Answer to Question #292858 in Physics for Shane

Question

A cylinder of mass m and radius R has a moment of inertia of 12mr2. The cylinder is released from rest at a height h on an inclined plane, and rolls down the plane without slipping. What is the velocity of the cylinder when it reaches the bottom of the incline?

Answer 1

Given:

$I=1/2mR^2$

The law of conservation of the energy says

$$E_i=E_f$$

$$E_{ki}+E_{pi}=E_{kf}+E_{pf}$$

$$mgh=\frac{mv_f^2}{2}+\frac{I\omega_f^2}{2}$$ $$\omega_f=v_f/R$$

Hence

$$mgh=\frac{mv_f^2}{2}+\frac{mv_f^2}{4}=\frac{3mv_f^2}{4}$$

Finally, the velocity of the cylinder when it reaches the bottom of the incline

$$v_f=\sqrt{4/3gh}$$