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I read through the question Minimum velocity of the particle at the highest point which is quite similar to my questions. But I still have some doubts in the answers -

  1. "The particle will complete the circle when at the highest point if the string doesn't slack at the highest point when θ=π." - Is it because if the rope slacks, then the mg component will pull it inwards and so the particle will not move as a circle?

  2. When we talk about conditions for the bob to oscillate (u <= sqrt(2gR)), why do we say that velocity must be equal to zero before tension is equal to zero? And likewise for conditions for leaving the circle, why should tension be zero and velocity not 0? What really is meant by leaving the circle? Is it like following a different parabolic path for some time? Please illustrate.

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/201452 & physics.stackexchange.com/q/193975 $\endgroup$
    – user36790
    Commented Sep 1, 2015 at 4:49
  • $\begingroup$ Ok, can you please explain this - "The particle will complete the circle when at the highest point if the string doesn't slack at the highest point when θ=π". Why is this? Is it because if the rope slacks, then the path won't be circular but something like a spiral because the radius decreases? What exactly is meant by slacking - In a way, compression? $\endgroup$
    – Shodai
    Commented Sep 1, 2015 at 5:00
  • $\begingroup$ The above was mentioned in the second link. And for the first link, can someone please explain the derivations for when T will be zero and when V will be 0? $\endgroup$
    – Shodai
    Commented Sep 1, 2015 at 5:10

2 Answers 2

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Condition for traversing the whole circle:

Let the particle of mass $m$ is attached to an inextensible string of length $R$ which provides the necessary centripetal force along with gravity.

Let the initial velocity be $u$ at the lowest point of the vertical circle. After time $dt$, it moves to some other point transversing angle $\theta$. The height at which it is now located is $$h = R(1 - \cos\theta)$$. Neglecting all non-conservative force, from work-kinetic energy theorem, we get the velocity at the point after $dt$ is $$v^2 = u^2 -2gh$$. The necessary centripetal force is $$T - mg\cos\theta = \dfrac{mv^2}{R} \, .$$

Now in order to move in circular motion, there must be an inward force i.e. centripetal force perpendicular to the instantaneous velocity. $$v = \sqrt{\frac{R}{m} \left(T - mg\cos\theta\right)}\big{|}_{\theta =\pi} \neq 0$$ In order to have the critical case or the minimum velocity for looping the circle $\left( T -mg\cos\pi\right)$ must be minimum which is possible when the tension $T=0$. Therefore, $$v_\text{min} = \sqrt{gR} \implies u_\text{min}^2 =v_\text{min}^2 +2gh \;\;\;[h = 2R]\, \implies u_\text{min}= \sqrt{5gR} $$.


As the particle moves against the gravity, kinetic energy starts to get converting into potential energy of the system thus reducing the linear velocity. If the velocity becomes zero at the top, it can't have any centripetal force available & it will slack under gravity. And if the initial velocity is less than $\sqrt{5gR}$, then all the kinetic energy would become zero either at the top or earlier.

As said above, if $u \lt \sqrt{5Rg}$, then before reaching the topmost point, tension & velocity become zero at different points. From $$T - mg\cos\theta = \frac{mv^2}{R} \; \text{when} \; T = 0, \\ \implies \cos\theta = \frac{2gh - u^2}{Rg}$$. Substituting the value of $\cos\theta $ in $h = R(1-\cos\theta)$, we get $$h= \frac{u^2 + Rg}{3g} = h_t$$; this is the height where $T = 0$ . Similarly, from $$v^2 = u^2 -2gh \; \text{when} \; v=0\\ \implies 0 = u^2 -2gh \\ \implies h = \frac{u^2}{2g} = h_v$$.

Condition for leaving the circle:

The particle would leave the circle when $h_t \lt h_v \notin h_t =h_v$. This is because when $h_t = h_v$, both tension $T$ & velocity $v$ become zero; but there is gravity pointing downwards. So, when the particle starts to fall under gravity, it gains velocity as well as centripetal force in the form of tension. Hence, the particle doesn't leave.Using the above inequality, we find $$\frac{u^2 + Rg}{3g} \lt \frac{u^2}{2g} \\ \implies 2u^2 + 2Rg \lt 3u^2 \\ \implies u\gt \sqrt{2Rg}$$.

Condition for oscillation:

The particle will oscillate when $h_t\gt h_v$, Why? Because, it is the component of the gravity $mg\sin\theta$ which acts as restoring force for oscillation & $T$ balances the other component $mg\cos\theta$ & if it were not so, then $mg\cos\theta$ would compel the particle to leave the circular trajectory. So, using the inequality above, we get $$\frac{u^2 + Rg}{3g} \gt \frac{u^2}{2g} \\ \implies 3u^2 \lt 2u^2 +2Rg \\ \implies u^2 \lt 2Rg \\ \implies u \lt \sqrt{2Rg} $$. Also, from the above argument, we do get that $u \leq \sqrt{2Rg}$.

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  • $\begingroup$ @user35828: Check this; I've tried to explain all the scenarios of vertical circular motion. So, have a feed on this & tell whether it helped you! If problem persists still, then ask without hesitation:) $\endgroup$
    – user36790
    Commented Sep 1, 2015 at 11:40
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1."The particle will complete the circle when at the highest point if the string doesn't slack at the highest point when θ=π." - Is it because if the rope slacks, then the mg component will pull it inwards and so the particle will not move as a circle?

Yes. Gravity (the weight $W=mg$) is then strong enough to pull it back from the "swing". The rope is slacked and the circle shape is not maintained anymore.

"Slacking" simply means that the rope is not tight anymore and then the circular shape is messed up. The object doesn't have to move in a spiral or so, since it automatically will fall down and gain speed in the lower part of the circle, where gravity helps the circle to be maintained. So as a result it might only be the top part of the circle that is more "flat" because the rope "slacks", and so the swing is not a perfect circle.

2.When we talk about conditions for the bob to oscillate ($u <= \sqrt{2gR}$), why do we say that velocity must be equal to zero before tension is equal to zero?

If you by "oscillate" mean that the object swings back and fourth like a pendulum (and doesn't complete a circular swing), then of course this will only happen if it stops in the end points and starting going backwards.

And likewise for conditions for leaving the circle, why should tension be zero and velocity not 0?

If we are back to the complete circular swing here then think of it like this:

As long as the object is being swung around, the tension in the string is what keeps it in place and keeps it in a circle. The tension in the string is what makes the body change its direction all the time so the path is circular. If in the highest point the rope is not tight, then the body was not swung outwards hard enough. So if the rope is not tight then we have the possibility of slacking.

This is the situation where the circular path can be "flattened". Only if the rope is not tight all the time can the body take another path. If the rope is tight always, then the only possible path is the circular path. So that is a clear condition to look for.

What really is meant by leaving the circle? Is it like following a different parabolic path for some time?

The path doesn't have to be parabolic. It could simply be a circle with its top cut off. The point is that the object can maybe follow the circular path all the way round and maybe only slack in the top, so only the top part is "cut off".

Leaving the circle is simply a way to say that the bocy doesn't follow the perfect circular path.

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  • $\begingroup$ I see. So in the third one, suppose the T became 0 at a height H, then body can move like a projectile until it is back to the height H on the other side, after which tension will come back - Is that right? 1st and 4th are pretty clear now (Thanks!), but in the second one I wanted to ask why "Tension must not be 0", is it because of the same reason - To keep it in a circle? Or some other reason? $\endgroup$
    – Shodai
    Commented Sep 1, 2015 at 7:48
  • $\begingroup$ Yes, the body will then fall as a projectile. Since the tension is gone, only gravity affects the body, so the body falls as if there was no string. Only with the string gets tight again and has tension, then the body will be force back into the cirular path. $\endgroup$
    – Steeven
    Commented Sep 1, 2015 at 7:57
  • $\begingroup$ That the tension in the rope must be 0 for this to happen is simply a consequence of the fact that as long as there is tension, the rope is apparently holding in the body. And when the rope is pulling in the body, the body can only move in a circular path. When the tension is 0 then the rope has just become slack. $\endgroup$
    – Steeven
    Commented Sep 1, 2015 at 7:58
  • $\begingroup$ I see, so basically for oscillation what we want is that v must be 0; T not equal to 0 only implies circular path? Thank you! $\endgroup$
    – Shodai
    Commented Sep 1, 2015 at 8:04

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