# A Sphere rolling down an incline. Find the speed at the bottom

## Homework Statement

A sphere of radius r_0 = 24.0 cm and mass m = 1.20{\rm{ kg}} starts from rest and rolls without slipping down a 37.0^\circ incline that is 15.0 m long.
Calculate its translational speed when it reaches the bottom.

ke=1/2mv^2
pe=mgh

## The Attempt at a Solution

Since PE=KE , i put the two equations together to form mgh=1/2mv^2. Canceling the m's we get gh=1/2v^2. Using triganomity i find that the height is 9.027m. Then pluging in all the values the answer comes out to be 13.3m/s. This isnt the right answer.

alphysicist
Homework Helper
Hi kavipach,

## Homework Statement

A sphere of radius r_0 = 24.0 cm and mass m = 1.20{\rm{ kg}} starts from rest and rolls without slipping down a 37.0^\circ incline that is 15.0 m long.
Calculate its translational speed when it reaches the bottom.

ke=1/2mv^2
pe=mgh

## The Attempt at a Solution

Since PE=KE , i put the two equations together to form mgh=1/2mv^2.

Your equation is neglecting the kinetic energy due to the rolling motion of the sphere. Once you add the rotational kinetic energy term you should get the right answer.

As mentioned above, what you have solved for is the situation of a sliding mass, not a rolling mass. In the sliding case, all of the kinetic energy of the mass is in its translational motion, so $$K_E$$ is simply $$K_E = K_{translational} = (1/2)mv^2$$, which is the equation you used.

However, in the rolling case, the kinetic energy of the mass is in its translational *and* rolling motion. Then your equation will become:

$$P_E = K_E$$
$$P_E = K_{translational} + K_{rotational}$$

The potential energy is now shared between the translational and rotational kinetic energies, so the translational speed at the bottom will be less than that of a sliding mass.

To actually solve this, you will need to know the equation for rotational kinetic energy and for the moment of inertia of a solid sphere... sound familiar?