Classic Angular momentum problem

[TheOriginalTy]

Homework Statement

Assume: This collision takes place in outer
space where there is no gravitational field and no friction.
A small piece of putty of mass 31 g and negligible size has a speed of 2.5 m/s. It makes a collision with a rod of length 3 cm and mass 83 g (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.

After the collisions the center-of-mass has a linear velocity V and an angular velocity ω about the center-of-mass “+ cm”.
What is the velocity V of the center-of-mass of the system after the collision?
Answer in units of m/s.

Homework Equations

Icm = 1/12 MR^2
Ipm = MR^2

The Attempt at a Solution

First I said that that the moment of inertia for the system is
I=1/12(M+m)R^2 ("M" being the mass of the rod and "m" being the mass of the putty)

Then using conservation of angular momentum i got:

mR^2w = (R^2/12)(M+m)w

substituting the respect v's (and Rs) in for w i got:
mv1R = (R/12)(M+m)v2

then

v2 = 12mv1/(M+m)

plugging in my numbers i got [12(31)(2.5)]/[(31+83)] = 8.16 m/s
Is this correct or did I leave something out? The answer doesn't look right to me just because it's 3 times bigger than the original velocity, but I'm not sure at all.

Feedback?

 

[tiny-tim]

Welcome to PF!

Hi TheOriginalTy! Welcome to PF!

(have an omega: ω and try using the X² tag just above the Reply box )

I=1/12(M+m)R^2 ("M" being the mass of the rod and "m" being the mass of the putty)

Nooo … m is a lump, not a rod!

 

[TheOriginalTy]

Hi TheOriginalTy! Welcome to PF!

(have an omega: ω and try using the X² tag just above the Reply box )


Nooo … m is a lump, not a rod!

Once it sticks to the rod doesn't the moment of inertia of the rod include the mass of the putty lump as well? Considering that it said that size is negligible so it doesn't change the shape of the rod.

 

[tiny-tim]

Considering that it said that size is negligible so it doesn't change the shape of the rod.

It does change the shape of the rod!

The rod used to be uniform, now it's got a lump on the end!

Only if the putty was spread out over the whole rod could you just add the masses.

1/12 MR² is only for a uniform rod.

 

[TheOriginalTy]

Okay thanks. I understand that now. I got 0.6798 as the velocity and that's correct. but now I have another problem.

I need to find the angular speed and the angular momentum right after the collision.

for angular speed i used that w= v/r and had 0.6798(m/s) /.015 (m) = 45.32 (rad/s)

But this is not correct. help?


Also. for angular momentum I got L= I*v/r and had (1/12M + M)*R*v

> = (.03792)(.015)(.6798) = .0003866 kg*m²/s

This is also wrong. Help?

 

[tiny-tim]

Hi TheOriginalTy!

(just got up :zzz: …)

for angular speed i used that w= v/r and had 0.6798(m/s) /.015 (m) = 45.32 (rad/s)

(what happened to that ω i gave you? )

no, ω = v/r only applies where one point is fixed, and another point a distance r away has speed v

in this case, v is the speed of the (new) centre of mass, and goodness-knows where the fixed point (the centre of rotation) is!

You need to find ω some other way.

I need to find … the angular momentum right after the collision.

Sorry, i don't understand the question …

the angular momentum about which point? ​