A solid cylinder of radius 0.85m is released from rest from a height of 1.8m and rolls down the incline as shown. What is the linear speed of the cylinder when it reaches the horizontal surface?

Answer 1

Use conservation of energy and be sure to include the rotational kinetic energy,
(1/2) I w^2

For a cylinder, the moment of inertia is
I = (1/2) M R^2

w is the angular velocity, which equals V/R. M and R will cancel out.

Total kinetic energy
= (1/2) M V^2 + (1/2) I w^2
= (1/2)M V^2 + (1/4)M V^2
= (3/4) M V^2

M g H = (3/4) M V^2
V = sqrt(4gH/3)

Answer 2

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