Particle like balls each of mass m are affixed at both the ends of a light rigid rod of length $l$ . The composite body thus formed in suspended from a nail O, with the help of two inextensible cords affixed on each ball length of each chord is also $l$ . Now arrangement in pulled aside bringing one ball in level with the nail and keeping both the chords straight and then released let $θ$ be the angular position of centre of mass of the system from horizontal at any instant of time, then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

If $T_{1}$ and $T_{2}$ are the tensions in the two threads then $T_{1} - T_{2} = m g \cos θ$

Maximum tension in a cord during subsequent motion = $\frac{m g}{2} (\sqrt{\frac{103}{3}} - \sqrt{3})$

Angular position $θ$ at which tension becomes maximum is given by $\tan θ = \frac{10}{\sqrt{3}}$

Its angular acceleration of the system as a function of $θ$ is $\frac{\sqrt{3}}{2} \frac{g}{l} \cos θ$

answer is A, B, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

At any instant, let $θ$ be the Angular position of center of mass of the system w.r.t horizontal.
Conservation of energy $\Rightarrow 2 m g \frac{\sqrt{3}}{2} l (\sin θ - \sin 30^{0}) = 2 (\frac{1}{2} m l {}^{2}w^{2})$
$\Rightarrow w^{2} = \sqrt{3} \frac{g}{l} (\sin θ - \frac{1}{2})$ ………………1)
$\Rightarrow α = w \frac{d w}{d θ} = \frac{\sqrt{3}}{2} \frac{g}{l} \cos θ$ ……………2)
CM moves in circular path about point O
$(T_{1} + T_{2}) \sin 60^{0} - 2 m g \sin θ = 2 m (\frac{\sqrt{3}}{2} l) w^{2}$ ………………3)
1) in 3) $\Rightarrow T_{1} + T_{2} = \frac{10}{\sqrt{3}} m g \sin θ - \sqrt{3} m g$
Torque about CM
$(T_{1} - T_{2}) \frac{\sqrt{3}}{2} \frac{l}{2} = 2 m {(\frac{l}{2})}^{2} α$
$\Rightarrow T_{1} - T_{2} = m g \cos θ$ ……………..4)
$T_{1}, \max \Rightarrow \frac{d T_{1}}{d θ} = 0,$ then we get $\tan θ = \frac{10}{\sqrt{3}}$
$T_{1}, \max = \frac{m g}{2} [\sqrt{\frac{103}{3}} - \sqrt{3}]$