What is the energy transfer of a battery rolling out of a rolling cup?

Question

A cup is on its side. inside the cup is a double A battery. An object hits the cup forcing the cup to roll. As the cup rolls the batter rolls out. What is the energy transfer of the battery rolling out as the cup was rolling?

Answer 1

Tom,

 

Your problem statement provides insufficient data to compute anything meaningful.  Following are some thoughts to consider.

 

Initially, the cup (a cylinder) is laying motionless on its side with a single AA battery laying on its side touching the cup's side. It is also motionless.  So the initial linear momentum of the system's CG is zero as is the angular momentum. We'll assume a horizontal surface so gravity is not a factor in the motion being vertical and opposed by the Normal force.

 

Then an external horizontal force jolts the cup with an impulse to cause the motion. How big is it? What is it's geometry since it creates linear momentum in the CG and angular motion of the cup to roll.  Obviously there must be rolling friction between the cup and the horizontal surface, otherwise the cup would have just slid with the AA battery still inside the cup.  What causes the battery to leave the cup? The impact spits it out or the cup undergoes some precession that allows the battery to slide out over time? What are the masses, and moments of inertia of the cup and the battery separately and together?  The angular rotation of the cup is related to the linear velocity by the cup's circumference if it is indeed rolling. etc., etc.

 

You get the idea?  Otherwise, how can you calculate how the energy imparted by the collision impulse gets split up between the cup and the battery, their linear and angular motions, etc?