A solid, uniform ball rolls without slipping up a hill, as shown in the figure (

Question

A solid, uniform ball rolls without slipping up a hill, as shown in the figure (Figure 1) . At the top of the hill, it is moving horizontally; then it goes over the vertical cliff. Take V = 29.0m/s and H = 25.0m .

Part A

How far from the foot of the cliff does the ball land?

Part B

How fast is it moving just before it lands?

Part C

Notice that when the ball lands, it has a larger translational speed than it had at the bottom of the hill. Does this mean that the ball somehow gained energy by going up the hill? Explain!

A solid, uniform ball rolls without slipping up a hill, as shown in the figure (Figure 1) . At the top of the hill, it is moving horizontally; then it goes over the vertical cliff. Take V = 29.0m/s and H = 25.0m . Part A How far from the foot of the cliff does the ball land? Part B How fast is it moving just before it lands? Part C Notice that when the ball lands, it has a larger translational speed than it had at the bottom of the hill. Does this mean that the ball somehow gained energy by going up the hill? Explain!

Explanation / Answer

You need to find the velocity at the top of the cliff. Use conservation of energy.

E.bottom = KE + PE
KE = 1/2 m v1^2 + 1/2 I ?^2
I = 2/5 m r^2
? = v/r
KE = 1/2m (v1^2 + 2/5 v1^2) = 7/10 m v1^2
v1= 29 m/s
PE = mgh
h = 0

E.top = KE + PE
KE = 7/10 m v2^2
PE = mgh
h = 25 m

E.bottom = E.top

7/10 m v1^2 + 0 = 7/10 m v2^2 + mg h
v2^2 = v1^2 -10/7gh
v2 = ?((29m/s) - 10/7*9.81m/s^2 * 25m)
v2 = 5.38 m/s

1. time to fall
h = 1/2 g t^2
t = ?(2h/g)
t = 2.25 s

Distance traveled:
L = t*v2
L = 65.34 m

2. v.vertical = ?(2gh)
v.vert = 22.14 m/s

total velocity = ? (v.hor^2 + v.vert^2)
= ? (5.38^2 + 22.14^2)
= 22.79m/s

3. No, It is not rotating as fast. It has lost rotational energy.

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