Question

A ball of mass $m$ is suspended by a thread of length $l$. With what minimum velocity has the point of suspension to be shifted in the horizontal direction for the ball to move along the circle about the point?

• A. $\sqrt{3gl}$
• B. $\sqrt{5gl}$
• C. $\sqrt{7gl}$
• D. $\sqrt{2gl}$

Answer 1

The correct option is B $\sqrt{5gl}$

The point of suspension is moving with velocity $u$ as shown in the figure.

From the frame of reference of point of suspension, ball will seem to be moving with same velocity in opposite direction. Hence the given case changes to

To complete the circle, tension in the thread when the ball is at the topmost point should be greater than zero.
Let the velocity of the ball at topmost point be $v$ and tension in the thread be $T$.
Centripetal force will also be acting in the downward direction.
Thus
$T+mg=\frac{m{v}^2}{l}$
$\Rightarrow T=\frac{m{v}^2}{l}-mg$
Condition for ball to complete the circle, $T>0$
$\Rightarrow \frac{m{v}^2}{l}-mg>0$
$\Rightarrow \frac{v^2}{l}>g$
$\begin{array}{}v>\sqrt{gl}& \text{(1)}\end{array}$
Hence minimum speed ball should have at the topmost point is $\sqrt{gl}$.
Now the change in kinetic energy when ball moves from bottom most point to the topmost point will be,
$\begin{array}{}\Delta K.E=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2& \text{(2)}\end{array}$
Work done by the gravity force when ball moves from bottom most point to the topmost point will be,
$W=-mg\times 2l$
(negative sign indicates that the direction of displacement and gravity force is opposite)
In the inertial frame of reference using work-energy theorem,
$\Delta K.E=W$
$\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=-2mgl$
Substituting the value of $v$ from (1),
$\frac{1}{2}(gl-u^2)=-2gl$
Solving the above equation, we get
$u=\sqrt{5gl}$

For detailed solution watch next video.