Question

A merry-go-round consists of a uniform solid disc of 225kg and a radius of 6.0m. A single 80kg person stands on the edge when it is coasting at 0.20 revolutions per sec. How fast would the device be rotating after the person has walked 3.5m toward the center. (The moments of inertia of compound objects add.)

Answer 1

Answer:

Angular speed of the disc becomes 0.305 rev/s

Explanation:

As we know that there is no torque on the system of disc and the person

so we can use angular momentum conservation

here we will have

$I_1\omega_1 = I_2\omega_2$

so we have

$I_1 = \frac{1}{2}MR^2 + mR^2$

$I_1 = \frac{1}{2}(225)(6^2) + 80(6^2)$

$I_1 = 6930 kg m^2$

now when man moves towards the center by distance 3.5 m

$I_2 = \frac{1}{2}MR^2 + mr^2$

r = 6 - 3.5 = 2.5 m

$I_2 = \frac{1}{2}(225)(6^2) + (80)(2.5^2)$

$I_2 = 4550 kg m^2$

now we have

$6930 \times 0.20 = 4550 (\omega)$

$\omega = 0.305 rev/s$