Mechanics Question... (1 Viewer)

[Dash]

Question:
A light inextensible string of length 3L is threaded through a smooth ring and carries a particle at each end. One particle A, of mass m, is at rest at a distance L below the ring while the other particle B, of mass M, is totating in a horizontal circle whose centre is A. Find m in terms of M

Verticle Component of A:
(T1) = mg ................................(1)

Verticle Component of B:
(T2)cos(theta) = Mg ................(2)

Horizontal Component of B:
(T2)sin(theta) = Mr(w^2).........(3)


Solution:
It goes something like this...

(1) / (2)

(T1) / (T2) = mg / 2Mg
............2g = m / M
.............m = 2gM

Now, I was wondering, does T1/T2 have a fixed value of g???
Because that is the only way this solution can play itself out...
Can someone plz explain this working to me?

 

[turtle_2468]

Hmm... dodgy working.
Correct soln:

Draw a diagram of B and A. Note that B is rotating at same level as A (as that is its horizontal centre of rotation). Therefore, letting the ring be X, as AX=L, and AX+BX=3L, BX=2L. Therefore, BA = root(3)*L (pythagoras), and so angle(BXA)=Pi/3.

So: Let the tension force be T. (the tension for both particles is same as it is an inextensible string)

Vertical for A: T=mg ... (1)
Vertical for B: T/2=Mg... (2)
This is because Tcos (Pi/3)=T/2.
Therefore (1)/(2) yields:
2=m/M
m=2M.

 

[Dash]

Awww crap!

I'm such an idiot for not picking that up!

This ain't my working, so yeh, I was trying to follow it...

Thanks turtle_2468!

 

[Dash]

Originally posted by turtle_2468
the tension for both particles is same as it is an inextensible string

By the way, arn't the tensions the same because the string is threaded through a "smooth" ring, not because it is an inextensible string???

The tensions in the two strings would be different if the ring was "rough". Am I right, or am I talking weird?

 

[turtle_2468]

No problem

Hmm. Yes and no.
Speaking in "common-sense" terms: The smooth ring means that no energy is lost on the contact point between ring and string, and hence that the force transmitted is the same. However, think about it this way: If you connect 2 pts by an _elastic_ string and apply a force on one side, the string will strech and hence not all the force will get transmitted.

Therefore both a smooth ring and inextensible string are needed for tensions to be the same.