An analysis of a rotating rod

Relevant wiki: Conservation of Angular Momentum

Conservation of angular momentum ( $\omega$ is the angular frequency immediately after the collision):

$\big(\frac{L}{2}\big) m v = \big(\frac{L}{2}\big)^2 m \omega + \frac{M L^2}{12} \omega \\ \implies \omega = \frac{2mv}{mL + \frac{ML}{3}}$

The gravitational potential energy of the rod is constant, regardless of its orientation. Therefore, the system comes to rest when the system turns around 180 degrees, and then the bead rises high enough to transform all of the initial kinetic energy into gravitational potential energy relative to the start height. In the equation below, the left side gives the initial kinetic energy, and the right side is the potential energy relative to the start height.

$\frac{1}{2} m \big(\frac{L}{2}\big)^2 \omega^2 + \frac{1}{2} \frac{M L^2}{12} \omega^2 = mg \frac{L}{2} sin\theta \\ (\frac{1}{2} m \big(\frac{L}{2}\big)^2 + \frac{1}{2} \frac{M L^2}{12}) \frac{4 m^2 v^2}{(mL + \frac{ML}{3})^2} = mg \frac{L}{2} sin\theta \\ (\frac{mL}{4} + \frac{M L}{12}) \frac{4 m^2 v^2}{(mL + \frac{ML}{3})^2} = mg sin\theta \\ (mL + \frac{M L}{3}) \frac{m^2 v^2}{(mL + \frac{ML}{3})^2} = mg sin\theta \\ \frac{m^2 v^2}{(mL + \frac{ML}{3})} = mg sin\theta \\ \frac{m v^2}{gL(m + \frac{M}{3})} = \frac{3m v^2}{gL(3m + M)} = sin\theta$

The full angle is therefore:

$180 + asin\big(\frac{3m v^2}{gL(3m + M)}\big)$

Relevant wiki: Applying Angular Kinematic Equations

In this solution I have used energy considerations and integral calculus.

First applying $L_{axis}$ $=$ $I_{axis,new}$ $ω_{initial}$ we get $ω_{initial}$ .

Now,The basic gist is that as the rod rotates there is a net external torque due to the "extra" mass of the bullet on one side acts and accelerates the rod from $0$ degrees to $90$ degrees.But the torque acting is variable and is dependent on the angle the rod has turned at a particular instant.So to find the total work done i have simply integrated the torque acting to avoid complex calculations if we would have had applied torque = $I_{axis}$ $\alpha$

Now what happens is that from $90$ degrees to $180$ degrees that same torque decelerates the rod.But then from $180$ degrees to $270$ degrees the torque again opposes the motion thus sometime between $180$ and $270$ degrees,the rod must have come to rest.So, support then oppose then again oppose.This causes it to finally stop.For rigour I have shown that the work done from $0$ degrees to $90$ degrees if work done is $W$ then from $90$ degrees to $180$ degrees work done is $-W$ .Hence net work done from $0$ degrees to $180$ degrees is $0$ .Thus it is moving with the same angular velocity that it has initially acquired.But as stated earlier now from $180$ degrees to $270$ degrees the torque is decelerating so it will finally stop somewhere along the interval $[180,270]$ .Our aim is to find this angle and add it to the original $180$ degrees.already we have got $\alpha$ $=$ $180$ degrees.

This is rigorously shown below:

So the answer is $180 + 3 + 3$ $=$ $186$

Question is original and made by me.Hope you liked it.

Relevant wiki: Rotational Kinetic Energy - Problem Solving

The rod + small bullet have rotational inertia $I = \tfrac1{12}M\ell^2 + m(\tfrac12\ell)^2 = \tfrac1{12}(M + 3m)\ell^2.$ The initial angular momentum of the system comes from the bullet, and is equal to $L = mvr = \tfrac12mv\ell.$ This momentum is conserved; it is transferred to the rod + bullet system. If we wanted, we could solve for the rotational speed $\omega = L/I$ . However, this is not necessary.

Consider now the conservation of energy during the swing. The rod's potential energy is constant, so we only consider the bullet's PE, which is equal to $mgy$ with $y = -\tfrac12\ell\sin\theta$ . We are interested in the situation where the KE of the system becomes zero. Thus $-\tfrac12 mg\ell\sin\theta = \tfrac12I\omega^2 = \tfrac12\frac{L^2}I;$ $\sin\theta = -\frac{L^2}{mg\ell I} = -\frac{(\tfrac12mv\ell)^2}{mg\ell\cdot \tfrac1{12}(M + 3m)\ell^2} = -\frac{3mv^2}{(M+3m)g\ell}.$ Since this $\sin\theta < 0$ and $\theta > 0$ , the first solution will have $\theta$ in the third quadrant; so $\theta = 180^\circ + \text{arc sin}\ \frac{3mv^2}{(M+3m)g\ell}.$ We submit $\alpha + \beta + \gamma = 180 + 3 + 3 = \boxed{186}.$