Homework Help: 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'
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