Question

Question:(II) A ball of radius rrolls on the inside of a track of radius R(see Fig. 8–53). If the ball starts from rest at the vertical edge of the track, what will be its speed when it reaches the lowest point of the track, rolling without slipping?

Short answer

The speed of the ball when it reaches the lowest point of the track is \(\sqrt {\frac{{10}}{7}g\left( {R - r} \right)} \).

Step-by-step solution

Identification of given data

The given data can be listed below as:

• The radius of the ball is r.
• The radius of the track is R.
• The acceleration due to gravity is \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

Understanding the motion of the ball on the track

The ball is rolling inside the track.Therefore, the total energy of the ball on the track is the sum of the energy of the ball in a vertical circle, its translational kinetic energy, and its rotational kinetic energy.The acceleration due to gravity also acts on the ball during its motion on the track.

Representation of the track along with the ball

The diagram of the track can be shown as:

Here, A is the highest position of the ball on the circular track as the motion can be considered in the vertical circle. B is the lowest position of the ball on the track.

Determination of the speed of the ball when it reaches the lowest point of the track

The moment of inertia of the spherical ball can be expressed as:

\(I = \frac{2}{5}m{r^2}\)

Here, m is the mass of the ball.

It is given that the ball is rolling without slipping. The speed of the ball can be expressed as:

\(v = r\omega \)

Here,\(\omega \)is the angular speed of the ball.

At the lowest position (B) of the track, the ball has both types of speed, namely angular and linear speed.

The total energy of the ball in the vertical circle can be expressed as:

\(\begin{aligned}{c}{E_T} &= {E_b} + K.{E_t} + K.{E_r}\\mgR &= mgr + \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}\\mg\left( {R - r} \right) &= \frac{1}{2}m{v^2} + \frac{1}{2} \times \frac{2}{5}m{r^2}{\omega ^2}\\g\left( {R - r} \right) &= \frac{1}{2}{v^2} + \frac{1}{5}{v^2}\end{aligned}\)

Here,\({E_T}\)is the total energy of the ball on the track,\({E_b}\)is the ball's energy in the vertical circle at point (A),\(K.{E_t}\)is the translational kinetic energy of the ball,\(K.{E_r}\)is the rotational kinetic energy of the ball, and g is the acceleration due to gravity.

The above equation can be further solved as:

\(\begin{aligned}{c}g\left( {R - r} \right) &= \frac{7}{{10}}{v^2}\\{v^2} &= \frac{{10}}{7}g\left( {R - r} \right)\\v &= \sqrt {\frac{{10}}{7}g\left( {R - r} \right)} \end{aligned}\)

Thus, the speed of the ball when it reaches the lowest point of the track is \(\sqrt {\frac{{10}}{7}g\left( {R - r} \right)} \).