1. **Problem statement:** A square block of mass $m$ and side length $d$ slides on a frictionless table at speed $v$. It hits a ledge and begins to tip over, rotating about the ledge edge. Given the moment of inertia $I=\frac{1}{6}md^2$ about its center, find: a) The initial angular velocity $\omega$ just after impact. b) The minimum speed $v$ for the block to tip over completely. 2. **Key concepts and formulas:** - The block rotates about the ledge edge after impact. - Use conservation of angular momentum about the pivot point (ledge edge) at the instant of impact. - Moment of inertia about the pivot (ledge edge) is found by the parallel axis theorem: $$I_{pivot} = I_{cm} + md^2 = \frac{1}{6}md^2 + md^2 = \frac{7}{6}md^2$$ - Angular momentum before impact about the pivot is: $$L = m v \times \frac{d}{2}$$ (since the velocity acts at the center of mass, which is $\frac{d}{2}$ from the pivot) - Angular velocity after impact: $$\omega = \frac{L}{I_{pivot}}$$ - For tipping, the block must have enough rotational kinetic energy to reach the vertical position where potential energy is maximized. 3. **Part (a) - Initial angular velocity $\omega$:** - Calculate angular momentum about pivot before impact: $$L = m v \frac{d}{2}$$ - Using conservation of angular momentum: $$\omega = \frac{L}{I_{pivot}} = \frac{m v \frac{d}{2}}{\frac{7}{6} m d^2} = \frac{v \frac{d}{2}}{\frac{7}{6} d^2} = \frac{v}{2} \times \frac{6}{7 d} = \frac{3 v}{7 d}$$ 4. **Part (b) - Minimum speed $v$ to tip over:** - The block must have enough rotational kinetic energy to raise its center of mass by $\frac{d}{2}$ (from horizontal to vertical). - Height increase of center of mass: $$h = \frac{d}{2}$$ - Potential energy increase: $$\Delta U = m g h = m g \frac{d}{2}$$ - Rotational kinetic energy just after impact: $$K = \frac{1}{2} I_{pivot} \omega^2$$ - Set $K = \Delta U$ for minimum speed: $$\frac{1}{2} \times \frac{7}{6} m d^2 \times \omega^2 = m g \frac{d}{2}$$ - Simplify: $$\frac{7}{12} m d^2 \omega^2 = \frac{m g d}{2}$$ - Cancel $m$ and one $d$: $$\frac{7}{12} d \omega^2 = \frac{g}{2}$$ - Solve for $\omega^2$: $$\omega^2 = \frac{6 g}{7 d}$$ - Recall from part (a) that $\omega = \frac{3 v}{7 d}$, so: $$\left(\frac{3 v}{7 d}\right)^2 = \frac{6 g}{7 d}$$ - Square and solve for $v^2$: $$\frac{9 v^2}{49 d^2} = \frac{6 g}{7 d}$$ $$v^2 = \frac{6 g}{7 d} \times \frac{49 d^2}{9} = \frac{6 g \times 49 d}{7 \times 9} = \frac{294 g d}{63} = \frac{14 g d}{3}$$ - Therefore: $$v = \sqrt{\frac{14 g d}{3}}$$ **Final answers:** - a) $\boxed{\omega = \frac{3 v}{7 d}}$ - b) $\boxed{v = \sqrt{\frac{14 g d}{3}}}$