A sphere of radius = 22.0 and mass = 1.20 starts from rest and rolls without sli

Question

A sphere of radius = 22.0 and mass = 1.20 starts from rest and rolls without slipping down a 37.0 incline that is 12.0 long.

a)Calculate its translational speed when it reaches the bottom.

b)Calculate its rotational speed when it reaches the bottom.

c)What is the ratio of translational to rotational kinetic energy at the bottom?

d)Does your answer in part A depend on mass or radius of the ball?

e)Does your answer in part B depend on mass or radius of the ball?

f)Does your answer in part C depend on mass or radius of the ball?

Explanation / Answer

Relevant equations ke=1/2mv^2 pe=mgh 3. 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. 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 KE is simply KE=Ktranslational=(1/2)mv2 , 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: PE=KE PE=Ktranslational+Krotational 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

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