A solid uniform spherical stone starts moving from rest at the top of a hill. At

Question

A solid uniform spherical stone starts moving from rest at the top of a hill. At the bottom of the hill the ground curves upward, and when it is a height H below the starting point the stone leaves the ground vertically straight upward. (see diagram)
a) If there is no friction (so basically the stone just slides like a block), what is the speed of the stone when it leaves the ground?
b) If there is no friction, how high will the stone go (relative to the point it leaves the ground)?
c) If there is enough friction for the stone to roll without slipping, what is the speed of the stone when it leaves the ground?
d) If there is enough friction for the stone to roll without slipping, how high will the stone go (relative to the point it leaves the ground)?

*hint: The only given quantity here is H, so that should be the only variable in your answers.

Explanation / Answer

a)

KE = PE

mgH = 1/2*mv^2

v = sqrt(2gH)

b)

Using v^2 = u^2 - 2gh we get

0 = [sqrt(2gH)]^2 - 2gh

Hence, h = H

c)

Moment of inertia of solid sphere = 2/5*mr^2

KE = KE rolling + KE translation = 1/2*Iw^2 + 1/2*mv^2 = 1/2*(2/5*mr^2)*(v/r)^2 + 1/2*mv^2 = 7/10*mv^2

PE = KE

mgH = 7/10*mv^2

v = sqrt(10/7*gH)

d)

Using v^2 = u^2 - 2gh we get

0 = [sqrt(10/7*gH)]^2 - 2gh

Hence, h =5/7* H

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