A particle of mass m is fixed to one end of a light rod of length l and rotated in a vertical circular path about its other end. What is the minimum speed of the particle at the highest point?
It occurs to me that the answer should be zero. But I am not getting a strong conceptual reasoning behind it. Can anyone please answer the question with proper explanation?
Thanks.
Answer
In order to understand this motion, let us check first the case of a particle's motion in a vertical circle; here 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 $$T - mg\cos\theta = \dfrac{mv^2}{R}$$
The particle will complete the circle when at the highest point if the string doesn't slack at the highest point when $\theta = \pi$. There must be centripetal force to make this happen. To find the minimum centripetal force, we make $T = 0$ so that our new centripetal force is $$mg = \dfrac{mv^2}{R} \implies v= \sqrt{gR}$$ . Now this is the minimum velocity at the top which the particle must have in order to cover the whole circle. Why?? Why in order to find the minimum centripetal force, I have made $T = 0$? Even if it is so, why doesn't the gravity bring down the particle as if it were a free-fall instead of acting as a centripetal force?? Because if there is velocity more than $\sqrt{gR}$, then only is tension required to provide the extra centripetal force for the extra velocity. Regarding free-fall, yes the particle would had a free-fall if it had zero velocity at the top; remember if the velocity vector is perpendicular to the force which is pointing radially to some point, it is a configuaration of __circular-motion. Using $v_\text{min}^2 = u_\text{min}^2 -2gh \implies u_\text{min} = \sqrt{5gR} $
Now come to your case where in place of the string a light rod is placed; the constraint has been changed. Now we need not bother about whether the particle would fall when at the top. So, from this point we can say the particle can have any velocity. It need not, thus, have the minimum initial velocity $\sqrt{5gR}$. It can have any velocity but WARNING the velocity CANNOT be equal to zero at the topmost point because if the velocity becomes zero, there would be no centripetal force to move the particle to complete the whole circle. But it can be very, very, very, close to zero ie, $v > 0$ but $v \approx 0$. To complete the case, I am adding the minimum initial velocity which we get from $$\dfrac{m(u^2 - v ^2)}{2} = mgh \implies u = 2\sqrt{gR}$$. So the minimum initial velocity must be infinitesimally larger than $2\sqrt{gR}$. The minimum velocity at the topmost point is thus infinitesimally greater than $0$.
Note of Apology: Yes, really sorry for writing some irrelevant stuffs but I deem it apt enough that in order to understand that the minimum topmost velocity cannot be $0$ in order to have centripetal force for covering the whole circle; here most of the texts baffle and write $v_\text{min} = 0$ at the topmost point but they forget that there must be centripetal force in order to transverse the whole circle & if the velocity is zero, from $\dfrac{mv^2}{R}$, we get the centripetal force also becomes $0$. However it can have the velocity at the top very, very, very close to zero as long as it doesn't nullify the centripetal force. Thus the minimum topmost-point velocity is just infinitesimally greater than zero.
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