Question 68: Work, Energy and Power & Circular Motion

Question

A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $$(	heta)$$ of thread deflection in the extreme position will be :

Accepted Answer

Loss in kinetic energy $=$ Gain in potential energy
$$\begin{aligned}
& \Rightarrow \frac{1}{2} \mathrm{mv}^2=\mathrm{mg} \ell(1-\cos 	heta) \
& \Rightarrow \frac{\mathrm{v}^2}{\ell}=2 \mathrm{~g}(1-\cos 	heta)
\end{aligned}$$
Acceleration at lowest point $$=\frac{\mathrm{v}^2}{\ell}$$
Acceleration at extreme point $$=g \sin 	heta$$
$$\begin{aligned}
& 	ext { Hence, } \frac{\mathrm{v}^2}{\ell}=\mathrm{g} \sin 	heta \
& 	herefore \sin 	heta=2(1-\cos 	heta) \
& \Rightarrow 	an \frac{	heta}{2}=\frac{1}{2} \Rightarrow 	heta=2 	an ^{-1}\left(\frac{1}{2}ight)
\end{aligned}$$

Answer

$$	an ^{-1}\left(\frac{1}{2}ight)$$

Answer

$$2 	an ^{-1}\left(\frac{1}{2}ight)$$

Answer

$$2 	an ^{-1}\left(\frac{1}{\sqrt{5}}ight)$$

Answer

$$	an ^{-1}(\sqrt{2})$$

Work, Energy and Power & Circular Motion

A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $(\theta)$ of thread deflection in the extreme position will be :

Choose the correct answer: