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

  • Thread starter Thread starter quickclick330
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves a playground merry-go-round with a specified moment of inertia and initial rotation rate. The scenario changes when a mass is added to the system, prompting a discussion on how this affects the rotation time.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the conservation of angular momentum and the implications of adding mass to the system. Questions arise about recalculating the moment of inertia and how it affects the rotation time.

Discussion Status

The discussion is active, with participants exploring the relationship between initial and final angular momentum. Guidance has been provided regarding the conservation principle, and there is an ongoing inquiry into the recalculation of moment of inertia.

Contextual Notes

Participants are working within the constraints of a homework problem, focusing on the effects of mass addition on rotational dynamics without external torque.

quickclick330
Messages
82
Reaction score
0
[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?
 
Physics news on Phys.org
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
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
2K
Replies
18
Views
7K
  • · Replies 27 ·
Replies
27
Views
6K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
15
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 44 ·
2
Replies
44
Views
7K