Question

a playground merry-go-round of radius r = 2.20 m has a moment of inertia i = 245 kg · m2 and is rotating at 11.0 rev/min about a frictionless vertical axle. facing the axle, a 26.0-kg child hops onto the merry-go-round and manages to sit down on the edge. what is the new angular speed of the merry-go-round?

Answer 1

8.92 rpm

Step-by-step explanation:

Given that

Radius of the merry go round, r = 2.2 m

Initial moment of inertia, I1 = 245 kgm²

Initial speed of rotation, w1 = 11 rpm

Mass of the child, m = 26 kg

To solve the problem, we use the law conservation of momentum

I1w1 = I2w2

I2 = mr² + I1

I2 = 245 + 26 * 2.2

I2 = 245 + 57.2

I2 = 302.2 kgm²

Now, applying the formula, we have

I1w1 = I2w2

245 * 11 = 302.2 * w2

w2 = 2695 / 302.2

w2 = 8.92 rpm

Thus, the new angular speed of the merry go round is 8.92 rpm