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

[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.

Homework Equations

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 isn't the right answer.

 

[alphysicist]

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.

Homework Equations

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.

 

[sir_manning]

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?