Problem 10.75 A 1.9 kg solid cylinder (radius 0.20 m length 0.50 m s released fr

Question

Problem 10.75 A 1.9 kg solid cylinder (radius 0.20 m length 0.50 m s released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.90 m high and 5.0 m long. Part A When the cylinder reaches the bottom of the ramp, what is its total kinetic energy? Express your answer using two significant figures Submit My Answers Give Up Part B When the cylinder reaches the bottom of the ramp, what is its rotational kinetic energy? Express your answer using two significant figures Submit My Answers Give Up

Explanation / Answer

For a cylinder, which rolls down a ramp and reaches the bottom,

Total kinetic energy, K.E. = 0.5*m*(Vcm)2 (1 + k2/r2)

Where k = radius of gyration and for solid cylinder k2 = r2/2

Hence K.E. = 0.75*m*(Vcm)2

Using energy conservation,

0.75*m*(Vcm)2 = mgh

m*(Vcm)2 = mgh/0.75 = 22.344 J

So

For part a

Total kinetic energy = 0.75*22.344 = 16.758 J = 17 J

Part b

Rotational kinetic energy = 0.5*m*(Vcm)2 ( k2/r2) = 5.586 J = 5.6 J

Part c

Translational kinetic energy = 0.5*m*(Vcm)2 = 11.172 J = 11 J

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