# Homework Help: Merry-go-round. angular velocity

1. Apr 11, 2010

### mybrohshi5

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*m2 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

$K_i = K_f$

$\frac{1}{2}*m*v^2 = \frac{1}{2}*I*\omega^2_{merry} + \frac{1}{2}*I*\omega^2_{child}$

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

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

$\omega = 0.7589 rad/s$

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*m2 + (40kg)(2.5m)2] * $\omega_f$

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

The answer is $\omega = 0.40 rad/s$

Thank you in advanced for any help :)

Last edited: Apr 11, 2010
2. Apr 11, 2010

### ehild

You cannot use the conservation of energy in his case, this is kind of a inelastic collision, the child and the merry-go round move together.

Neither you can use conservation of momentum, as the merry go round is attached to the axle. But the conservation of the angular momentum is valid. What is the initial angular momentum of the child with respect to the axle? (Consider the child a point mass. ) What is the angular momentum of the child when he sits on the merry go round? What is the relation between angular momentum, angular velocity and moment of inertia?

ehild

3. Apr 11, 2010

### mybrohshi5

I get it now.

So it uses the conservation of angular momentum which is

$L_i = L_f$

$m(v)(rsin\theta) = I(\omega) + m(r^2)(\omega)$

$(40)(3)(2.5) = (500 + 40(2.5^2))*(\omega)$

I forgot this was going to be on the test haha

Thanks for reminding me and for the help :)

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook