Solving Centrifugal Forces in a Rotating Coordinate System

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In summary, the figure has a centre of mass, two masses at the edges of the figure, and a force acting between them. The equation for the force is F(A)-T=m(A)*r''(A), where r''(A) is the distance between the two masses at the edge of the figure. The equation for the centrifugal force is F(B)-T=m(B)*r''(B), where r(B) is the distance between the centres of mass of the two masses. The sign convention for these radially symmetric problems is 'radially outwards is positive'.
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warhammer
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Homework Statement
A disk rotates with constant angular velocity ω. Two masses A & B slide without friction in a groove passing through center of disk. They are connected by string of length 'l' and are initially held in position by catch with mass A at a distance r(A) from the center. Neglect gravity. At t=0 the catch is removed and masses are free to slide. Find r''(A) immediately after catch is removed in terms of A,B,l,r(A) & ω.
Relevant Equations
F(A)-T=m(A)*r''(A)

T-F(B)=m(B)*r''(B)

r(A)+r(B)=l
I am also attaching picture of the figure as well as my diagram to showcase the forces.

I have devised these equations because as per the question, r(A) is acting like a reference.

The constraint r(A)+r(B)=l--------(i) transforms to r''(B)= - r''(A) ------- (ii)

Using-

F(A)-T=m(A)*r''(A)

T-F(B)=m(B)*r''(B)

Adding both & using (i) & (ii), F(A)-F(B)= r''(A) {m(A)-m(B)}

r(A)*ω^2 [m(A)+m(B)] - m(B)*l*ω^2= r''(A) {m(A)-m(B)}

{ r(A)*ω^2 [m(A)+m(B)] - m(B)*l*ω^2} /{m(A)-m(B)}

This answer is not coming out to be correct. I sense I have taken force directions wrong but unable to understand why.
 

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  • #2
warhammer said:
T-F(B)=m(B)*r''(B)
Can you see the sign error there?
 
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  • #3
haruspex said:
Can you see the sign error there?

I used this sign convention because r(A) would increase if A tended to move away from the center. Since I used that as reference, I also applied the similar convention for B, T is 'towards' A and F(B) away from it. Can I not do that?
 
  • #4
warhammer said:
I used this sign convention because r(A) would increase if A tended to move away from the center. Since I used that as reference, I also applied the similar convention for B, T is 'towards' A and F(B) away from it. Can I not do that?
But elsewhere you used r(B) for B's distance from the centre. The tension tends to reduce that and the centrifugal force to increase it.
 
  • #5
haruspex said:
But elsewhere you used r(B) for B's distance from the centre. The tension tends to reduce that and the centrifugal force to increase it.

I replaced it with (l-r(A)) and r''(A).

(I'm sorry if my questions seem silly sir. It's all new to me and I've hit an understanding block).
 
  • #6
EDIT: Changed answer to reflect working in the rotating coordinate system.

Looks like F(A) and F(B) are the (fictitious) centrifugal forces. So you are working in the non-inertial rotating frame-of-reference of the disc.

The usual sign-convention for these radially symmetric problems is 'radially outwards is positive'.

Using this sign convention:
F(A) - T = m(A)*r''(A) is fine because F(A) is outwards and T in inwards
So you need to address the equation T - F(B) = m(B)*r''(B).
 
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  • #7
Steve4Physics said:
EDIT: Changed answer to reflect working in the rotating coordinate system.

Looks like F(A) and F(B) are the (fictitious) centrifugal forces. So you are working in the non-inertial rotating frame-of-reference of the disc.

The usual sign-convention for these radially symmetric problems is 'radially outwards is positive'.

Using this sign convention:
F(A) - T = m(A)*r''(A) is fine because F(A) is outwards and T in inwards
So you need to address the equation T - F(B) = m(B)*r''(B).

Thank you so much for your help sir.
 

FAQ: Solving Centrifugal Forces in a Rotating Coordinate System

1. What is the formula for solving forces in r(A)+r(B)=l?

The formula for solving forces in r(A)+r(B)=l is to first identify the forces acting on the objects A and B, then use vector addition to find the resultant force, which should equal the length l. This can be represented as r(A)+r(B)=l.

2. How do I determine the direction of the resultant force?

The direction of the resultant force can be determined by using the direction of the individual forces and applying vector addition. The resultant force will be in the same direction as the sum of the individual forces.

3. Can I solve for the individual forces if I know the resultant force and the length l?

Yes, you can solve for the individual forces by using the formula r(A)+r(B)=l and rearranging it to solve for either r(A) or r(B). Then, using the direction of the resultant force, you can determine the direction of the individual forces.

4. What units should I use when solving forces in r(A)+r(B)=l?

The units used for solving forces in r(A)+r(B)=l will depend on the units used for the individual forces and the length l. It is important to make sure that all units are consistent and in the same system (e.g. SI units).

5. Can this formula be applied to any type of force, such as gravitational or electromagnetic forces?

Yes, this formula can be applied to any type of force as long as the forces are represented as vectors. This means that the magnitude and direction of the forces should be known and can be added using vector addition.

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