A solid, uniform, horizontal disk with a diameter of 2.0 m and a mass of 4.0 kg

Question

A solid, uniform, horizontal disk with a diameter of 2.0 m and a mass of 4.0 kg freely rotates at 36 rpm about a vertical axis through its center. A small 0.50-kg stone is suddenly dropped onto the disk and sticks to the disk at a distance of 80 cm from the axis of rotation. The figure shows before and after views.

(a) Before the stone fell, what was the moment of inertia of the disk about its central axis?

(b) After the stone stuck, what was the moment of inertia of the system about the same axis?

(c)What is the angular velocity of the disk (in rpm) after the stone fell onto the disk?

Explanation / Answer

a) I1 = (1/2)*M*R^2 = (1/2)*4*1*1 = 2 kg m^2

b) I2 = I1 + m*r^2 = 2+ (0.5*0.8*0.8) = 2.32 kg m^2

c) initial angular momentum = final angular momentum

Li = Lf

I1*w1 = I2*w2

w2 = i1*w1/I2 = (2*136)/2.32

w2 = 117.24 rpm

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