A uniform rod of mass M and length d is initially at rest on a horizontal and frictionless table contained in the xy plane, the plane of the screen. The rod is free to rotate about an axis perpendicular to the plane and passing through the pivot point at a distance d/3 measured from one of its ends. A small point mass m, moving with speed v0, hits the rod and stick to it at the point of impact at a distance d/3 from the pivot.
a. If the mass of the rod is M=4m, what is the magnitude of the angular velocity of the rod+small mass system after the collision?
b. Using again M=4m. What is the speed of the center of mass of the rod right after collision?
I know I need conservation of angular momentum, but how do I set up the equation -- i.e what is the initial moment of inertia expression, and what is the final?
To solve this problem, we can use the conservation of angular momentum. In this case, the initial angular momentum of the system is equal to the final angular momentum after the collision.
Let's calculate the initial and final angular momenta:
a. Initial Angular Momentum (before the collision):
The rod is initially at rest, so its initial angular momentum is zero.
b. Final Angular Momentum (after the collision):
After the collision, the small mass m sticks to the rod. The system now consists of the rod and the small mass rotating together about the pivot point.
The moment of inertia of a rod rotating about an axis perpendicular to it and passing through its center of mass is given by:
I_rod = (1/12) * M * L^2
Since the axis of rotation is passing through the pivot point, which is at a distance d/3 from one end, we need to use the parallel axis theorem to calculate the moment of inertia about this axis. The parallel axis theorem states that the moment of inertia about an axis parallel to and at a distance d from an axis through the center of mass is given by:
I = I_cm + M * d^2
Using the parallel axis theorem, the moment of inertia of the rod about the pivot point is:
I_rod_pivot = (1/12) * M * L^2 + M * [(2d/3)^2]
The moment of inertia for the small mass m about the pivot point is:
I_small_mass = m * [(d/3)^2]
The final angular momentum is given by the sum of the angular momenta of the rod and the small mass:
L_final = I_rod_pivot * ω + I_small_mass * ω
Since the small mass sticks to the rod, they have the same angular velocity ω after the collision. We can substitute the expressions for I_rod_pivot and I_small_mass into the equation:
L_final = [(1/12) * M * L^2 + M * (4d^2/9)] * ω + m * (d^2/9) * ω
Now, equating the initial angular momentum to the final angular momentum, which is zero, we have:
0 = [(1/12) * M * L^2 + M * (4d^2/9)] * ω_final + m * (d^2/9) * ω_final
Simplifying this equation, we can solve for the magnitude of the angular velocity ω_final.
For part a of the question, with M = 4m, substitute the given values and solve for ω_final.
For part b, we need to find the speed of the center of mass of the rod right after the collision. The center of mass speed is given by the formula:
v_cm = R * ω
where R is the distance of the center of mass from the pivot point. Since the center of mass of the rod is in the middle, R = (d/2).
Substitute the values of ω_final and R into the equation to find the speed of the center of mass of the rod right after the collision.