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We can derive easily the expression for tension and velocity for a particle in vertical circular motion. We know that in a string tension occurs when the attached object pulls the rope away from the point where the rope is rigidly attached. enter image description here

We see that when $\theta$ is acute then there is a radial component of $mg$ which pulls the object, which in turn the rope away from the center of the circle. Thus it produces tension in the rope. But when the angle is obtuse (like $90^o+\theta$), then we see that the component of $mg$ also act radially inwards, so there is nothing which pulls the rope away from the center and also the velocity is tangential.

$T=\frac{mu^2}{r}-2mg+3mgcos\theta$
So, $T=0$, when $cos\phi=\frac{2gl-u^2}{3gl}$
So, when $u=\sqrt{2gl}$, then $T=0$ at $\theta=90^o$
But when $\sqrt{2gl}<u<\sqrt{5gl}$ at $90^o<\theta<180^o$
If we think intuitively, $T$ should be $0$ at all $\theta$ such that $90^o<\theta<180^o$, irrespective of any initial speed greater than $\sqrt{2gl}$. Because at $\theta>90$, the radial component of $mg$,i.e., $mgsin\theta$ also act inwards, so there is nothing which pulls the rope outwards and so alone this $mgsin\theta$ should provide the centripetal acceleration.
We can think $T$ to exist when $\theta>90^o$, when there is some external agency which holds the rope and gives the object a vertical circular motion by pulling the rope to provide centripetal acceleration, but if we think the rope is attached to the nail, there should not be any tension at $\theta>90^o$.
Please help me in explaining intuitively why tension exists in rope at $90^0<\theta<180^o$ when there is nothing which pulls the rope.

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  • $\begingroup$ I don't think I understand your point about $T$ being intuitively 0 for $0^\circ \leq \theta \leq 90^\circ.$ The free-body diagram clearly indicates that $T = mu^2/r - mg \sin{\theta}.$ If it helps, you can think about the $mg \sin{\theta}$ component 'pushing' on the rope to provide part of the tension. $\endgroup$
    – Yejus
    Jul 31, 2020 at 15:12
  • $\begingroup$ Why we don't write it as $mu^2/r=mgsin\theta$. $\endgroup$
    – Manu
    Jul 31, 2020 at 15:14
  • $\begingroup$ Because there is tension in the string. Think about the string from the string's frame of reference. It is being pulled outward in its motion around the center due to the centrifugal force of its motion. It doesn't matter what the angle is, as long as it is moving in circles, it is going to experience an outward force. That, along with the component of the weight (although it is in the opposite direction) means there is tension in the string. I don't see any sense in assuming $T = 0$ except at specific points. $\endgroup$
    – Yejus
    Jul 31, 2020 at 15:18
  • $\begingroup$ Basically the question is tension is producesd in a rope when it gets pulled, but if we see here the radial component of g basically not pulling the rope at an obtuse angle, so intuitively we should not take into account the tension when $\theta$ is obtuse and only that radial component of g should provide required centripetal acceleration if the partic le moves in circular motion? $\endgroup$
    – Manu
    Jul 31, 2020 at 15:18
  • $\begingroup$ @Yejus in second comment, yes the centrifugal force acts on the object in its rotating reference frame but shouldn't it is the incomplete picture, because if we say as long as the radial component of g balances the centrifugal force then the tension should be zero? $\endgroup$
    – Manu
    Jul 31, 2020 at 15:22

1 Answer 1

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You have to understand how the rope affects the motion of the particle.

Imagine to cut the rope at an instant, e.g. when the angle is obtuse. What would the particle do? It would no more move along the circle. Rather, it would follow a parabolic trajectory (because of gravity alone) which is tangent to the circle at the point in which the particle was when the rope was cut.

If the particle goes fast enough, this parabola would bring the particle further from the center to which the rope is attached. This means that, without the rope, the particle would move away from the center of the circular motion. The rope "wants" to keep its length unchanged, so it prevents the particle to move away from the center, and it does so by means of the tension. It is this "natural" motion of the particle away from the center, that keeps the rope under tension.

On the other hand, if the motion of the particle is not fast enough, the parabola would not bring the particle further from the center. So there is no need for tension in the rope. In this scenario, the particle simply follows the parabola, even if there is a rope, and it "falls down" with respect to the circular trajectory (this can happen only during the part of the trajectory where the angle is obtuse).

This is why the particle has to be faster than a minimal speed in order to complete a full loop.

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  • $\begingroup$ I have read your answer. I tried your explanation of cutting the rope but at $\theta=90°$. If I cut the rope at 90°, then the object will have some velocity(tangentially/upward) and g is acting downward so particle will follow a straight line motion. But suppose if it is still connected to rope at 90°, then at that moment also g is downward. So my question is that how tension arises because now onwards no component of g is acting radially outwards to pull the rope thus no tension. May you please explain how justify the existence of tension if we don't know that object exhibits circular motion $\endgroup$
    – Manu
    Jul 31, 2020 at 16:16
  • $\begingroup$ for $\theta>90$. $\endgroup$
    – Manu
    Jul 31, 2020 at 16:17
  • $\begingroup$ @Manu If you cut at $\theta=90º$, the object goes along a straight line, and as it moves it gets further away from the center. The point is that without rope, the object would "naturally" move further from the center. But the rope constrains it to be no further than the circle's radius. To do so, it has to pull the object. $\endgroup$
    – HicHaecHoc
    Jul 31, 2020 at 16:26
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    $\begingroup$ Tension arises when the object "wants" to get further from the center. This could be because of $\vec{g}$, but not necessarily. It could also be simply because of the object's inertia. $\endgroup$
    – HicHaecHoc
    Jul 31, 2020 at 16:28
  • $\begingroup$ When $\theta>90º$ it is only because of inertia. $\endgroup$
    – HicHaecHoc
    Jul 31, 2020 at 16:31

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