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A light inextensible string of length $a$ is threaded through a fixed smooth ring $R$. One end of the string is attached to a particle $A$ of mass $3m$. The other end of the string is attached to a particle $B$ of mass $m$. The particle $A$ hangs in equilibrium at a distance $x$ vertically below the ring. The angle between $AR$ and $BR$ is $\theta$ (see diagram). The particle $B$ moves in a horizontal circle with constant angular velocity $2\sqrt{\frac{g}{a}}$.

Show that $\cos \theta = \frac{1}{3}$ and find $x$ in terms of $a$.

Diagram

What I tried:

B experiences two external forces: its weight and the inward radial acceleration. These two forces are balanced by the tension in the string. So I wrote down the following: $$T = 3mg \\ T = \sqrt{\left(\frac{mg}{\text{cos }\theta}\right)^2+\left(\frac{m\cdot a_{radial}}{\text{sin }\theta}\right)^2}$$ But when I equate the two values of T together, I'm not really able to get anywhere. You're left with three unknowns: the radius $r$, $\theta$, and $a$.

$r$ can be substituted with $(a-x)\sin \theta$, but you'd still have three unknowns.

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  • $\begingroup$ There are only two forces acting on A. The weight and the tension. $\endgroup$
    – nasu
    Sep 26, 2021 at 20:30
  • $\begingroup$ @nasu Yes, therefore T = 3mg $\endgroup$ Sep 26, 2021 at 21:48
  • $\begingroup$ And the same is true for B. Just two forces $\endgroup$
    – nasu
    Sep 26, 2021 at 22:11
  • $\begingroup$ @nasu How come? What about the centripetal force? $\endgroup$ Sep 26, 2021 at 22:14
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    $\begingroup$ You are correct that there are two forces. But those are tension and the weight. There's no real force as centripetal force. A component of tension force acts as centripetal force here. $\endgroup$
    – ACB
    Sep 27, 2021 at 5:07

2 Answers 2

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The tension in the string provides the centripetal force. It does not come out of nowhere. Reevaluate the problem. You can easily get the angle $\theta$ by balancing the vertical component of the tension with the particle's weight at B. Then, the remaining part of the problem is just equating the horizontal component of the tension with the centripetal force.

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You have to just balance the forces acting on masses and maintain their equilibrium. The tension is generated in string because of radail acceleration of the small mass $B$.

So, for $A$, we have $$T=3mg$$

For $B$, by making F.B.D: $$Tcos(\theta)=mg$$ $$Tsin(\theta)=\frac{mv^2}{r}$$

From simple geometry, we can find, $r=(a-x)sin(theta)$.

Now from first and second equations, we can find $cos\theta$, and doing simple algebrae, we can find $x$

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