What is the value of vp for a particle ejected from a rotating rod?

[snoggerT]

The figure below is of an overhead view of a thin uniform rod of length 0.790 m and mass M rotating horizontally at angular speed 22.0 rad/s about an axis through its center. A particle of mass M/3.00 attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particle's speed vp is 6.00 m/s greater than the speed of the end of the rod just after ejection, what is the value of vp?

Li=Lf, L=Iw, l=rmv

The Attempt at a Solution

- I haven't been able to attempt this problem because I'm having problems setting it up. I'm trying to set up the initial side right now and I am using 1/12ML^2 for the rod, but unsure what to use for the particle since it's inside of the rod. I think I can solve the problem if I can get past this step.

 

[learningphysics]

Use conservation of angular momentum. The angular momentum of the particle about the center is mvr

 

[ogb p]

I would suggest using the conservation of angular momentum to solve this problem. The initial angular momentum of the system is equal to the final angular momentum, which can be represented as Li = Lf. The initial angular momentum can be calculated using the formula L = Iw, where I is the moment of inertia and w is the angular velocity.

For the initial side of the equation, you are correct in using 1/12ML^2 for the rod's moment of inertia. As for the particle, since it is attached to the end of the rod, its moment of inertia can be calculated as I = (1/3)M(L/2)^2 = (1/12)ML^2. This is because the particle's distance from the axis of rotation is half the length of the rod.

Now, for the final side of the equation, we need to consider the change in angular momentum due to the ejection of the particle. The particle's initial angular momentum is zero since it is attached to the rod. However, after ejection, its angular momentum will be Mvp, where vp is the particle's linear velocity. The rod's final angular momentum will be (1/12)ML^2 * 22, as it is still rotating at the same angular velocity.

Setting the initial and final sides of the equation equal to each other, we get (1/12)ML^2 * 22 = (1/12)ML^2 * 22 + Mvp. Solving for vp, we get vp = (1/12)ML^2 * 22 / M = (1/12)L^2 * 22. Plugging in the given values of M = 3M/3 and L = 0.790 m, we get vp = 6.00 m/s, which is the particle's speed.