Question

A composite rod of mass '2m' $\&$ length '2l' consists of two identical rods joined end to end at P. The composite rod is hinged at one of its ends and is kept horizontal. If it is released from rest, ·Find its angular speed when it becomes vertical

• A. $\sqrt{\frac{3}{2}\frac{g}{l}},\sqrt{\frac{g}{2l}}$
• B. $\sqrt{\frac{3g}{2l}},\sqrt{\frac{3g}{2l}}$
• C. $\sqrt{\frac{3g}{2l}},\sqrt{\frac{g}{l}}$
• D. $\sqrt{\frac{3g}{2l}},\sqrt{\frac{g}{3l}}$

Answer 1

The correct option is B

$\sqrt{\frac{3g}{2l}},\sqrt{\frac{3g}{2l}}$

(a)

Suppose that the composite rod acquired an angular speed ω when it reaches its vertical position. Its centre of mass G moves from $G_{1}to{G}_2$. Therefore, the potential energy of the composite rod decreases by 2mgh where h = l.

Applying law of conservation of energy, $\left(\frac{1}{2}\right)l\omega =2mgl$

Where I = M.I. of the composite rod about O

=$\frac{\left(2m\right)(2l{)}^2}{3}=\frac{8m{l}^2}{3}$

$\Rightarrow \omega =\sqrt{\frac{3g}{2l}}$

(b)

Referring to fig. we can see that, just at the vertical position, during the break, the weights of the component rods 1 $\&$ 2, the reaction force R at the pivot and the reaction forces N at the joint of the rods, pass through the pivot O. Therefore these forces can not produce any moment about O; that means the rods do not experience any horizontal force and torque during breaking. At vertical position, the angular momentum of the system about O remains same as just after the break-up

We can also argue that, the angular momentum of each rod remains constant just before and after breaking. The radial forces cannot produce any moment about the centre of mass of the rods 1 $\&$ 2 in the vertical position. The linear velocity of the c.m. of the rods remains constant.