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?
angular momentum of the entire system (rod and particle) around the pivot point is the same before and after. There are no external torques about that point. (Of course before the crash it is just m Vo times distance d/3 between pivot point and crash point.)
To solve this problem, we need to use conservation of linear momentum and conservation of angular momentum.
a. To find the magnitude of the angular velocity of the rod+small mass system after the collision, we can use conservation of angular momentum. Before the collision, the angular momentum is zero since the system is at rest. After the collision, the angular momentum is given by the equation:
Iω = mvr,
where I is the moment of inertia of the rod+small mass system, ω is the angular velocity, m is the mass of the small point mass, v is its velocity, and r is the distance of the collision point from the pivot.
The moment of inertia of the system can be calculated as the sum of the moment of inertia of the rod and the moment of inertia of the small point mass:
I = (1/3)Md^2 + mr^2,
where M is the mass of the rod, d is the length of the rod, and r is the distance of the collision point from the pivot.
Since M = 4m, and r = d/3, the equation becomes:
I = (1/3)(4m)d^2 + md^2 = (4/3)md^2 + md^2 = (16/3)md^2.
Substituting this into the equation for angular momentum:
(16/3)md^2ω = mvr,
ω = vr / ((16/3)d^2).
Substituting the given values, the magnitude of the angular velocity is:
ω = (v0 * (d/3)) / ((16/3)d^2),
ω = (v0 / 16d).
b. To find the speed of the center of mass of the rod right after the collision, we can use conservation of linear momentum. Before the collision, the total momentum is zero since the system is at rest. After the collision, the total momentum is given by:
Mvcm = mv,
where M is the mass of the rod, vcm is the speed of the center of mass, m is the mass of the small point mass, and v is its velocity.
Since M = 4m, the equation becomes:
4mvcm = mv.
Simplifying, we find:
vcm = v/4.
Substituting the given value of v, the speed of the center of mass of the rod after the collision is:
vcm = v0/4.