What is the ball's linear and angular velocity at the bottom of the hill? Thanks its a physic question?
Wednesday, March 10th, 2010 at
2:21 pm
A solid sphere (I= 2/5mr^2, radius .8m) is perched precariously at the top of a 5 meter (measured vertically) high hill. Chuck (the cat who delights in doing such things) pushes it down. Assuming no loss of energy what is the balls linear and angular velocity at the bottom of the hill?
Thank you so much i just couldnt get past this one problem thanks so much.
Filed under: Cat Furniture
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The way I’d attack this is to argue that the gravitational potential energy the sphere has at the top of the hill will all be converted into kinetic energy of motion and rotation at the bottom of the hill.
So,
mgh = 1/2mv^2 + 1/2 I w^2
Where v is the speed of the center of mass of the sphere at the bottom and w is the sphere’s angular velocity.
I’m assuming that the problem intends that the sphere rolls without slipping. If so the angular velocity and linear velocity are related by
v = wr
Plugging values and solving should get you the answers for v and w, I hope.
One pitfall I almost fell into just now is realizing that h = 5m, not 5m + 0.8m, because even at the bottom the CM of the sphere is still 0.8m above zero height.