# Relative Speed of Two Masses Connected by a Rod

1. Oct 28, 2014

### B3NR4Y

1. The problem statement, all variables and given/known data
The system shown in consists of two balls A and B connected by a thin rod of negligible mass. Ball A has three times the inertia of ball B and the distance between the two balls is ℓ. The system has a translational velocity of v in the x direction and is spinning counterclockwise at an angular speed of ω=2v/ℓ.
Determine the ratio of the instantaneous speeds of the two balls vA/vB at the moment shown.
The diagram is below.

2. Relevant equations
I don't know any...

3. The attempt at a solution
Since ball A is travelling in the direction of motion, I added the angular speed and translational speed.
Since ball B is travelling in the opposite direction, I subtracted angular from translational.

$v_{a} = \frac{2v+vℓ}{ℓ}$ and $v_{b} = \frac{vℓ-2v}{ℓ}$ and then when I divided vA by v B , I get $\frac{2v+vℓ}{vℓ-2v}$. Not the right answer. The hint I was given by my TA was to get the both in the same units, so I converted angular speed to translational speed and got v as the answer, which is obviously no help because I get 0 in the denominator. Not good...

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2. Oct 28, 2014

### Simon Bridge

This is not true - you have had 74 (at time or writing) other posts and some of them have relevant equations in them.
You should have some relative velocity equations and notes somewhere.

... with respect to what?
This is important because it's a relative velocity problem.

... here, let's see if I can help:
$$v_{a} = \frac{2v+vℓ}{ℓ},\qquad v_{b} = \frac{vℓ-2v}{ℓ},\\ \implies \frac{v_a}{v_b} = \frac{2v+vℓ}{vℓ-2v}$$
... that what you meant?
Doesn't make any sense to me - what was your reasoning behind all that?

The hint from the TA is OK as it goes - but to get the linear velocity from the angular one, you need to know the center of rotation.

3. Oct 28, 2014

### B3NR4Y

And the moment I posted this I figured out the solution. I forgot everything I knew about the problem, and worked from scratch and got an answer of 3. Looking at your post, however, I you would have been massive help. Thank you for your potential help, U(x) ;)

75 posts now.

4. Oct 28, 2014

### Simon Bridge

Working from scratch (just using physics) is the way to go. Well done.