Question

A ball suspended by a thread swings in a vertical plane so that its acceleration values in the extreme and the lowest position are equal. Find the thread deflection angle in the extreme position (approximate).

• A. ${53}^{\circ }$
• B. ${37}^{\circ }$
• C. ${68}^{\circ }$
• D. ${22}^{\circ }$

Answer 1

The correct option is A ${53}^{\circ }$

The ball has only normal acceleration at the lowest position and only tangential acceleration at any of the extreme position. Let $v$ be the speed of the ball at its lowest position and $l$ be the length of the thread, then according to the problem
$\frac{v^2}{l}=g\sin \alpha$ (1)
where $\alpha$ is the maximum deflection angle
From Newton's law in projection form: $F_t=m{w}_t$
$-mg\sin \theta =mv\frac{dv}{ld\theta }$
or, $-gl\sin \theta d\theta =vdv$
On integrating both the sides within their limits.
$-gl{\int }_0^{\alpha }\sin \theta d\theta ={\int }_v^{0}vdv$
or, $v^2=2gl(1-\cos \alpha )$ (2)
Note: Equation (2) can easily be obtained by the conservation of mechanical energy of the ball in the uniform field of gravity.
From equations (1) and (2) with $\theta =\alpha$
$2gl(1-\cos \alpha )=lg\sin \alpha$
or
$sin\alpha =2[1-cos\alpha ]=2[1-1+2si{n}^2\frac{\alpha }{2}]$
or
$2sin\frac{\alpha }{2}\cos \frac{\alpha }{2}=4si{n}^2\frac{\alpha }{2}$
or
$cot\frac{\alpha }{2}=2$
or
$\alpha =2co{t}^{-1}2$
so, $\alpha ={53}^{\circ }$