Rotational Inertia of merry go round

[bphysics]

Homework Statement

a. A merry go round is rotating in the counter-clockwise direction. Initially, a 50 kg child is sitting on the edge of the merry-go-around, which rotates at 0.5 radians per second. The moment of inertia of the merry-go-round is 2150 kg-m2. The radius is 2.50 meters. The child can be treated as a point mass. What is the total angular momentum?

b. If the child crawls to the center of the merry-go-round, will the speed at which it rotates change? If so, what is the new rotation speed?

Homework Equations

[tex]\vec{L}[/tex] = I[tex]\vec{w}[/tex]
I = m₁r₁² + m₂r₂²

The Attempt at a Solution

The moment of inertia of the merry-go-round is 2150 kg-m². We need to add to that value the moment of inertia of the child.

I_child = (2.50 m) * (50 kg) = 125 kg-m²
I_total = 2150 + 125 = 2275 kg-m²

[tex]\vec{L}[/tex] = I[tex]\vec{w}[/tex] = (2275)(0.5) = 1137.5

 

[rock.freak667]

I_child = (2.50 m) * (50 kg) = 125 kg-m²

If the child can be considered a point mass, then the child's moment of inertia is given by mr². Recalculate this, then the total moment of inertia and then the total angular momentum.

For part b, if the child moves closer to the centre of rotation, what property of the child is changing?

 

[nlightner]

kg-m2/s

b. The speed at which the merry-go-round rotates will change when the child moves towards the center. This is because the moment of inertia of the system decreases as the child moves closer to the center, resulting in an increase in angular velocity to maintain the same angular momentum.

Using the equation \vec{L} = I\vec{w}, we can solve for the new angular velocity:

1137.5 kg-m2/s = (2275 kg-m2)(\vec{w}')
\vec{w}' = 0.5 radians per second * (2275/1137.5) = 1.0 radians per second

Therefore, the new rotation speed will be 1.0 radians per second. This is twice the initial rotation speed, as expected since the moment of inertia has been halved.