0
$\begingroup$

Suppose a particle of mass $m$ is attached to an inextensible string of length $R$ and is undergoing vertical - circular motion.

The centripetal force is given by the tension & its weight:

$$T - mg\cos\theta = m\dfrac{v^2}{R}$$. Now, the velocity is decreasing since it is being retarded by $mg\sin\theta$. So, $T - mg\cos\theta$ will change.

Now, my book says

The particle will complete the circle if the string does not slack even at the highest point ie. at $\theta = \pi$. Thus tension in the string should be greater than or equal to zero at $\theta = \pi$.

Now, how can tension be zero upward? Even if it is zero, why doesn't it slack then?

Then the book jot down an equation:

So, $$T \geq 0 \quad \text{at} \quad \theta = \pi.$$ In critical case, substituting $\ T = 0\ $ and $\ \theta = \pi\ $ in the above mentioned equation, we get $$mg = m\dfrac{{v_{min}}^2}{R} \implies v_{min} = \sqrt{gR}.$$ After some kinematic calculation, we get the intitial velocity to be $\sqrt{5gR}$. Thus, to complete the circle, the particle must have velocity $\geq \sqrt{5gR}$.

Now, here the author always wrote $v_{min}$. What is $\text{min}$? Meant to say, why it should be minimum?? I'm not getting the sense.

$\endgroup$
2
$\begingroup$

There are two senses to $v_{\mathrm{min}}$. Firstly it means that the speed of the particle must be greater or equal to $v_{\mathrm{min}}$ at $\theta = \pi$ or else it will not complete the cycle. Instead it will follow a ballistic trajectory until such time that tension comes back into the string. Secondly, the speed of the particle is not constant: it is fastest at the very bottom and slowest at the very top. Thus $v_{\mathrm{min}}$ it is the slowest speed of the particle at any point around the cycle for a particle that just completes the cycle.

EDIT: The reason that there is no tension in the string is that the centripetal acceleration is perfectly balanced by the particle's weight when $v=v_{min}$ at the very top. If the particle were travelling faster here then the "excess" centripetal acceleration is taken up by tension in the string.

$\endgroup$
  • $\begingroup$ Sir, can you tell me why the tension should be zero. I'm not understanding it. And sir, can you please elaborate, how it would follow a ballistic trajectory and how tension would reclaim it? Please help. $\endgroup$ – user36790 Apr 15 '15 at 8:55
0
$\begingroup$

The reason the mass completes it’s circular path even if tension is zero at the highest point is gravitational force.

If tension is momentarily zero at highest point the body would still be able to complete the loop.

But it should have a minimum speed of $\sqrt{gR}$ at the topmost point. If the speed is lower than $\sqrt{gR}$, the centripetal force of gravity which will always be present will overcome the centrifugal force.

If speed is greater than $\sqrt{gR}$, centrifugal force will be $$F_{centrifugal} \geq\frac{m\sqrt{{gR}}^2}{R}$$ ie $$F_{centrifugal} \geq{mg}$$

In these conditions, tension will come into action, and provide the necessary centripetal force.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy