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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?

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lublana

Answer:

0.325 m/s

Explanation:

We are given that

Moment of inertia,[tex]I_1=\frac{1}{2}MR^2[/tex]

M=250 kg

R=2 m

[tex]I_1=\frac{1}{2}(250)(2^2)=500 kgm^2[/tex]

Angular speed,[tex]\omega_1=0.4rad/s[/tex]

m=40 kg

Distance,r=1.7 m

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

[tex]I_2=I_1+40r^2=500+40(1.7)^2=615.6kgm^2[/tex]

Using conservation of momentum

[tex]I_1\omega_1=I_2\omega_2[/tex]

[tex]500\times 0.4=615.6\omega_2[/tex]

[tex]\omega_2=\frac{500\times 0.4}{615.6}=0.325 rad/s[/tex]

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