A disk-shaped merry go round (I=1/2 MR2 M=250 kg R=2 m) rotates initially at 0.4 rad/s. A 40 kg mass is dropped onto the merry go round and lands at a distance of 1.7 m from the axis of rotation. What is the final angular velocity of the merry go round with the added mass?

0.325 m/s

Explanation:

We are given that

Moment of inertia,

M=250 kg

R=2 m

Angular speed,

m=40 kg

Distance,r=1.7 m

We have to find the angular velocity of the merry go wound with added mass.

Using conservation of momentum

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg