A wheel of mass M and radius R comes off a moving truck and rolls without slippi

Question

A wheel of mass M and radius R comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is moving at 10 m/s. Its moment of inertia about its rotation axis is 0.8 MR2. The wheel rolls up the hill to a stop, a height h above the bottom of the hill. An irreversible frictional loss of energy on this path is equal to one third of the total kinetic energy that the wheel had at the bottom of the hill.

(a) Calculate h.
(b) What would be the height h for a wheel in the form of a solid homogeneous disk?

Explanation / Answer

Moment of inertia = 0.8MR2

Initial velocity = 10 m/s

(a). Initial kinetic energy = (1/2)MV2(1 + 0.8) = 90M joule.

Energy lost to friction = (1/3)(90M) = 30M joule

Potential energy gained = Mgh

Now . 90M = 30M + Mgh

Or. h = 6.12 m

(b).for solid homogenious disc

Moment of inertia = 0.5MR2

Initial energy = (1/2)MV2 (1+ 0.5) = 75M

Frictional loss of energy = 25M

Now. 75M = 25M + Mgh

Or. h = 5.10 m

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