Small mass attached to a string sliding between two circles

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SUMMARY

A small mass M attached to a string slides between two circles on a frictionless table, with circle X having a radius twice that of circle Y. As M transitions from circle Y to circle X, the work done by the force F is zero, and M's kinetic energy at circle X is one quarter that at circle Y. There is no torque acting on M during this transition, and M's angular momentum at circle X is equal to that at circle Y. Additionally, M's angular velocity at circle X is one quarter that at circle Y.

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sharkmanic
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Homework Statement



A small mass M attached to a string slides in a circle (Y) on a frictionless horizontal table, with the force F providing the necessary tension (see figure). The force is then decreased slowly and then maintained constant when M travels around in circle (X). The radius of circle (X) is twice the radius of circle (Y).

http://imgur.com/w970m

Possible answers (T,F,>,<,=)

As M moves from Y to X, the work done by F is ... 0.
M's kinetic energy at X is one quarter that at Y.
While going from Y to X, there is no torque on M
M's angular momentum at X is ... that at Y.
M's angular velocity at X is one quarter that at Y.

Homework Equations



I=mr^2
x=2y
etc..

The Attempt at a Solution



The main problem I run into, is that the HW system does not show you what you got right and what you got wrong. It's an all or nothing question. Anyway.

I've done some calculations and I think that the Wy = 4*Wx, and Iy = (1/4)Ix but I am not sure about the others, or if those are even correct. If someone could walk me through the reasoning behind these that would be great!
 
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sharkmanic said:

Homework Statement



A small mass M attached to a string slides in a circle (Y) on a frictionless horizontal table, with the force F providing the necessary tension (see figure). The force is then decreased slowly and then maintained constant when M travels around in circle (X). The radius of circle (X) is twice the radius of circle (Y).

Possible answers (T,F,>,<,=)
Haven't seen a question yet!
(1) As M moves from Y to X, the work done by F is ... 0.
(2) M's kinetic energy at X is one quarter that at Y.
(3) While going from Y to X, there is no torque on M
(4) M's angular momentum at X is ... that at Y.
(5) M's angular velocity at X is one quarter that at Y.

The main problem I run into, is that the HW system does not show you what you got right and what you got wrong. It's an all or nothing question. Anyway.

I've done some calculations and I think that the Wy = 4*Wx, and Iy = (1/4)Ix but I am not sure about the others, or if those are even correct. If someone could walk me through the reasoning behind these that would be great!
I suspect you've been over-thinking it.
Tackling them one at a time:
(1) W=Fd - which distance is important for F?
(2) How do you calculate kinetic energy?
(3) What would a torque do to the motion? Has this happened?
(4) You know the formula for angular momentum!
(5) You know the relationship between angular velocity and angular momentum!
 

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