A child of mass m is playing on a merry-go-round which has mass 6m and radius r. Assume the merry-go-round is a uniform disk. The child is standing on the edge of the merry-go-round, which is spinning with an angular speed ω1.
a) What is the moment of inertia, I1, of the compound system (child + merry-go-round) about the center of the merry-go-round?
b) The child now walks halfway to the center of the merry-go-round. What is the new angular speed, ω2, of the compound system?
c) How much work did the child do in walking halfway to the center of the merry-go-round? Express your answer as a numerical fraction of the original kinetic energy, K1, in the compound
system.
d) A second child, also of mass m, initially standing at rest next to the merry-go-round, steps on to the edge of the merry-go-round (and nearly has her arms torn off in the process). What is the new angular speed, ω3, of the compound system? Express your answer as a numerical fraction of the original angular speed, ω1.
e) The second child now walks to the center of the merry-go-round. What is the new angular speed,ω4, of the compound system? Express your answer as a numerical fraction of the original angular speed, ω1.
a) To find the moment of inertia, I1, of the compound system (child + merry-go-round) about the center of the merry-go-round, we need to consider the moment of inertia of the merry-go-round and the child separately and then add them together.
The moment of inertia of a uniform disk rotating about its center is given by the equation:
I_mg = (1/2) * m_mg * r^2
where m_mg is the mass of the merry-go-round and r is its radius.
The moment of inertia of a point mass rotating about an axis passing through its center of mass is given by:
I_c = m_c * r^2
where m_c is the mass of the child.
Since the child is standing on the edge of the merry-go-round, the distance between the child and the center of the merry-go-round is r. Therefore, we can write the moment of inertia of the child as:
I_c = m_c * r^2
To find the moment of inertia of the compound system, we add the moments of inertia of the merry-go-round and the child:
I1 = I_mg + I_c
Substituting the values given in the problem, we have:
I1 = (1/2) * (6m) * r^2 + m * r^2
= (3m) * r^2 + m * r^2
= 4m * r^2
b) When the child walks halfway to the center of the merry-go-round, the moment of inertia of the compound system changes. Since the child has moved closer to the center, the moment of inertia decreases.
To find the new moment of inertia, I2, we need to consider the contribution of both the child and the merry-go-round.
The new moment of inertia is given by:
I2 = (1/2) * m_mg * r^2 + (1/2) * m_c * (r/2)^2
Substituting the values given in the problem, we have:
I2 = (1/2) * (6m) * r^2 + (1/2) * m * (r/2)^2
= 3m * r^2 + (1/2) * m * (r/2)^2
= 3m * r^2 + (1/8) * m * r^2
= (25/8) * m * r^2.
c) The work done by the child in walking halfway to the center of the merry-go-round can be calculated using the work-energy principle. The work done is equal to the change in kinetic energy.
The original kinetic energy of the compound system, K1, can be expressed as:
K1 = (1/2) * I1 * ω1^2
where ω1 is the angular speed of the compound system.
When the child walks halfway to the center, the new angular speed of the compound system is ω2. The new kinetic energy, K2, can be expressed as:
K2 = (1/2) * I2 * ω2^2
The work done, W, is given by:
W = K2 - K1
Substituting the expressions for K1 and K2, we have:
W = (1/2) * I2 * ω2^2 - (1/2) * I1 * ω1^2
Substituting the values of I2 and I1 from parts (a) and (b), and the given values of ω1 and r, we can calculate the work done by the child.
d) When the second child steps on to the edge of the merry-go-round, the angular momentum of the compound system is conserved.
The initial angular momentum of the compound system is given by:
L1 = I1 * ω1
where I1 is the moment of inertia of the compound system and ω1 is the initial angular speed.
Since the second child initially stands at rest, their contribution to the initial angular momentum is zero.
The final angular momentum of the compound system is given by:
L2 = I2 * ω3
where I2 is the new moment of inertia of the compound system and ω3 is the final angular speed.
Since angular momentum is conserved, we have:
L1 = L2
Substituting the expressions for L1 and L2, we have:
I1 * ω1 = I2 * ω3
Substituting the values of I1 and I2 from parts (a) and (b), and the given value of ω1, we can solve for ω3.
e) Similar to part (b), the new angular speed, ω4, when the second child walks to the center of the merry-go-round, can be found using the moment of inertia of the compound system.
The moment of inertia of the compound system, I3, when the second child is at the center is given by:
I3 = (1/2) * m_mg * (r/2)^2 + m_c * 0^2
where m_mg is the mass of the merry-go-round, r is its radius, and m_c is the mass of the child.
Substituting the values given in the problem, we have:
I3 = (1/2) * (6m) * (r/2)^2 + m * 0^2
= (3/8) * m * r^2
The new angular speed, ω4, is related to the moment of inertia of the compound system, I3, and the original angular speed, ω1, through the conservation of angular momentum:
I1 * ω1 = I3 * ω4
Substituting the values of I1 and I3 from parts (a) and (e), and the given value of ω1, we can solve for ω4.
a. add the two moments to gat combined.
b. momentum is conserved. compute the final moment.
final moment*wf=iniital moment(wi)
c. compute the final energy, and the initial energy, the kid did that as work.