Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Required torque on a motor

  1. Oct 26, 2005 #1
    A professor of mine has stated that when designing a martian rover, you want it to be able to apply enough torque to the wheels to climb a verticle wall (providing there was an arbitrary normal force).

    I was able to do that, but the problem I was having was that if the torque provided by the motor, exceeds the torque due to friction, the wheel will slip. According to our equation, our wheel will always slip. Since we want m*g*r = m*g*r*u (u=coefficient of static friction). U is almost always a decimal, meaning the friction force will always be less than torque force. What am i missing here. To go up a steeper hill, we need less torque? doesnt make sense.

    I also tried to think of it in terms of energy and power output. Moving along the ground, the force over a distance, or work done by the motor must be more than the work done by friction. Also, when going uphil you also have to fight gravity, so it makes sense that steeper angles require more energy. However, I haven't been able to get much further than that.

    Any help is greatly appreciated.
     
  2. jcsd
  3. Oct 28, 2005 #2
    I dont understand how it can climb a verticle wall, the normal force will be zero unless there is a normally applied force. In which case, the force due to friction will have to balance the force due to gravity. mg = F_fr = N*u

    and the radius of the wheel is r, and the number of wheels are x, then you will have F_fr/x friction force on each wheel, requiring a torque of,

    T = (F_fr)(r)/(x) = (N*u*r)/(x) on each wheel.

    >>Im probably wrong<<

    This is assuming you climb with constant velocity, and are on the verge of slipping. And I dont see how the normal can be "arbitrary". A normal of zero wont work.

    As for an incline, you no longer have mg, but now you have mgsin(theta)=N and mgcos(theta) = F_x. Where theta is measured from the upper slope of the ramp to the normal force. But that case is easy, the maxium angle you can climb will be equal to arctan(mu), and so you will know the maxium torque required to climb given those conditions.

    The wheels are static in contact with the ground, so I dont think its possible to accelerate up an incline greater than this angle, as the relative motion wont be static at the wheels/ramp anymore.

    So as theta increases, the normal force becomes less, and you have less traction, but you require more and more force in the +x direction. So you need more torque as the angle increases, but you cant get it, becasue friction wont allow it.

    (Im calling +x the direction along the ramp, not the horizontal axis.)

    According to your equation, m*g*r = m*g*r*u your taking the moment about some point, but it does not seem correct. If the cars climbing vertically, then there is a torque if you sum the moments about the vertical wall. Mgs causes a torque, where s is the distance from the ceneter of mass to the vertical wall. And since the friction forces act along this wall, they contribute no moment. (Even if they did, they would help to create more moment in the same direction as gravity). So the car is going to try to loose contact with the top wheels first, and rotate around the bottom wheels. This is why you need a normal force to counteract that moment as well. If the car were climbing up a ramp, friction again would cause a moment lifting up the front of the car, which means there would be an increased normal on the back wheels to counter the moment, and thus more traction on the back wheels as well.

    *I probably have alot of errors in what I said* :-(
     
    Last edited: Oct 28, 2005
  4. Oct 28, 2005 #3

    Danger

    User Avatar
    Gold Member

    At first the mention of climbing a vertical wall seemed kinda dumb to me too, but then I got to thinking...
    If the wheels were covered with those nanotube artificial gecko hairs as mentioned in the 'Improving on Nature' thread, vehicles would actually be able to do that.:bugeye:
     
  5. Oct 28, 2005 #4

    russ_watters

    User Avatar

    Staff: Mentor

    Since each wheel has its own motor (I think), individual wheels might be going vertically up a small rock. Since most of the other wheels will be on the ground, the force they can apply to the one or two wheels going up a rock will almost certainly be enough to hold it to the rock. The torque then just needs to be enough to lift the fraction of the rover's weight that the wheel needs to support.
     
  6. Oct 28, 2005 #5

    Danger

    User Avatar
    Gold Member

    I can't help thinking that it would be more complicated than that. If one or more wheels are going vertically up a surface, the rest will have to almost stop until they get over the top, or the thing will just spin around the obstruction and/or spin its other wheels. If the surface is high enough, the whole thing will tip over as the climbing wheels exceed the stability factor. Am I missing something?
     
  7. Oct 29, 2005 #6

    russ_watters

    User Avatar

    Staff: Mentor

    Dunno - you know electric motors can produce a significant torque while stationary (unlike gas/steam engines), right?

    That's not to say I didn't misunderstood the question...
     
  8. Oct 29, 2005 #7

    Danger

    User Avatar
    Gold Member

    I was under the impression that steam engines can exert considerable torque while stationary as well, at least until the steam in the cylinder condenses. Anyhow, that's not the point that I was making. If you work your way up from horizontal travel through increasing ramp angles to vertical, you can see that for the same rpm's a steeper angle results in less horizontal movement. When the incline reaches vertical, forward movement would be zero. The wheels that remain on flat ground would have to quit turning and the whole vehicle twist on its horizontal axis while the wheels in question climb straight up. That would entail a pretty weird suspension system as well, because the horizontal twist would angle the climbing wheels and they'd end up describing an arc that would flip the thing over. You'd have to have steering correction to keep them going straight up, and the thing would still eventually turn turtle.
     
    Last edited: Oct 29, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Required torque on a motor
  1. Torque required (Replies: 1)

  2. Motor Torque (Replies: 0)

Loading...