How can torque on the rotating wheel be calculated using the given parameters?

In summary, the wheel is spinning and a current is flowing down to the brush. The wheel has a magnetic field, which causes a torque.
  • #1
because789
4
0
FOR PICTURE of CIRCUIT: community.webshots.com/photo/2479536900103460081IWuFAY
A wheel consiting of a large number of thin conducting spokes is free to rotate about its axle. The brush makes contact one spoke at a time. There is a uniform B field pointing into the page. The wheel is initially at rest.
R = radius of wheel
J = moment of inertia of wheel
L = inductance of circuit
V = voltage

QUESTION:
Find an expression relating ω(t) with I(t) by calculating the torque on the wheel.

ATTEMPT:
I tried using τ = dL/dt and L=Jω to get that τ = J*dω/dt. For the other half of the expression I use τ = r x F. I am unclear how to do that part. Any help would be awesome, thanks!
 
Physics news on Phys.org
  • #2
bump b/c I am still confused
also, picture is at community.webshots.com/photo/2479536900103460081IWuFAY
 
  • #3
You know there's a current going from the center of the wheel down to the brush. I think the inductor is just there to hint at the fact that the current remains constant (dI/dt for a conductor is always finite). Then you can use the force law:

[tex]\vec{F}=I\vec{L} \times \vec{B}[/tex]

In this case, L would be the radius of the spinning thing and points down. Then the torque is:

[tex]\vec{\tau} = \int_0^L{d \vec{r} \times \vec{F}}[/tex]

F is constant here, so this pretty easy. (NOTE: I don't know why, but I keep looking at the above equation, and it seems wrong, but I'm pretty sure it's right. Anyway, if something doesn't work out, maybe I just lied to you above)

That's probably enough to get you started.

NOTE: these are reducible to scalar equations (e.g. F=ILB).
 
  • #4
This might help you but I am also little confused about that:

Divide the thing into two parts (wheel and wire) - two wings of the wheel close the wire connection (so they are just switches that are turned on and off after every few seconds)

And as misho said because V = L.di/dt so the current doesn't make sudden jumps - stay constant

Problem that I am facing:
one force arrow goes up (for the wing that is connected to the top part of the circuit), and one force arrow points left (for the left side wire)
so, torque = 0?
 
  • #5
Ok, when I use the above equations I get a constant torque, but the torque is not constant along the spoke (I got τ = I*B*R^2). Or is that not constant b/c I(t) changes w/ time?
 
  • #6
Well, the current shouldn't change with time. I'm pretty sure about that.
 
  • #7
Problem that I am facing:
one force arrow goes up (for the wing that is connected to the top part of the circuit), and one force arrow points left (for the left side wire)
so, torque = 0?

I think you're confused just because the diagram isn't very good. I think that there is a wire that goes to the middle of the spinner and then the current goes down to the brush (please tell me if I'm wrong about that).
 
  • #8
misho said:
I think you're confused just because the diagram isn't very good. I think that there is a wire that goes to the middle of the spinner and then the current goes down to the brush (please tell me if I'm wrong about that).

You are right - the current travels from the wire through the spoke, and to the brush. Sorry for the crappy diagram.

EDIT: also, i think the current does change with time b/c on a latter part of this question it asks for me to solve for I(t).
 
  • #9
rootX said:
And as misho said because V = L.di/dt so the current doesn't make sudden jumps - stay constant

Yes, V = L di/dt
V is constant, as is L.
But i is not constant, according to the equation.
 
  • #10
because789, you wouldn't happen to be taking a take-home test for a certain Physics 217 course at Cornell, would you? Keep in mind that your email address is linked to your forum account.
 
  • #11
Redbelly98 said:
Yes, V = L di/dt
V is constant, as is L.
But i is not constant, according to the equation.

V = L di/dt <- the V here is the voltage across the inductor, which could be zero. Having said that, dI/dt is not zero, I was wrong about that.

I decided to look at this in terms of conservation of power:

Source power: [tex]P_s (t) = VI(t)[/tex]

Inductor power: [tex]P_L (t) = V_L I_L = L I(t) \frac{dI}{dt}[/tex]

Wheel power: [tex]P_w = \tau (t) \omega (t)[/tex]

At some point, it hit me that if we have a force acting on an object the way it does on a spoke of the wheel here, we would take it to act at the center of mass, so:

[tex]\tau = \frac{1}{2}RF[/tex]

And since the force is magnetic:

[tex]F = RBI(t)[/tex]

So: [tex]\tau = \frac{1}{2}R^2 B I(t)[/tex]

Conservation of power says:

[tex]VI(t) = L I(t) \frac{dI}{dt} + \frac{1}{2}R^2 B I(t) \times \omega (t) [/tex]
[tex]V = L \frac{dI}{dt} + \frac{1}{2}R^2 B \omega (t) [/tex]


Hopefully, I didn't screw anything up this time.
 

1. What is the "Weird Circuit with Wheel"?

The "Weird Circuit with Wheel" is a circuit design that incorporates a wheel or rotating component into its function. It is often used in mechanical or electrical systems to achieve specific tasks or operations.

2. How does the "Weird Circuit with Wheel" work?

The specific function of the "Weird Circuit with Wheel" will vary depending on the design and purpose. However, in general, the wheel or rotating component is used to transfer energy or motion within the circuit, allowing for the desired operation to be carried out.

3. What are the advantages of using the "Weird Circuit with Wheel"?

One of the main advantages of the "Weird Circuit with Wheel" is its versatility. It can be used in a wide range of applications and can be designed to perform different tasks. It also allows for efficient transfer of energy and can help reduce friction and wear in a system.

4. Are there any limitations to the "Weird Circuit with Wheel"?

As with any circuit design, there may be limitations to the "Weird Circuit with Wheel" depending on the specific application. For example, it may not be suitable for high-speed or high-torque operations. Additionally, proper maintenance and care may be required to ensure the longevity of the wheel or rotating component.

5. Can the "Weird Circuit with Wheel" be used in different fields of science?

Yes, the "Weird Circuit with Wheel" can be applied in various fields of science such as physics, engineering, and robotics. It can be used in different types of machines and systems, from simple household appliances to complex industrial equipment.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
4
Replies
121
Views
9K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
Back
Top