleration $\alpha = \frac{\tau}{I} = \frac{10}{2} = 5\text{ rad/s}^2$. Time $t = \frac{\omega_0}{\alpha} = \frac{50}{5} = 10\text{ s}$. Angle $\theta = \frac{\omega_0^2}{2\alpha} = \frac{2500}{10} = 250\text{ rad} \implies N = \frac{250}{2\pi} \approx 39.8\text{ rev}$.

State Principle of Conservation of Angular Momentum ($L = I\omega = \text{const}$). Explain ballet dancer.

Sol: When net external torque $\tau_{ext} = 0$, $L = I \omega = \text{constant}$. A ballet dancer folds her arms to decrease $I$, thereby increasing $\omega$ to spin faster.

Section C: Pure Rolling Solutions

A solid cylinder rolls down an incline $\theta$ without slipping. Derive $a = \frac{2}{3} g \sin\theta$.

Sol: $a = \frac{g \sin\theta}{1 + K^2/R^2}$. For solid cylinder, $K^2/R^2 = 1/2 \implies a = \frac{g \sin\theta}{1 + 1/2} = \frac{2}{3} g \sin\theta$.

Find total kinetic energy of a rolling sphere of mass $M$ and velocity $v$.

Sol: $K_{total} = \frac{1}{2} M v^2 \left(1 + \frac{K^2}{R^2}\right)$. For solid sphere $K^2/R^2 = 2/5 \implies K_{total} = \frac{1}{2} M v^2 \left(1 + \frac{2}{5}\right) = \frac{7}{10} M v^2$.

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