In the figure here, a 34 kg child stands on the edge of a stationary merry-go-ro

Question

In the figure here, a 34 kg child stands on the edge of a stationary merry-go-round of radius 2.1 m. The rotational inertia of the merry-go-round about its rotation axis is 200 kg-m^2. The child catches a ball of mass 1.0 kg thrown by a friend. Just before the ball is caught, it has a horizontal velocity of magnitude 8 m/s, at angle psi = 11 " with a line tangent to the outer edge of the merry-go-round, as shown. What is the angular speed of the merry-go-round just after the ball is caught? omega =

Explanation / Answer

To answer this question, use conservation of momentum:

The ball has momentum of mv = (1kg)(8 m/s) = 8 kg m/s

After the boy on the merry-go-around catches the ball, that momentum is transfered to the moments of inertia of the merry-go-around, plus the boy and the ball:

mv = I(total)

where I(total) = I(merry-go-around) + I(boy) + I(ball)

I(merry-go-around) = 200 kg m^2

I(boy) = m(boy)r^2 = (34)(2.1)^2 = 149.94 kg m^2

I(ball) = m(ball)(rsin)^2 = (1)(2.1)^2(sin 11°)^2 = 0.160 kg m^2

I(total) = (200) + (149.94) + (0.160) = 350.1 kg m^2

Now = mv/I(total) = (8)/(350.1) = 0.0228 rad/sec

or 0.217723962 revolutions per minute

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.