A 1.8-kg cylinder (radius = 0.08 m, length = 0.50 m) is released from rest at th
ID: 1676355 • Letter: A
Question
A 1.8-kg cylinder (radius = 0.08 m, length = 0.50 m) is released from rest at thetop of a ramp and allowed to roll without slipping. The ramp is0.82 m high and 5.0 m long. (a) When the cylinder reaches the bottom of theramp, what is its total kinetic energy?1 J
(b) What is its rotational kinetic energy?
2 J
(c) What is its translational kinetic energy?
3 J (a) When the cylinder reaches the bottom of theramp, what is its total kinetic energy?
1 J
(b) What is its rotational kinetic energy?
2 J
(c) What is its translational kinetic energy?
3 J
Explanation / Answer
initial energy =mgh=1.8*9.8*0.82 =14.46J (=total energy) All this potential energy is converted into a combination ofkinetic energy(1/2mv2) and rotationalenergy(1/2I2) I=1/2mR2 since there is rolling without slipping v=r Total finalenergy=1/2m(R22+1/2R22) =0.5*1.8*1.5*0.0822 Equating the energies: 0.5*1.8*1.5*0.0822=14.46 =40.91 rad/s and v=3.27 m/s So, KE(translational)=9.64J and KE(rotational)=4.82J
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