Answer to Question #240506 in Mechanics | Relativity for Sachi

Question

(a) A merry-go-round with 5 m radius is rotating around its vertical axis with a rate of 0.30 rev/s. A 90 kg man jump on to the merry-go-round and he lands and remains at a distance of 3 m from the axis of rotation. If the moment of inertia of the merry-go-round is 10,000

kgm2, find the new rate of rotation after the person has landed.

The same merry-go-round mentioned in the part (a) is at rest and a 90 kg man is standing at the edge of the merry-go-round. Then he starts to walk around the perimeter at a speed of 0.8 m/s relative to the ground. What rotation rate does this motion give to the merry-go-round?

Answer 1

Given:

$r=5\:\rm m$

$n=30\:\rm rev/s$

$I=10000\:\rm kg\cdot m^2$

$m=90\;\rm kg$

$d=3\:\rm m$

(a) The law of conservation of angular momentum says

$$In=I'n'$$

$$I'=I+md^2$$

Hence

$$n'=\frac{I}{I+md^2}n=\frac{10000}{10000+90*3^2}*30=28\:\rm rev/s$$

(b)

$$mvr=I\omega$$

$$\omega=mvr/I=90*0.8*5/10000=0.036\:\rm rad/s$$