55.Starting from rest, a basketball rolls from the top of a hill to the bottom,

Question

55.Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of 6.6m/s. Ignore the frictional losses (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translation speed of the frozen juice can when it reaches the bottom?

57. A bowling bal encounters a 0.760-m vertical rise on the way back to the ball rack. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 3.50m/s at the bottom of the rise. Find the translational speed at the top.

Explanation / Answer

56.

(a) Using conservation of energy, the potential energy at the top will convert into K.E. and rotational energy of the basketbaall

P.E. @ top = K.E. @ bottom + R.E. @ bottom
mgh = mv²/2 + I²/2
mgh = mv²/2 + (2mr²/3)(v/r)²/2
gh = 5v²/6
h = 5v²/6g
= 5(6.6)²/6(9.81)
= 3.70m

(b) supposing the juice to cylinder (like pepsi can), same way

mgh = mv²/2 + (1/2)(mr²/2)(v/r)²
v = [4gh/3]
= [4(9.81)(3.7m)/3]
= 6.96 m/s.

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