Challenges in mechanics by Ronak Agarwal(Part 3)

I have assume that the particle collides little bit(consider it as negligible) to the right of the edge.

Let the velocity of the CoM of the rod just after the collision to be $V$ and that of the particle be $v_{0}$ .

Now

The velocity of the CoM of the rod will be in vertical direction because the impulse acts in vertical direction only. As the collision is elastic so $e=1$ . So we can equate the Velocity of approach and velocity of separation in common normal direction.

$V_{app}=v cos\theta$ and $V_{sep}=V+\frac{l\omega}{2}+v_{0} cos\alpha$ ( $\alpha$ is the angle between $v_{0}$ and the common normal direction).

So $v cos\theta=V+\frac{l\omega}{2}+v_{0} cos\alpha$ ................(1)

As no force in common tangent direction so

$v sin\theta=v_{0} sin\alpha$ ..........................(2)

Now equal and opposite impulses act on the particle and the rod.

So this impulse is equal to the change in the velocity of the particle and change in the velocity of the CoM of the rod.

So

$MV=m(v_{0}cos\alpha+vcos\theta)$ .........(3) Be care full with the signs.

Now the impulsive torque which acts on the rod is

${ \tau }_{ imp }=\frac{l}{2}J$

This impulsive torque is equal to the change in the angular momentum of the rod.

So $\frac{l}{2}J=I\omega$ ...................(4)

Now by energy conservation

$\frac { 1 }{ 2 } m{ v }^{ 2 }=\frac { 1 }{ 2 } M{ V }^{ 2 }+\frac { 1 }{ 2 } I{ \omega }^{ 2 }+\frac { 1 }{ 2 } m{ v }_{ 0 }^{ 2 }$ ..........................(5)

On simplifying these equation I got

$V=\frac { 2mvcos\theta }{ 4m+M }$

and $\omega =\frac { 6 }{ l } (\frac { 2mvcos\theta }{ 4m+M } )$

Now I considered a reference frame fixed to the CoM of the rod.

So velocity of the particle wrt the CoM of the rod is

${ v }_{ 0 }sin\alpha \hat { i } -{ (v }_{ 0 }cos\alpha +V)\hat { j }$

In this figure I have drawn the trajectories of the point B and the particle wrt CoM.

In order to collide the particle and point B must have to be at the same position at a particular time.

So time required by the particle to travel AD in horizontal direction and DC in vertical direction is equal to time required by the rod to rotate by angle $\beta$ . Note that $\beta=2\phi$

Considering speed in horizontal direction I got--

$\frac { \frac { l }{ 2 } -\frac { l }{ 2 } cos\left\{ -\left( 180-2\phi \right) \right\} }{ v_{0}sin\alpha } =\frac { 2\phi }{ \omega }$ ............(6)

Considering speed in vertical direction I got--

$\frac { 0.5lsin2\phi }{ { v }_{ 0 }cos\alpha +V } =\frac { 2\phi }{ \omega }$ ........(7)

On simplifying eq(6) and eq7 using eq1 and eq2

I got

$\frac{\pi sin(2\phi)}{4}=\phi$

So $\phi=\frac{\pi}{4}$

and $\frac { { sin }^{ 2 }\phi }{ sin\theta } =\frac { \left( 4m+M \right) \phi }{ 6mcos\theta }$

On putting value in this equation I got

$\theta ={ tan }^{ -1 }\left( \frac { 2 }{ \pi +2 } \right)$ .

We can also use caonservation of Momentum and Angular momentum to find equations. But that will lead us ultimately to the same equation which I have written above.