Question

A sphere of radius 'R' is rolling with angular velocity 'w'. Find 'w' so that it can just climb the step of height 'h'.

Answer 1

A sphere of radius 'R' is rolling with angular velocity 'w'. Find 'w' so that it can just climb the step of height 'h'.

The law of conservation of energy:

$$T + U = \text{const}$$

$T$ - kinetic energy

$$T = \frac{m v^{2}}{2} + \frac{l w^{2}}{2}$$

where:

- m - mass of the body

- v - speed of center mass

- $l$ - moment of inertia (for sphere equals $\frac{2}{3} m R^2$)

- $w$ - angular velocity

If it rolling $= wR$, therefore

$$T = \frac{1}{2} m w^{2} R^{2} + \frac{1}{3} m w^{2} R^{2} = \frac{5}{6} m w^{2} R^{2}$$

$U = mgh$ - potential energy

where:

- g - gravitational acceleration

- h - high

$$T_{1} + U_{1} = T_{2} + U_{2}$$

- 1 - initial state

- 2 - final state

$$U_{1} = 0, T_{2} = 0:$$ $$T_{1} = U_{2}$$

Therefore:

$$\frac{5}{6} m w^{2} R^{2} = m g h$$

Finally:

$$w = \sqrt{\frac{6 g h}{5 R^{2}}}$$

Answer: $w = \sqrt{\frac{6 g h}{5 R^{2}}}$