Question

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?

Answer 1

0.325 m/s

Step-by-step explanation:

We are given that

Moment of inertia,
`I_1=(1)/(2)MR^2`

M=250 kg

R=2 m

`I_1=(1)/(2)(250)(2^2)=500 kgm^2`

Angular speed,
`\omega_1=0.4rad/s`

m=40 kg

Distance,r=1.7 m

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

`I_2=I_1+40r^2=500+40(1.7)^2=615.6kgm^2`

Using conservation of momentum

`I_1\omega_1=I_2\omega_2`

`500* 0.4=615.6\omega_2`

`\omega_2=(500* 0.4)/(615.6)=0.325 rad/s`