Problem 10.76 A 1.6kg solid sphere (radius = 0.10m ) is released from rest at th

Question

Problem 10.76

A 1.6kg solid sphere (radius = 0.10m ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.90m high and 5.7m long.

Part A

When the sphere reaches the bottom of the ramp, what is its total kinetic energy?

Express your answer using two significant figures.

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Part B

When the sphere reaches the bottom of the ramp, what is its rotational kinetic energy?

Express your answer using two significant figures.

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Part C

When the sphere reaches the bottom of the ramp, what is its translational kinetic energy?

Express your answer using two significant figures.

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Explanation / Answer

total energy= rotation energy + translational energy

final velocity=

v2=2a*S (a=gsin(theta))

v= (2*9.8*(0.9/5.7)*5.7)1/2 = 4.2 m/s

finnal w = v/r = 4.2/0.1 = 42 rad/s

rotational energy = 0.5*I*w2 = 0.5*((2/5)*1.6*(0.10)2)*422 =0.1344 J

translational kinetic energy = 0.5*m*v2 = 0.5*1.6*(4.2)2 = 14.112 J

total energy= 14.2464 J

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