Name: Inertial Frames of Reference
Uploaded: 2019-07-19
Duration: 7 min 10 s
Description: Imagine if you are on a bus or a train, can you say you are in an inertial frame of reference? If yes, how can you predict it? Let us explore more in this vi...

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 that point? What will be the tension of the thread at the moment it will be passing the horizontal position?

Accepted Answer

Here, according to the question, the point of suspension be shifted with velocity $v_{A}$ in the horizontal direction towards the left. Then, in the frame of the point of the suspension, the ball starts with the same velocity horizontally towards the right. Let us work in this frame.
From Newton's second law at the uppermost point (say $B$ ),

$m g + T = \frac{m v_{B}^{2}}{l}$
or $T = \frac{m v_{B}^{2}}{l} - m g . . . . . . (1)$

The condition required to complete the vertical circle is that $T \geq 0 . . . . (2)$
But,

$\frac{1}{2} m v_{A}^{2} = m g (2 l) + \frac{1}{2} m v_{B}^{2}$ ( Assuming point $A$ as reference, zero height/potential energy )
So, $v_{B}^{2} = v_{A}^{2} - 4 g l . . . . . . (3)$

From $(1), (2)$ and $(3)$ ,

$T = \frac{m (v_{A}^{2} - 4 g l)}{l} - m g \geq 0$
or $v_{A} = \sqrt{5 g l}$ .

Thus, $v_{A (\min)} = \sqrt{5 g l}$ .

From the equation $F_{n} = m w_{n}$ at point $C$ ,

where $C$ is the point at same horizon of the centre.

Normal acceleration, $w_{n}$ and $v_{c}$ is the velocity at point $C$ .

$∴ w_{n} = \frac{{v^{2}}_{c}}{l}$ (Normal or centripetal acceleration.)

$T^{'}$ be the tension in the string at the moment when the ball crosses $C$ .

As $T^{'}$ provides the necessary centripetal force,

$T^{'} = \frac{m {v_{c}}^{2}}{l} . . . . . (4)$

Again from the energy conservation, between points $A$ and $C$ ,

Assuming bottom-most point $A$ as reference (zero height / zero potential energy),

$\frac{1}{2} m v_{A}^{2} = \frac{1}{2} m v_{c}^{2} + m g l . . . . . . (5)$

From $(4)$ and $(5)$ ,

$T = 3 m g$ .

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