Angular momentum merry go round

Click For Summary

Homework Help Overview

The problem involves a person running on the edge of a merry-go-round, affecting its angular momentum. The scenario includes a person with a mass of 74 kg on a 6.6 m diameter turntable with a moment of inertia of 1820 kg.m², initially at rest. The person runs at a speed of 4.1 m/s relative to the turntable, causing the turntable to rotate in the opposite direction.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conservation of angular momentum and the relationship between the linear speed of the person and the angular velocity of both the person and the turntable. Questions arise about how to correctly relate these velocities and the implications of the reference frame used.

Discussion Status

Some participants have provided hints regarding the need to convert linear speed to angular velocity and have pointed out the distinction between speeds relative to the turntable and the ground. There is ongoing exploration of how to calculate the angular velocities involved, but no consensus has been reached on the correct approach.

Contextual Notes

Participants express urgency due to a deadline, indicating a time constraint that may affect the depth of the discussion. The original poster's initial calculations have been questioned, leading to further exploration of assumptions and definitions related to angular momentum.

sumitmanhas
Messages
12
Reaction score
0

Homework Statement


Suppose a 74 kg person stands at the edge of a 6.6 m diameter merry-go-round turntable that is mounted on frictionless bearings and has a moment of inertia of 1820 kg.m^2. The turntable is at rest initially, but when the person begins running at a speed of 4.1 m/s (with respect to the turntable) around its edge, the turntable begins to rotate in the opposite direction.
Part A
Calculate the magnitude of the angular velocity of the turntable.


Homework Equations


conservation of angular momentum


The Attempt at a Solution


this is how i approached it.
initial angular momentum = 0
final anglar momentum = I([tex]\omega[/tex]P) - I([tex]\omega[/tex]T)

while [tex]\omega[/tex]p = v/r and IP = MR2

MR2V/R - I([tex]\omega[/tex]T) = 0

[tex]\omega[/tex]T = MRV/IT

[tex]\omega[/tex]T = (74)(3.3)(4.1)/ 1820

[tex]\omega[/tex]T = 0.55 RAD/S

but due to some reason turns out to be the wrong answer , someone please help ??
 
Physics news on Phys.org
The speed of the person is given with respect to the turntable, not with respect to the ground (which is what you need).

Hint: If the angular speed of the turntable (with respect to the ground) is ω, what would be the angular speed of the runner with respect to the ground?
 
what i think is that angular velocity of the person with respect to ground = 4.1 + angular velocity of table
can be wrong though...someone reply as soon as possible this question is due within an hour
 
sumitmanhas said:
what i think is that angular velocity of the person with respect to ground = 4.1 + angular velocity of table
can be wrong though...someone reply as soon as possible this question is due within an hour
Almost. But 4.1 m/s is a linear speed, not an angular velocity. Convert it to an angular velocity first.
 
but what would be the angular velocity of the turntable then for finding the angular velocity of the person with respect to ground . do i use then one i calculated at my first attempt . and then plug it into equation for finding the angular velocity of person with respect to ground.??

and then after finding the angular velocity of the person with respect to ground u plug it back into the very first equation for finding angular velocity of the table.
tried this whole thing but never worked.

help ??
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
18
Views
7K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
10K
  • · Replies 9 ·
Replies
9
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
16K