physics

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?

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  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)

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  2. Thank you for your help. I appreciate it.

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