Question:

A circular cylinder of radius \(r\) rolls down on an inclined plane from height \(h\). Compare the velocities at the bottom of this object with that of a point object.

Answer:

The moment of inertia of a cylinder (disk) is given by

\[   I = \frac{1}{2}mr^2  \]

The derivation of this moment of inertia is provided on this webpage: http://hirophysics.com/Study/moment-of-inertia.pdf Use conservation energy. For this cylinder, we need to include the rotational kinetic energy in addition to linear kinetic energy. We suppose that the entire potential energy is transferred into all the kinetic energy.

\[  \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh  \]

Since \(\omega=\frac{v}{r}\) and \(I = \frac{1}{2}mr^2\), we substitute it in the above, then solve for the terminal velocity.

\[  \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2 = mgh  \]

\[   \frac{3}{4}mv^2=mgh \]

\[   v_C = \sqrt{\frac{4gh}{3}}  \]

\(v_C\) is the final velocity of the cylinder. Let us find the final velocity of a point object. Likewise, we use conservation of energy, but the rotational kinetic energy is excluded in this case.

\[  \frac{1}{2}mv^2 = mgh  \]

\[ v_P = \sqrt{2gh}  \]

Now compare both of them.

\[  \frac{v_C}{v_P}=\frac{\sqrt{\frac{4gh}{3}}}{\sqrt{2gh}}=\sqrt{\frac{2}{3}}\sim 0.816  \]

The final velocity of the cylinder is slower than that of point object by the factor of 0.816.