Question

A small, solid sphere of mass 0.9 kg and radius 47 cm rolls without slipping along the track consisting of slope and loop-the-loop with radius 4.75 m at the end of the slope. It starts from rest near the top of the track at a height h, where h is large compared to 47 cm. If the g = 9.8 m/s^2 and I(solid sphere) = 2/5 mr^2, what is the minimum value of h such that the sphere completes the loop?

Answer 1

Answer:

$h_{min} = 12.825~m$

Explanation:

The condition that the sphere completes the loop is that it doesn't fall off at the top. In order to do that, the sphere has to have enough velocity to beat the centripetal acceleration.

The forces acting on the sphere at the top of the loop:

1- Weight of the sphere to the downwards direction.

2- The normal force to the downwards direction.

So Newton's Second Law gives the following equation

$mg + N = \frac{mv^2}{R}$

At the minimum velocity, which corresponds to minimum height 'h', the normal force is equal to zero. That means, if the sphere moves any less than the minimum velocity, its weight will be greater than the centripetal force, hence it will fall off.

Therefore,

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

Now, we can use the conservation of energy to find the minimum height.

$K_1 + U_1 = K_2 + U_2\\0 + mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 + mg2R\\mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 + 2mgR\\gh = \frac{v^2}{2} + \frac{v^2}{5} + 2gR\\gh = \frac{7v^2}{10} + 2gR = \frac{7gR}{10} + 2gR = \frac{27gR}{10}\\h = 2.7R\\h = 12.82~m$