A playground merry-go-round of radius R = 1.92m has a moment of inertia of I = 221kgm^2 and is rotating at 14.0rev/min about a frictionless vertical axle. Facing the axle, a 29.8kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed (in revolutions per minute) of the merry-go-round?
How much work is done by the child?
I got 9.35 rev/min or 0.979 rad/s but I have no clue how to approach the child's work
find ke before child jumps
(1/2)I omega^2
find new I (which you did already)
find new omega (which you did already)
Find new ke = (1/2) Inew (omega new)^2
now
work done = new ke - old ke
I went up a little but omega went down so I suspect the work will be negative
In other words the ride did work on the child to bring the kid up to speed.
To find the new angular speed of the merry-go-round, you can use the principle of conservation of angular momentum.
The initial angular momentum of the merry-go-round is given by L_initial = I_initial * ω_initial, where I_initial is the moment of inertia and ω_initial is the initial angular speed in radians per second.
The final angular momentum of the system is given by L_final = I_final * ω_final, where I_final is the moment of inertia of the system with the child on it and ω_final is the final angular speed in radians per second.
According to the conservation of angular momentum, L_initial = L_final.
Since the merry-go-round and the child are considered as a system, their total moment of inertia is the sum of the moment of inertia of the merry-go-round alone and the moment of inertia of the child.
I_final = I_merry-go-round + I_child.
Given that moment of inertia I_merry-go-round = 221 kg*m^2 and the radius R = 1.92 m, you can calculate the moment of inertia using the formula: I_merry-go-round = (1/2) * m * R^2, where m is the mass of the merry-go-round.
For the child, the moment of inertia is calculated using a similar formula: I_child = m_child * R^2.
Given that the mass of the child is 29.8 kg, you can calculate the moment of inertia of the child.
Now, substitute the values into the conservation of angular momentum equation:
I_initial * ω_initial = (I_merry-go-round + I_child) * ω_final.
Solve for ω_final to find the new angular speed in radians per second.
To convert the angular speed from radians per second to revolutions per minute, use the following conversion factor: 1 rev/min = 2π rad/s.
To calculate the work done by the child, you need to use the relationship between work and change in kinetic energy.
The work done by the child is equal to the change in kinetic energy of the system.
The initial kinetic energy of the merry-go-round is given by KE_initial = (1/2) * I_initial * ω_initial^2.
The final kinetic energy of the system is given by KE_final = (1/2) * (I_merry-go-round + I_child) * ω_final^2.
Calculate the final and initial kinetic energies, and then subtract the initial kinetic energy from the final kinetic energy to get the change in kinetic energy.
This difference will give you the amount of work done by the child.
I hope this explanation helps!
To find the new angular speed of the merry-go-round, we can apply the conservation of angular momentum. Initially, the merry-go-round is rotating at 14.0 rev/min, which can be converted to radians per second as follows:
Initial angular speed (ωi) = 14.0 rev/min * (2π rad/rev) * (1 min/60 s)
= 14.0 * 2π / 60 rad/s
The moment of inertia (I) is given as 221 kgm^2, and the radius of the merry-go-round (R) is 1.92 m.
When the child hops on, the moment of inertia of the system changes to include the child. The moment of inertia of a point mass is given by the formula I = mr^2, where m is the mass and r is the radius. So, the moment of inertia of the child on the merry-go-round is:
Ichild = mchild * R^2
Substituting the given values, the moment of inertia of the child is:
Ichild = 29.8 kg * (1.92 m)^2
Now, we can use the conservation of angular momentum to find the new angular speed (ωf) of the merry-go-round and child system. Angular momentum is given by the product of moment of inertia (I) and angular speed (ω):
Initial angular momentum (Li) = Ii * ωi
Final angular momentum (Lf) = If * ωf
Since angular momentum is conserved, we have Li = Lf.
Therefore, Ii * ωi = If * ωf
Plugging in the given values, we have:
221 kgm^2 * (14.0 * 2π / 60 rad/s) = (221 kgm^2 + 29.8 kg * (1.92 m)^2) * ωf
Solving for ωf, we get:
ωf = [221 kgm^2 * (14.0 * 2π / 60 rad/s)] / [221 kgm^2 + 29.8 kg * (1.92 m)^2]
Calculating this expression will give us the new angular speed (ωf) of the merry-go-round in radians per second. Converting this to revolutions per minute, we can multiply ωf by (60 s / 2π rad) and divide by (1 min / 1 s).
To find the work done by the child, we need to calculate the change in kinetic energy of the system. The work done (W) is equal to the change in kinetic energy, given by:
W = (1/2) * If * (ωf^2 - ωi^2)
Substituting the given values and the calculated value of ωf, we can find the work done by the child.