Calculating Torque for Constrained Wheel Rotation

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The discussion centers on calculating the torque for a wheel constrained to rotate about the z-axis, with a force vector acting on its edge. The initial misunderstanding involved incorrectly using the equation Torque = R*F instead of the correct cross-product formula Torque = r x F. Participants clarified that the torque is calculated using the cross product, which results in a vector perpendicular to both the radius and the force. The importance of distinguishing between dot and cross products was emphasized, as well as the need to apply the right-hand rule for direction. Ultimately, the problem was resolved by correctly applying the cross-product, leading to a successful calculation of torque.
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Homework Statement


A wheel of diameter 34.0 cm is constrained to rotate in the xy plane, about the z axis, which passes through its center. A force vector F = (-30.0 i-hat + 39.6 j-hat) N acts at a point on the edge of the wheel that lies exactly on the x-axis at a particular instant. What is the torque about the rotation axis at this instant?

Homework Equations


Torque=R*F

The Attempt at a Solution


Tx= 0
Ty= 0
Tz= 6.732
*EDIT* SOLVED *EDIT*
 
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Hi, Stryder_SW!

What component of the force creates the torque? I can tell you that it's not the vector given.
 
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Hi Stryder_SW! :smile:
Stryder_SW said:
Torque=R*F

Nooo! :cry:

torque = r x F :smile:
 
You have written Torque = R*F but it should be T = R x F. This is important.
 
Um...what part of that is important?
 
Stryder_SW said:
Um...what part of that is important?

Dr.D :smile: and I may be wrong, but we think you're trying to multiply F by the number r,

instead of cross-producting it with the vector r :wink:
 
The x is important.
 
No, you're right. by cross-producting it do you mean like dot product? and even if you did this how is the torque in the Tz/k-hat direction not zero when there is no force applied in that direction
 
Most specifically the cross product is not a dot product. A cross product is a vector, perpendicular to the two original vectors, in the direction given by rotating the first vector to the second vector and applying the right hand rule. The magnitude of the cross product is mag(r x F) = r*F*sin(r,F).
 
  • #10
Stryder_SW said:
No, you're right. by cross-producting it do you mean like dot product? and even if you did this how is the torque in the Tz/k-hat direction not zero when there is no force applied in that direction

still can't tell whether you've got it :redface:

dot product is almost the opposite

have a look at the PF library entry on cross product :wink:

EDIT: oooh, the doctor beat me this time! :biggrin:
 
  • #11
Well I've never seen cross-product before, and the PF library explanation...kinda confuses me. So I looked it up in my physics book, apparently its not discussed until the chapter AFTER the one that this homework is in. I'm going to take a look at the book now and hopefully I'll get somewhere.
 
  • #12
Awesome, the explanation in my book was surprisingly good. Got the problem using the cross-product. And I think I get why that works, anyways thank you very much for your help.
 
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