What is the new rotation time of the merry-go-round?

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The discussion focuses on calculating the new rotation time of a playground merry-go-round after adding a block of mass. Initially, the merry-go-round has a moment of inertia of 50 kg m² and completes one revolution in 0.5 seconds. When a 5.0 kg block is placed at a distance of 1.0 m from the center, the conservation of angular momentum principle is applied to determine the new rotation time. The final rotation time is calculated to be approximately 0.63 seconds.

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[SOLVED] Not sure where to start...

10. A playground merry-go-round is mounted on a frictionless axle, so it can only rotate
in the horizontal plane. This object has a moment of inertia about the axle of I=50 kg m2,
and it has a diameter of 2.2 meters. Initially it is turning at a constant rate so that it
completes one revolution in 0.5 seconds. Standing next to it, you carefully place a block
of mass m= 5.0 kg on the rotating table, so that it sits a distance r=1.0 m from the center.
How long does it take to complete one rotation now?
1. 0.5 s
2. 0.55 s
3. 0.59 s
4. 0.63 s
5. 0.67 s
6. 0.71 s

all i have is that i found wi = 12.57 rad/s...but i don't know what to do after that?
 
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Can you think of any quantity which will remain conserved?
 
The moment of inertia?
 
No, because the mass distribution has changed after you put the mass on it.

If there's no external torque on a body, then the angular momentum remains the same. So, you should find the initial and final angular momenta.
 
so initial and final angular momentum should equal since there is no external torque on the system right?
 
You said it.
 
Thanks!
 
I do have one more question...how do u recalculate the new moment of inertia?
 
MI of the mass = mass*(dist from axis)^2
Final MI = MI of disk + MI of the mass
 
  • #10
okay, thanks
 

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