A playground merry-go-round is essentially a uniform disk that rotates about a v

Question

A playground merry-go-round is essentially a uniform disk that rotates about a vertical axis through its center. Suppose the disk has a radius of r = 2.0 m and a mass of 166 kg; a child of mass m = 17.6 kg sits at the edge of the merry-go-round. [Hint: Treat the child as a point mass at the edge of the disk.]

Consider the case where the child is initially standing on the ground when the merry-go-round is rotating at 0.79 rev/s. The child then steps on the merry-go-round. How fast is the merry-go-round rotating now? rev/s

By how much did the rotational kinetic energy of the merry-go-round and child change? J

The answer for part b is NOT 18.59 J or 19 J

Explanation / Answer

a)

by conservation of angular momentum ,

I1 w1 = I2 w2, where w2 is the required angular speed

hence, m1 r^2/2 x w1 = ( m1r^2/2 + m2 r^2 ) w2

ie., 166 x 4/2 x 0.79 = (166 x4/2 + 17.6 x 4) w2

ie., w2 =420.25 /402.4 = 1.02rev /sec.

b)

change in kentic energy dEk = 1/2 I1 w1^2 - 1 /2 I2 w2^2 = 1/2 x 332 x 0.79^2 - 1/2 x 70.4 x1.02^2

ie., dEk = 66.97 J.

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