A merry-go-round is rotating at 4.775 revs/min with a 45 kg child on the outer edge. The radius is 2.0 m. The child moves to the centre of the merry go round. What is the final angular velocity of the merry go round. The moment of inertia of the merry-go-round is 360kgm^s
First thing, w initial=0.5 rad/s
So L initial= 90kgm^2/s
I'm going to use conservation of momentum to find the final angular velocity, but I'm not sure what the final momentum of inertia is (ie. when the child moves to the centre?)
I tried taking 360 and adding it to 1/2MR^2 (to find the contribution of the moment of inertia of the child) trying to account for the child being at the centre.. then using conservation of momentum and solving for Wf.. but i didn't get the correct answer
Consider the angular moment of inertia to be I. Then the moment of inertia due to m at some r is 1/2 m r^2
conservation of momentum:
Iinitial wi= Ifinal * Wf
(I+1/2mr^2)w=(I + 1/2 m 0^2)wf
(I+1/2 45*2).5=(I wf)
you are given I as 360
405*.5=360 wf
wf=.563 rad/sec
check me.
Yes!
That's right!
Except the moment of inertia of the person is just mr^2 (not 1/2mr^2), making the final answer 0.75, aka the correct answer in the solutions manual!
Thanks so much :)
To find the final angular velocity of the merry-go-round when the child moves to the center, you can use the principle of conservation of angular momentum. However, it seems like there might be some confusion in your approach.
Let's break down the steps to find the final angular velocity:
1. Initial angular velocity (ω_initial): You correctly mentioned that the initial angular velocity is 0.5 rad/s. This value represents the angular velocity of the merry-go-round when the child is on the outer edge.
2. Initial angular momentum (L_initial): Given that the mass of the child is 45 kg and the radius is 2.0 m, we can calculate the initial angular momentum using the formula:
L_initial = moment of inertia * ω_initial
The initial angular momentum is given as 90 kg·m²/s, so using the above formula, we can determine the moment of inertia of the merry-go-round:
90 kg·m²/s = 360 kg·m² + (1/2) * 45 kg * (2.0 m)²
90 kg·m²/s = 360 kg·m² + 90 kg·m²
Thus, the moment of inertia of the merry-go-round is indeed 360 kg·m².
3. Final moment of inertia (I_final): When the child moves to the center, their contribution to the moment of inertia decreases. The moment of inertia of the merry-go-round remains the same, but now without the child's contribution.
I_final = moment of inertia (merry-go-round) - moment of inertia (child)
Since the moment of inertia of the merry-go-round is 360 kg·m² and the moment of inertia of the child is (1/2) * mass * radius², we can calculate:
I_final = 360 kg·m² - (1/2) * 45 kg * (2.0 m)²
I_final = 360 kg·m² - 90 kg·m²
I_final = 270 kg·m²
Now we have the final moment of inertia of the merry-go-round.
4. Final angular velocity (ω_final): To find the final angular velocity, we use the conservation of angular momentum:
L_initial = L_final
moment of inertia (merry-go-round) * ω_initial = I_final * ω_final
Substituting the known values:
360 kg·m² * 0.5 rad/s = 270 kg·m² * ω_final
Simplifying the equation:
ω_final = (360 kg·m² * 0.5 rad/s) / 270 kg·m²
ω_final = 0.6667 rad/s
Therefore, the final angular velocity of the merry-go-round, when the child moves to the center, is approximately 0.6667 rad/s.