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. Given the moment of inertia $I=\frac{1}{6}md^2$ about its center of mass, find: (a) The initial angular velocity $\omega$ just after impact. (b) The minimum speed $v$ for the block to fall off the table. 2. **Key Concepts and Formulas:** - Angular velocity $\omega$ relates to linear velocity $v$ by conservation of angular momentum about the pivot point. - Moment of inertia about pivot (ledge) uses the parallel axis theorem: $$I_{pivot} = I + md^2_{pivot}$$ where $d_{pivot}$ is the distance from center of mass to pivot. - Energy conservation for tipping condition: block falls if rotational kinetic energy is enough to raise center of mass over pivot. 3. **Step (a): Find initial angular velocity $\omega$** - The block pivots about the ledge at the table's edge. - Distance from center of mass to pivot is $\frac{d}{2}$ (half the side length). - Moment of inertia about pivot: $$I_{pivot} = \frac{1}{6}md^2 + m\left(\frac{d}{2}\right)^2 = \frac{1}{6}md^2 + \frac{1}{4}md^2 = \frac{5}{12}md^2$$ - Initial angular momentum about pivot before impact: $$L = m v \times \frac{d}{2}$$ - After impact, angular momentum is: $$L = I_{pivot} \omega$$ - Equate and solve for $\omega$: $$m v \frac{d}{2} = \frac{5}{12} m d^2 \omega \implies \omega = \frac{m v \frac{d}{2}}{\frac{5}{12} m d^2} = \frac{v \frac{d}{2}}{\frac{5}{12} d^2} = \frac{v}{d} \times \frac{\frac{1}{2}}{\frac{5}{12}} = \frac{v}{d} \times \frac{6}{5} = \frac{6v}{5d}$$ 4. **Step (b): Find minimum speed $v$ to fall off** - The block will fall if it can rotate enough to tip over the pivot. - The center of mass must rise by $\frac{d}{2}$ vertically. - Potential energy increase: $$\Delta U = mg \frac{d}{2}$$ - Initial rotational kinetic energy: $$K = \frac{1}{2} I_{pivot} \omega^2$$ - Using $\omega$ from (a), substitute: $$\frac{1}{2} \times \frac{5}{12} m d^2 \times \left(\frac{6v}{5d}\right)^2 \geq mg \frac{d}{2}$$ - Simplify kinetic energy term: $$= \frac{5}{24} m d^2 \times \frac{36 v^2}{25 d^2} = \frac{5}{24} m \times \frac{36 v^2}{25} = m v^2 \times \frac{5 \times 36}{24 \times 25} = m v^2 \times \frac{180}{600} = m v^2 \times \frac{3}{10}$$ - Set kinetic energy $\geq$ potential energy: $$m v^2 \times \frac{3}{10} \geq mg \frac{d}{2}$$ - Cancel $m$ and solve for $v^2$: $$v^2 \geq g d \times \frac{5}{3}$$ - Therefore, $$v \geq \sqrt{\frac{5}{3} g d}$$ **Final answers:** (a) $$\boxed{\omega = \frac{6v}{5d}}$$ (b) $$\boxed{v \geq \sqrt{\frac{5}{3} g d}}$$