# Terminal velocity problem

1. Apr 16, 2006

### Punchlinegirl

A bar of negligible resistance and mass of 17 kg is pulled horizontally across frictionless parallel rails, also of negligible resistance, by a massless string that passes over an ideal pulley and is attached to a suspended mass of 190 g. The uniform magnetic field has a magnitude of 600 mT, and the distance between the rails is 33 cm. The rails are connected at one end by a load resistor of 34 m$$\omega$$.What is the magnitude of the terminal velocity reached by the bar? Answer in units of m/s.

First I drew a free body diagram on the weight, and found that $$F_g= Mg= F_m= ILB$$
so I= Mg/LB
Then I used Ohm's Law.
I= E/R= $$1/R * d\phi /dt$$
and flux= BA
so $$d\phi/dt$$ = B (dA/dt)= BLV
then I= BLV/R , where V is the terminal velocity
so Mg/LB = BLV/R
Solving for V gives MgR/L^2 *B^2.
Then I plugged in my numbers
V= (.19)(9.8)(.034)/(.33^2)(.6^2)
and I got 1.61 m/s
The second part is: What is the acceleration when the velocity v= 1.1 m/s?

2. Apr 17, 2006

### Andrew Mason

Where does the 17kg mass of the bar fit into this? The dropping weight provides the force that accelerates both itself and the bar.

AM

3. Apr 17, 2006

### Punchlinegirl

so would I just do
1.1= (.19+17)(.034)g/(.33^2)(.6^2) ?
Thats the only place that I think the mass can go.

4. Apr 17, 2006

### Andrew Mason

You are using a condition for v that applies only where the acceleration is 0.

Go back to the beginning:

$$Mg - T = Ma$$ and

$$T - F_m = m_{bar}a$$ so:

$$Mg - F_m = (m_{bar} + M)a$$

so:

$$a = (Mg - F_m)/(m_{bar} + M)$$

so:

$$a = (Mg - B^2L^2v/R)/(m_{bar} + M)$$

Just plug in the numbers.

AM