Question

A ball suspended by a thread swing in a vertical pane so that its acceleration value in the extreme and the lowest position are equal. Find the thread deflection angle in the extreme position.

• A. $2{\tan }^{-1}2$
• B. $2{\tan }^{-1}\frac{1}{2}$
• C. ${\tan }^{-1}2$
• D. ${\tan }^{-1}\frac{1}{2}$

Answer 1

The correct option is B $2{\tan }^{-1}\frac{1}{2}$

According to figure,

Velocity at extreme positions is zero so, centripetal acceleration is zero.

Now, only tangential acceleration is $g\sin \theta$acts due to $mg\sin \theta$

Now, in the lowest position the object acquires velocities due to h

$h=R-R\cos \theta$

Now, energy conservation between $P_1$ and $P_0$

$mgh=\frac{1}{2}m{v}^2$

$mg(R-R\cos \theta )=\frac{1}{2}m{v}^2$

$2g(1-\cos \theta )=\frac{v^2}{R}$

We know that,

$a_0=\frac{v^2}{R}$

So, $2g(1-\cos \theta )=a_0$

Now, total acceleration at $P_2$

$a_0=2g(1-\cos \theta )$

Now, total acceleration at $P_1$

$a_1=g\sin \theta$

But, $a_1=a_0$

So, $g\sin \theta =2g(1-\cos \theta )$

$\sin \theta =2(1-\cos \theta )$

$\sin \theta =2[1-1+2{\sin }^2\frac{\theta }{2}]$

$2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=4{\sin }^2\frac{\theta }{2}$

$\theta =2{\tan }^{-1}\frac{1}{2}$

Hence, the angle in the extreme position is $2{\tan }^{-1}\frac{1}{2}$