(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I am studying for my physics test on tuesday and i came across this problem and i thought i got it right after i worked through it but it was wrong.

A playground merry-go-round is at rest, pivoted about a frictionless axle. A child of mass 40 kg runs along a path tangential to the rim with initial speed 3 m/s and jumps onto the merry-go-round. The rotational inertia of the merry-go-round is 500 kg*m^{2}and the radius of the merry-go-round is 2.5m.

Find the magnitude of the resulting angular velocity of the merry-go-round.

3. The attempt at a solution

Conservation of Energy

[itex] K_i = K_f [/itex]

[itex] \frac{1}{2}*m*v^2 = \frac{1}{2}*I*\omega^2_{merry} + \frac{1}{2}*I*\omega^2_{child} [/itex]

[itex] \frac{1}{2}(40kg)(3^2m/s) = \frac{1}{2}(500kg*m^2)(\omega^2) + \frac{1}{2}[ \frac{1}{2}(40kg)(2.5m)^2](\omega^2) [/itex]

[itex] 180 kg*m/s = [\frac{1}{2}(500kg*m^2) + (\frac{1}{2})(\frac{1}{2})(40kg)(2.5m)^2] (\omega^2) [/itex]

[itex] \omega = 0.7589 rad/s [/itex]

This is wrong but i can't figure out why or where i went wrong :(

This is what the solution to the practice test says to do.

(40kg)(3m/s)(2.5m) = [500kg*m^{2}+ (40kg)(2.5m)^{2}] * [itex]\omega_f[/itex]

I can't figure out what formula this is from or why this solution works but it does.

The answer is [itex] \omega = 0.40 rad/s [/itex]

Thank you in advanced for any help :)

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# Merry-go-round. angular velocity

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