Angular Momentum and Angular Speed?

Question

Assume no gravity and no friction. A small piece of putty of mass 40 grams with neglible size has a speed of 1.3 m/s. It makes a collision with a rod of length 6 cm and mass 6g ( initially at rest) such that the putty hits the very end of the rod. The putty sticks to the end of the rod and spins around after the collision. The rod has a moment of inertia of (1/12)* m *r^2.

A.) What is the angular momentum of the system relative to the center-of-mass after the collision. Answer in kg*m^2/s

B.) What ist he system's angular speed about the center-of-mass after the collision. Answer in rad/s

C.) What is the percent ratio of Ef / Ei % of the energy of the system after the collision to the energy of the system before the collision.

Each question is based on the previous and I am stuck on A. From linear momentum conservation I get the final mass to be going .509804 m /s.I then converted this to angular speed by V/R, to be .509/.06 = 8.496. Then plugged into L=I*angV, but this didn't work

Yeah sorry, I am stuck on B, which is leaving me stuck on A.

How did you find the distance of the center of mass from the end of the Rod?

Correction the mass of the rod is 62 g

Answer 1

[Edit: Noticed the new value for the mass of the rod - also had to correct a mistake I made for the moment of inertia of the rod, see revised calculations below. I'd thought the r in (1/12)mr^2 was half the length of the rod, but when I decided to check that formula I realised it was the full length of the rod.]

You say you are stuck on A, but it sounds like you're trying to do B first. It won't work the way you suggest; ω = v/r only applies where r is the distance from the moving point to the axis of rotation, and v is the perpendicular velocity. Here different parts of the object are at different distances, and the velocity of an individual point is difficult to write an expression for because of the rotation.

Note that angular as well as linear momentum is conserved. So all you need for A is the angular momentum of the system at the point of collision, about the centre of mass of the combined system.

To find the centre of mass of the combined system, measuring from the end of the rod where the putty is we have a 40g mass at 0cm and effectively a 62g mass at 3cm. So the centre of mass is at (40×0 + 62×3)/(40+62) = 31/17 cm from the end of the rod. So the putty is 31/17 cm on one side of it and the centre of mass of the rod is 3 - 31/17 = 20/17 cm on the other side of it.

Now at the instant of collision we have no angular momentum from the rod, and 0.040(1.3)(0.31/17) = 9.48×10^-4 kg m^2/s from the putty. This is therefore the total angular momentum of the system.

B) The moment of inertia of the rod about the new centre of mass is (1/12)(0.06)^2(0.062) + 0.062(0.20/17)^2
= 2.72×10^-5 kg m^2. The moment of inertia of the putty is 0.04(0.31/17)^2 = 1.33×10^-5 kg m^2. So the total moment of inertia is 4.05×10^-5 kg m^2.
Now L = Iω
<=> 9.48×10^-4 = 4.05×10^-5 ω
<=> ω = 23.4 rad/s.

C) Ef = linear KE + angular KE
= 1/2 (0.102) (1.30*0.04 / 0.102)^2 + 1/2 (4.05×10^-5) (23.4)^2
= 1.33×10^-2 + 1.11×10^-2
= 0.0244 J
Ei = 1/2 (0.04)(1.30)^2 = 0.0338 J
So the ratio is 0.0244/0.0338 * 100 = 72.1%.