Question

Find the angle $\theta$ between the thread and the vertical at the moment when the total acceleration vector of the sphere is directed horizontally.

• A. $\theta ={53.7}^{\circ }$
• B. $\theta ={55.7}^{\circ }$
• C. $\theta ={54.7}^{\circ }$
• D. $\theta ={45.7}^{\circ }$

Answer 1

Answer: (C)

$\theta ={54.7}^{\circ }$

Let the thread make angle $\theta$ with the vertical at time $t$.

Using the conservation of mechanical energy, $E_i+U_i=E_f+U_f$

so, $0+mgL=\frac{1}{2}m{v}^2+mgL(1-\cos \theta )$ where L is the length of thread.

$\Rightarrow v^2=2gL\cos \theta$

Centripetal acceleration at this moment is $a_n=\frac{v^2}{L}=2g\cos \theta ....\left(1\right)$

Tangential acceleration at this moment, $m{a}_t=mg\sin \theta \Rightarrow a_t=g\sin \theta ...\left(2\right)$

If $\varphi$ is the angle between normal acceleration and the total acceleration then,

$\tan \varphi =\frac{a_t}{a_n}=\frac{g\sin \theta }{2g\cos \theta }=\frac{1}{2}\tan \theta ...\left(3\right)$

When total acceleration is directed horizontally, $\varphi +\theta =90$

$\tan \varphi =\tan (90-\theta )=\cot \theta ....\left(4\right)$

$\left(3\right)/\left(4\right),1=\frac{1}{2}{\tan }^2\theta$

$\tan \theta =\sqrt{2}\Rightarrow \theta ={54.7}^o$