Question: A hollow sphere has a radius of 6.4 m. Minimum velocity required by a motorcyclist at the bottom to ...

Question

A hollow sphere has a radius of 6.4 m. Minimum velocity required by a motorcyclist at the bottom to complete the circle will be

Answer 1

Solution

To complete a vertical circle, the motorcyclist must have a minimum velocity such that they just maintain contact with the sphere at the top of the circle. At this point, the normal force exerted by the sphere on the motorcyclist becomes zero, and the centripetal force required is provided solely by gravity.

Let $R$ be the radius of the hollow sphere and $m$ be the mass of the motorcyclist.

At the top of the circle, for minimum velocity $v_{top}$ : The gravitational force $mg$ provides the necessary centripetal force.

$mg = \frac{mv_{top}^2}{R}$ $v_{top}^2 = gR$

Now, we use the principle of conservation of mechanical energy between the bottom and the top of the circle. Let $v_{bottom}$ be the velocity at the bottom. Taking the bottom of the sphere as the reference level for potential energy (PE = 0). The height of the top of the sphere from the bottom is $2R$ .

Energy at the bottom: $E_{bottom} = \frac{1}{2}mv_{bottom}^2 + 0$

Energy at the top: $E_{top} = \frac{1}{2}mv_{top}^2 + mg(2R)$

By conservation of energy, $E_{bottom} = E_{top}$ : $\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv_{top}^2 + mg(2R)$ Divide by $m$ : $\frac{1}{2}v_{bottom}^2 = \frac{1}{2}v_{top}^2 + 2gR$ Substitute $v_{top}^2 = gR$ : $\frac{1}{2}v_{bottom}^2 = \frac{1}{2}(gR) + 2gR$ $\frac{1}{2}v_{bottom}^2 = \left(\frac{1}{2} + 2\right)gR$ $\frac{1}{2}v_{bottom}^2 = \frac{5}{2}gR$ $v_{bottom}^2 = 5gR$ $v_{bottom} = \sqrt{5gR}$

Given: Radius $R = 6.4\,m$ Gravitational acceleration $g \approx 9.8\,m/s^2$

Calculate $v_{bottom}$ : $v_{bottom} = \sqrt{5 \times 9.8 \times 6.4}$ $v_{bottom} = \sqrt{49 \times 6.4}$ $v_{bottom} = \sqrt{313.6}$ $v_{bottom} \approx 17.708\,m/s$

Rounding to one decimal place, $v_{bottom} \approx 17.7\,m/s$ .