# Calculate velocity

1. Oct 23, 2015

### Granger

1. The problem statement, all variables and given/known data
A wheel drive in the horizontal plane with angular velocity 13 rad / s. Two bodies mass mA = 1 kg and mB = 1.4 kg slide without friction along a rail fixed to the disk and passing through its center. The two bodies are connected by a bar inextensible (negligible mass) and 0.3m long.

Knowing that the body A is initially 15 cm from the center, determine the initial radial velocity that should print the body so that his movement tends to break even.

2. Relevant equations

aA = xA w^2 - (mB l w^2) / (mA+mB)
xA(eq) = (mBl/(mA+mB))

3. The attempt at a solution
I determine the equations above, but I don't how to use them to calculate what I need to...
Te correct answer is 0,325 m/s

2. Oct 23, 2015

### haruspex

You need to write your first equation as a differential equation and solve it. Then use the information that x is bounded as time tends to infinity.

3. Oct 24, 2015

### Granger

Hi!
So I did it and I found out that

rA = C1 e^-wt + C2 e^wt
rA (t=0) = C1 + C2
vA (t=0) = -C1 w + C2 w

we need C2=0 so I told that v (t=0) was also -wC1 -wC2

Then I wrote

rA = C1 e^-wt
vA = -wC1 e^-wt

rA(t=0) = r0 = C1

So

rA = r0 e^-wt
vA = -r0 w e^-wt

and then vA (t=0) - 1,95 m/s

But theres something I'm missing because the result is wrong... There's something about rAeq that I'm not guessing.... Can you please help me?

4. Oct 24, 2015

### haruspex

How are you defining rA?

5. Oct 24, 2015

### Granger

What do you mean?
rA = r0 e^-wt

6. Oct 24, 2015

### haruspex

That's an equation, not a definition. What does you variable rA represent? If it represents the distance of mass A from the axis then your differential equation solution is clearly wrong. It would imply that the mass reaches the axis at infinite time.

7. Oct 24, 2015

### Granger

Oh yes now I'm understanding I guess... Thanks! So How can I fix the solution? Can you explain?

8. Oct 24, 2015

### haruspex

Please post the differential equation you obtained and the steps up to the solution you posted.

9. Oct 25, 2015

### Granger

rA = C1 e^-wt + C2 e^wt
rA (t=0) = C1 + C2
vA (t=0) = -C1 w + C2 w

we need C2=0 so I told that v (t=0) was also -wC1 -wC2

-C1w +C2w = -wC1 - wC2
C2 = 0

Then I wrote

rA = C1 e^-wt
vA = -wC1 e^-wt
vA(t=0)= -C1w - C2w
rA(t=0) = r0 = C1

So

rA = r0 e^-wt
vA = -r0 w e^-wt

vA (t=0) = - r0 w e^-wt

and then vA (t=0) - 1,95 m/s

10. Oct 25, 2015

### haruspex

No, I asked for the differential equation. Your very first line there is an incorrect solution to the equation.

11. Oct 25, 2015

### Granger

Ah

aA - w^2 rA = - w^2 (req - X)

if X = rAeq

aA - w^2rA = 0

12. Oct 25, 2015

### haruspex

We're not looking for the equilibrium position. We need the general equation of motion of this system. Consider forces.

13. Oct 25, 2015

### Granger

I'm really not understanding what i need to do

14. Oct 25, 2015

### haruspex

Consider the forces acting on the system and write the equation for the resulting acceleration along the rod.
E.g. let x be the distance of mass A from the axis. Find an equation relating $\ddot x$ to x, $\omega$, length of rod L, and the two masses.

Edit: it's the first of your two 'relevant equations'