Question

Find the maximum angular displacement of the rod if a particle of mass $m$ moving with velocity $6\text{m/s}$ sticks to it after collision.

• A. ${63}^{\circ }$
• B. ${60}^{\circ }$
• C. ${53}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is A ${63}^{\circ }$

On applying law of conservation of angular momentum before and after collision,

$L_i=L_f$

Substituting the expression for initial and final angular momentum,

$\Rightarrow mv{r}_{\perp }=I\omega$

$\Rightarrow mv{r}_{\perp }=(\frac{2m{l}^2}{3}+m{l}^2)\omega$

$\Rightarrow 6\times 1=(\frac{2\times 1^2}{3}+1^2)\omega$

$\Rightarrow \omega =\frac{18}{5}\text{rad/s}$

Now, applying law of conservation of mechanical energy between vertical position and maximum angular displacement,

$M.E_i=M.E_f$

Substituting the values with the help of figure,

$\Rightarrow \frac{1}{2}I{\omega }^2=2mg\frac{l}{2}(1-\cos \theta )+mgl(1-\cos \theta )$


$\Rightarrow \frac{1}{2}(\frac{2m{l}^2}{3}+m{l}^2){\omega }^2=2mg\frac{l}{2}(1-\cos \theta )+mgl(1-\cos \theta )$

Substituting the values of $m$ and $l$,

$\Rightarrow \frac{1}{2}(\frac{2\times 1^2}{3}+1^2){\omega }^2=2\times 10\times \frac{1}{2}(1-\cos \theta )+10\times 1(1-\cos \theta )$

Substituting the value of $\omega$,

$\Rightarrow \frac{5}{6}\times {\left(\frac{18}{5}\right)}^2=2(10-10\cos \theta )$

$\Rightarrow 20\cos \theta =\frac{46}{5}$

$\Rightarrow \cos \theta =\frac{23}{50}=0.46$

Since, $\cos {60}^{\circ }=0.5$, we can deduce that $\theta ={63}^{\circ }$ will be correct.

Hence, option (a) is the correct answer.