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  • A bob of heavy mass $m$ is suspended by a light string of length $l$. The bgb is given a horizontal velocity $v_0$ as shown in figure.

Q.19) A bob of heavy mass $m$ is suspended by a light string of length $l$. The bgb is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point $P$ making an angle $\theta$ from the horizontal, the ratio? the speed $v$ of the bob at point $P$ to its inttial speed $v_0$ is:

A) $\left(\frac{\sin \theta}{2+3 \sin \theta}\right)^{1 / 2}$
 

B) $(\sin \theta)^{1 / 2}$
 

C) $\left(\frac{1}{2+3 \sin \theta}\right)^{1 / 2}$
 

D) $\left(\frac{\cos \theta}{2+3 \sin \theta}\right)^{1 / 2}$

 

Answers (1)

best_answer

Using energy conservation:

$$
v_0^2=v^2+2 g l(1-\cos \theta)
$$


From radial force condition at point P (when string just goes slack):

$$
\frac{m v^2}{l}=m g \cos \theta \Rightarrow v^2=g l \cos \theta
$$


Substitute (2) into (1):

$$
v_0^2=g l \cos \theta+2 g l(1-\cos \theta)=g l(2+\cos \theta-2 \cos \theta)=g l(2+3 \sin \theta)
$$


So,

$$
\frac{v^2}{v_0^2}=\frac{g l \cos \theta}{g l(2+3 \sin \theta)}=\frac{\cos \theta}{2+3 \cdot \downarrow} \Rightarrow \frac{v}{v_0}=\left(\frac{\cos \theta}{2+3 \sin \theta}\right)^{1 / 2}
$$
Answer: Option (4).

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Dimpy

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