Masses Moving Radially on a Rotating Disk

AI Thread Summary
The discussion centers on the dynamics of a rotating disk with masses moving radially. When four objects are moved inward, the conservation of angular momentum increases the disk's angular velocity, potentially causing the fifth object to slide off if it exceeds its maximum speed. The calculations reveal that if the fifth object remains at the outer position, it will exceed the maximum angular velocity and slide off. Participants express concerns about the exercise's clarity, suggesting that the problem may not have been well-constructed, as it overlooks the effects of static and kinetic friction. Ultimately, the consensus is that the fifth object cannot remain on the disk under the given conditions.
hquang001
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Homework Statement
Four objects (each mass m=60 kg) are attached to a disk, which has a radius of r= 1.5 m and a moment of Inertia of I=130 kg m2. The objects can be moved by a motor from an outer position (ro=1.5 m) to an inner position (ri= 0.3 m). A fifth object (mass m5=60kg) is placed onto the disc. This object is not fixed to the disc like the other ones. But it has an unbelievable high static friction coefficient with the disc of μs=3.0. The kinetic friction coefficient has a value of μk=0.9. The disc starts to rotate at a modest 20 rev/min. All objects are at their outer positions. Then suddenly the motor moves the four objects to the inner positions. What is happening to the fifth object? Explain why and prove it by calculation. Keep in mind that g=9.81 m/s2
Relevant Equations
I1 w1 = I2 w2
F = ma
22.png

m = 60kg, ω0 = 2.094 rad/s, I of disk = 130 kgm^2 , outer position ro = 1.5m, inner position ri = 0.3m
∴Fifth object :
Ffriction = m.ac
μ.m.g = m. v^2 / R
=> vmax = √ 3. (1.5m) . (9.81 m/s^2 ) = 6.64 m/s => ωmax = 4.43 rad/s
so when the fifth object move with greater speed than vmax =6.64m/s , it will slide off

When four objects are moved to inner position, due to conservation of angular momentum, the final angular velocity of the disk will increase. and if it exceed the maximum angular speed, it slides off
Using conservation of angular momentum:I1ω1 = I2ω2
(130 + 5.m.r0^2). (2.094 rad/s ) = (130 + 4.m.ri^2 + m.x^2) .ω2

I got stuck here. I can't find final angular velocity and therefore can't find the position of the fifth object
 
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hquang001 said:
can't find the position of the fifth object
Assume it stays in place at ro and see if that leads to a problem or not.
 
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BvU said:
Assume it stays in place at ro and see if that leads to a problem or not.
Assume fifth object stay at ro :
=> w2 = 5.88 rad/s
it exceeds the maximum and will slide off but the position of the object is 1.5m, so is it wrong ?
 
So it will slide off.
And that answers the question of the exercise.
 
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BvU said:
So it will slide off.
And that answers the question of the exercise.
Ohh i got it
Thank you
 
This is a strange question.
We are told the motor suddenly moves the four blocks to a smaller radius. That would generate a large acceleration of the disk, and the tangential acceleration surely would defeat the static friction anyway.
That in itself would not jettison the mass; it would simply find itself a bit further round the disk. So whether it then stays on the disk depends on the kinetic friction, which we are given.
Seems to me that the question setter has made a mistake, not having intended that the new radial acceleration alone would be enough to overcome static friction.
 
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When I re-read the probem statement, the fifth object starting at the outer position has no chance to move further round the disk !

I don't think the exercise composer intended anything more complicated than ##\ m\omega^2/r_o > f_s\,N ##

##\ ##
 
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BvU said:
When I re-read the probem statement, the fifth object starting at the outer position has no chance to move further round the disk !

I don't think the exercise composer intended anything more complicated than ##\ m\omega^2/r_o > f_s\,N ##

##\ ##
Just allow a small margin between the object and the very edge of the disk. A sudden increase in spin would mean that the initial trajectory of the object relative to the disk is tangential. Only after that slippage starts, robbing the object of most of its frictional force, will it slide radially.
The two suspicious facts are
  • The movement of the other blocks is described as sudden
  • We are given a much lower kinetic friction coefficient
Why do either of those if they are irrelevant?
 
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I see,

Can I make a case that the slippage occurs instantaneously ?
So the block no longer contributes to the moment of inertia - meaning ##\omega## steps from 2.094 not to 5.883 but to 11.12 rad/s

Kinetic friction is enough to maintain circular trajectory up to ##\omega ## = 2.43 rad/s which takes about 0.04 s.

That would mean: even without margin the object would do about 5 degrees on the rim and then continue in a straight line ? I think this is one of those exercises that hasn't been thought through completely by the composer; average to good students have no problem but very good students run into time-wasting complications.

##\ ##
 
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