Conservation of Angular Momentum

[Lomion]

A double-pendulum, initially in a vertical position, is attached to a pin. The distance from the mass m (3.2 kg) at the top of the pendulum to the pin is 0.2 m, the distance from the mass m (3.2 kg) at the bottom of the pendulum to the pin is 0.4 m. A bullet of mass 0.05 kg is fired into the bottom mass at 300 m/s, 20 degrees up from the horizontal. What is the maximum angle that the pendulum reaches before it swings back?

EDIT: I forgot this info. The bullet becomes embedded in the bottom mass after it hits. And the initial ω=6 rad/s (counter-clockwise)

I'm not too sure how to approach this problem. I used the principle of Conservation of Angular Momentum and figured out ω=-2.78 rad/s after the impact. (Hpendulum + Hbullet = Htotal)

I know that ω=0 rad/s when the pendulum reaches the maximum angle. However, I'm not sure how to calculate that angle, since if ω is 0, then H=0?

I think that maybe I should use energy methods, but I'm not sure how to do so exactly.

Any help would be appreciated!

 

[gnome]

Why not try using linear momentum. Using the angular momenta, you can calculate the linear momentum of each of the masses immediately after the collision. From that you can calculate their velocities. And that will give you the kinetic energy immediately after the collision. When the rotation stops, all of the kinetic energy will have been converted to potential, which will give you heights of the two masses, and that in turn will give you the angle. Remember that the initial potential energy is NOT zero -- the masses start out at different heights.

Also, I think you need to re-check your calculation of the angular momentum. I'm getting a different result.

 

[M98Ranger]

The Conservation of Angular Momentum states that the total angular momentum of a system remains constant unless acted upon by an external torque. In this case, the system consists of the double-pendulum and the bullet, and the external torque is the force of the bullet hitting the bottom mass.

To solve this problem, we can use the principle of Conservation of Angular Momentum, as you have already done. However, there is a missing piece of information in the problem - the initial angular velocity of the pendulum. Without this, we cannot accurately calculate the maximum angle that the pendulum will reach.

Assuming that the initial angular velocity is zero, we can use energy methods to solve for the maximum angle. Initially, the system has potential energy due to the height of the pendulum and kinetic energy due to the initial angular velocity of the pendulum. After the bullet hits and becomes embedded in the bottom mass, the system will have a different amount of potential and kinetic energy. We can equate the initial and final energies to solve for the maximum angle.

Using the conservation of energy equation, we have:

Initial energy = Final energy

(0.5 * I * ω^2) + (mgh) = (0.5 * I * ω'^2) + (mgH)

Where I is the moment of inertia of the pendulum, ω is the initial angular velocity, ω' is the final angular velocity, m is the mass of the pendulum, g is the acceleration due to gravity, h is the initial height of the pendulum, and H is the maximum height reached by the pendulum.

Since the bullet becomes embedded in the bottom mass, the moment of inertia, I, will change. We can use the parallel axis theorem to calculate the new moment of inertia:

I' = I + md^2

Where m is the mass of the bullet and d is the distance from the bottom mass to the pin. Substituting this into the energy equation, we have:

(0.5 * (I + md^2) * ω^2) + (mgh) = (0.5 * (I + md^2) * ω'^2) + (mgH)

Solving for H, we get:

H = h + (I/m) * (ω^2 - ω'^2)

Plugging in the given values, we get:

H =