A solid cylinder with mass m = 2.0 kg and radius R = 0.08 m is released from the

Question

A solid cylinder with mass m = 2.0 kg and radius R = 0.08 m is released from the top of a 3.0 m tall hill, as shown in the picture. Its moment of inertia about its center of mass is I = 1/2 mR^2. a) If the cylinder rolls from rest without slipping, what is its speed at the bottom of the hill? b) At the bottom of the hill, what is the cylinder's rotational kinetic energy? The cylinder from part (b) travels up a second hill, which is covered in ice. There is no friction between the cylinder and the hill. c) To what vertical height can the cylinder travel on the second, frictionless hill?

Explanation / Answer

(a) initial energy of the cylinder = mgh = 2*9.81*3 =58.86 J
At the bottom of the hill the cylinder will have only kinetic energy = rotational kinetic energy + translational kinetic energy
= (1/2)Iw2 + (1/2)mv2
we know that v = wR
where v is linear velocity and w is angular velocity
=(1/2)I(v/R)2 + (1/2)mv2
= (1/2)(mR2/2)*(v/R)2 + (1/2)mv2 = (1/4)mv2 + (1/2)mv2 = (3/4)mv2
Equating the two energy
(3/4)mv2 = 58.86
v = 6.2641 m/s
(b)Cylinder rotational kinetic energy = (1/2)Iw2 = (1/4)mv2 = (1/4)*2*6.26412 = 19.62 J
(c) Again we will apply the conservation of energy. Since there is no friction loss therefore
the height will be 3m .

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