Physics - Rotation Question with Solution | TestHub

Sphere rolls down height $h$ without slipping:

$$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{R}^2\right)\frac{v^2}{R^2}=\frac{7}{10}m{v}^2$$

$$v=\sqrt{\frac{10gh}{7}}$$

horizontal at the edge.

Fall through the second height $h$:

$$t=\sqrt{\frac{2h}{g}}$$

(no horizontal force, spin does not change the path).

$$x=vt=\sqrt{\frac{10gh}{7}\cdot \frac{2h}{g}}=h\sqrt{\frac{20}{7}}=2h\sqrt{\frac{5}{7}}$$

$$\frac{x}{h}\sqrt{\frac{7}{5}}=2$$

Answer: 2.