Question

A particle is moving in a circular path in a vertical plane. It is attached at one end of a string of length $l$ whose other end is fixed. The velocity at the lowest point is $u$. The tension in the string is $\vec T$ and acceleration of the particle is $\vec a$ at any position. Then $\vec T \cdot \vec a$ is zero at highest point if
(A) $u > \sqrt {5gl}$
(B) $u = \sqrt {5gl}$
(C) Both (a) and (b) are correct
(D) Both (a) and (b) are wrong

Answer 1

Hint : Since acceleration will always exist irrespective of tension because of gravity, the dot product of tension and acceleration signifies that tension is zero. The total energy of the system is always constant.

Formula used: In this solution we will be using the following formula;
$KE + PE = {\text{constant}}$(principle of conservation of energy) where $KE$ is the kinetic energy of a body, $PE$ is the potential energy.
${F_c} = \dfrac{{m{v^2}}}{r}$ where ${F_c}$ is the centripetal acceleration of a circulating body, $m$is the mass, $v$is the linear velocity of the at a point, and $r$ is the radius of the circle.
${F_{NET}} = ma$ where ${F_{NET}}$ is the net force on a body and $a$ is the acceleration of that body.

Complete step by step answer
The body in the question is rotation on a vertical plane. The net force in the bottom is given by
$T - mg = \dfrac{{m{u^2}}}{r}$(since tension and weight are in opposite directions, and also, velocity at the bottom is given as $u$)
But for the top, it is given as
$T + mg = \dfrac{{m{v^2}}}{r}$
Now, as mentioned in the question $\vec T \cdot \vec a = Ta\cos \theta = 0$at the top. In general, this implies that $T = 0$or $a = 0$ or $\theta = 90$. However, for the question, since gravity acts downward and thus can provide the centripetal acceleration, it must imply that $T = 0$ at the top.
Hence, the equation above becomes
$mg = \dfrac{{m{v^2}}}{r} = \dfrac{{m{v^2}}}{l}$ since the length is the radius
Now, since the tension is zero at the top, this means that no extra effort was applied to stretch the string, hence the total energy in the system is constant.
Hence,
$KE + PE = {\text{constant}}$ where $KE$ is the kinetic energy of a body, $PE$ is the potential energy. Hence,
$K{E_b} + P{E_b} = K{E_t} + P{E_t}$ where the subscript b and t signifies bottom and top respectively.
Making the centre the reference point and picking the potential below the centre is negative and above the centre positive, we have that
$\dfrac{1}{2}m{u^2} - mgl = \dfrac{1}{2}m{v^2} + mgl$,
Now, from $mg = \dfrac{{m{v^2}}}{l}$, ${v^2} = gl$
Hence, $\dfrac{1}{2}m{u^2} - mgl = \dfrac{1}{2}mgl + mgl$
Adding $mgl$ to both sides, we have
$\dfrac{1}{2}m{u^2} = \dfrac{1}{2}mgl + mgl + mgl = \dfrac{5}{2}mgl$
Making $u$ subject of formula we get
$u = \sqrt {5gl}$
Hence, the correct answer is B.

Note
To avoid confusions, the fact that we chose the reference point of the potential to be the centre is a matter of choice to make the equation simpler. It could be chosen from the ground as common, as in
$\dfrac{1}{2}m{u^2} + mgs = \dfrac{1}{2}m{v^2} + 2mgl + mgs$, where $s$ is the height of the stone from ground when at the bottom. We observe that $mgs$ cancels and leaves us with something similar as in the solution.