Calculating Torque for Constrained Wheel Rotation

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In summary, the torque is created by the force vector F which is in the -30.0 i-hat + 39.6 j-hat direction.
  • #1
Stryder_SW
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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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  • #2
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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  • #3
Hi Stryder_SW! :smile:
Stryder_SW said:
Torque=R*F

Nooo! :cry:

torque = r x F :smile:
 
  • #4
You have written Torque = R*F but it should be T = R x F. This is important.
 
  • #5
Um...what part of that is important?
 
  • #6
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:
 
  • #7
The x is important.
 
  • #8
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
 
  • #9
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.
 

FAQ: Calculating Torque for Constrained Wheel Rotation

What is torque?

Torque is a measure of the rotational force applied to an object. It is calculated by multiplying the force applied by the distance from the axis of rotation.

How is torque represented using i, j, and k-hat?

Torque is represented using the three dimensional vector notation, where i, j, and k-hat represent the x, y, and z components respectively. This notation helps in calculating the direction and magnitude of the torque.

How do you calculate torque using i, j, and k-hat?

To calculate torque using i, j, and k-hat, you need to first determine the force vector and the position vector. Then, you can use the cross product formula (T = r x F) to calculate the torque vector, where r is the position vector and F is the force vector.

What are some real-life examples of torque?

Some real-life examples of torque include the tightening of a bolt using a wrench, the rotation of a doorknob, and the movement of a bicycle pedal.

How does the direction of torque affect an object?

The direction of torque affects the rotational motion of an object. If the direction of torque is in the same direction as the object's rotation, it causes the object to speed up. If the direction of torque is opposite to the object's rotation, it causes the object to slow down or stop.

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