91Ó°ÊÓ

The rotational kinetic energy is obtained by evaluating the value of the moment of inertia and angular velocity of the object. Its value is altered linearly to the value of the object's MOI.

Draw a diagram of both inclines.

Let \(h\) be the height of each incline; \(\theta \) is the angle of one incline; \(\phi \) is the angle of another incline; the mass of the steel ball is \(M\); the radius is \(R\); \(v\) is the speed of the ball at the bottom, and \(\theta > \phi \).

The expression for the moment of inertia of the solid ball is as follows:

\(I = \frac{2}{5}M{R^2}\)

The expression for the angular velocity is as follows:

\(\omega = \frac{v}{R}\)

Now apply the conservation of energy at the top and bottom of the incline.

\(\begin{aligned}{c}Mgh &= \frac{1}{2}M{v^2} + \frac{1}{2}I{\omega ^2}\\Mgh &= \frac{1}{2}M{v^2} + \frac{1}{2}\left( {\frac{2}{5}M{R^2}} \right){\left( {\frac{v}{R}} \right)^2}\\Mgh &= \frac{1}{2}M{v^2} + \frac{2}{{10}}M{v^2}\\v &= \sqrt {\frac{{10gh}}{7}} \end{aligned}\)

From the above equation, the equation is independent of \(\theta \) and \(\phi \). Therefore, the steel ball will have the same speed at the bottom. The ball on the incline with the lower angle will take longer to reach the bottom than the ball on the incline with the higher angle.