A 900-kg merry-go-round (a flat, solid cylinder) supports 10 children, each with a mass of 50.0 kg, located at the axis of rotation (thus you may assume the children have no angular momentum at that location). You need to describe a plan to move the children such that the angular velocity of the merry-goround decreases to one-half its initial value. ou decided to redistribute the children around the merry-go-round so that each one is the same distance, R(Child), from the rotation axis.

Find R(Child) in terms of R(Merry-Go-Round).

well, one wants to change the moment of inertia to half. Flat cylinder? should be more like a disk, to me.

Initial I=1/2 900* r^2
final I= 2x of that initial

final I= 1/2 900*r^2+500c^2 where c is the radius of the child from center.
900*r^2=1/2 900*r^2+500c^2
450 r^2=500c^2

solve for c,

To find the relationship between R(Child) and R(Merry-Go-Round), we can start by considering the conservation of angular momentum.

The initial angular momentum (L_initial) of the merry-go-round and the children can be calculated as the product of the moment of inertia (I) and the angular velocity (ω):

L_initial = I_initial * ω_initial

The moment of inertia of a flat, solid cylinder can be calculated using the equation:

I_initial = 1/2 * M * R(Merry-Go-Round)^2

where M is the total mass of the merry-go-round.

Similarly, the moment of inertia of each child around the axis of rotation can be approximated as a point mass rotating at a distance R(Child) from the axis. Therefore, the moment of inertia of each child (I_child) is given by:

I_child = m_child * R(Child)^2

where m_child is the mass of each child.

Since there are 10 children, the total moment of inertia of the children (I_total_children) is:

I_total_children = 10 * I_child = 10 * m_child * R(Child)^2

Now, if the children are redistributed such that they are equidistant from the axis of rotation, the total moment of inertia of the children becomes:

I_total_children = 10 * m_child * R(Child_equiposed)^2

where R(Child_equiposed) is the distance of each child from the axis after redistribution.

Since angular momentum is conserved, the final angular momentum (L_final) of the system should be one-half the initial angular momentum (L_initial):

L_final = L_initial / 2

Therefore, we can set up the equation:

L_final = I_initial * ω_final

Replacing the expressions for the initial angular momentum and moment of inertia, we get:

I_initial * ω_final = (1/2) * (I_initial * ω_initial)

Simplifying the equation:

ω_final = ω_initial / 2

Now, we can substitute the expressions for the moment of inertia of the merry-go-round and the children:

(1/2) * M * R(Merry-Go-Round)^2 * ω_final = (1/2) * (M * R(Merry-Go-Round)^2) * ω_initial - 10 * m_child * R(Child_equiposed)^2 * ω_initial

Cancelling out similar terms and solving for R(Child_equiposed):

R(Child_equiposed)^2 = (R(Merry-Go-Round)^2 * ω_initial) / (20 * m_child * ω_initial)

Simplifying further:

R(Child_equiposed) = sqrt( R(Merry-Go-Round)^2 / (20 * m_child) )

Therefore, the relationship between R(Child) and R(Merry-Go-Round) is:

R(Child) = sqrt( R(Merry-Go-Round)^2 / (20 * m_child) )