Merry-go-round. angular velocity

In summary, the child has an initial angular momentum of 180 kg*m/s with respect to the merry-go-round. The angular momentum of the child when he sits on the merry go round is 180 kg*m/s. The relation between angular momentum, angular velocity and moment of inertia is L_i = L_f.
  • #1
mybrohshi5
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Homework Statement



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.

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*m2 + (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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  • #2
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
I get it now.

So it uses the conservation of angular momentum which is

[itex] L_i = L_f [/itex]

[itex] m(v)(rsin\theta) = I(\omega) + m(r^2)(\omega) [/itex]

[itex] (40)(3)(2.5) = (500 + 40(2.5^2))*(\omega) [/itex]

I forgot this was going to be on the test haha

Thanks for reminding me and for the help :)
 

FAQ: Merry-go-round. angular velocity

What is a merry-go-round?

A merry-go-round is a type of amusement ride consisting of a circular platform with seats for riders. It rotates at a constant speed, allowing riders to experience the feeling of being on a carousel.

What is angular velocity?

Angular velocity is a measure of how fast an object is rotating around a fixed point, such as the center of a merry-go-round. It is typically measured in radians per second.

How is angular velocity related to the speed of a merry-go-round?

The angular velocity of a merry-go-round is directly related to its linear speed. As the angular velocity increases, so does the speed at which riders are moving.

What factors affect the angular velocity of a merry-go-round?

The angular velocity of a merry-go-round can be affected by the radius of the circular platform, the mass of the riders and the platform, and any external forces such as friction or air resistance.

How is angular velocity calculated?

Angular velocity is calculated by dividing the change in angle by the change in time. It can also be calculated by dividing the linear speed by the radius of the circular path.

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