# Rolling wheel problem

Doc Al
Mentor
Another question I would like to ask is, consider the free body diagram there is a translational force in the direction of the smack, why isn't the whole stick move translationally according to the force provided?
It does! The force creates a translational acceleration of the center of mass.

From your example it seems that there is a portion of force used to cause rotational motion and the other portion causes translational motion.
No. The same force does two things: It accelerates the center of mass and it creates a rotational acceleration about the center of mass.

What factors determine the the fraction of force that is used to move the stick translationally and that used to cause the torque?
The entire force is used for both. (Of course the amount of torque produced depends on the point of application.)

Gold Member
Example: Imagine a stick lying on a frictionless table. Give it a smack perpendicular to one end. The stick will rotate about its center of mass as well as translate.
Thanks for the answers. I would like to ask why the torque on a body is always the force times perpendicular distance to the center of mass if there is no supporting point on that body? What's the principle behind that?

Gold Member
If I calculate the torque and moment of inertia about the other point, will I get exactly the same result?

Doc Al
Mentor
I would like to ask why the torque on a body is always the force times perpendicular distance to the center of mass if there is no supporting point on that body? What's the principle behind that?
The definition of torque about any point is force times perpendicular distance to that point. (No need for a support point.) But using the center of mass makes describing the dynamics of the object much simpler to describe. For example, regardless of where the force is applied, the acceleration of the center of mass is the same. And with torque, using the center of mass as your reference allows you to simply calculate the angular acceleration.

If I calculate the torque and moment of inertia about the other point, will I get exactly the same result?
In general, no. Using a point other than the center of mass to calculate torques makes it more difficult to calculate the resulting motion. (You can do it, and of course you'll get the same answer, but the dynamics is more complicated to describe. These details are generally covered in a 2nd level mechanics course.)

1 person
The rolling resistance force ( deemed to be constant regardless of velocity ) depends on the materials of the wheel and the surface it is rolling on (meriting a rolling resistance co-efficient (Crr)) and the weight on the wheel.

A typical example is a motorcycle with a rolling resistance co-efficient of 0.03 ( tyre on ashphalt )

If the Crr is the same for both wheels :

Say an all in weight of 250 kg

The rolling resistance force (Frr) from :
Frr = 250 (kg) * 9.81 (local g rate) * 0.03 (Crr)
Frr = 73.575 Newtons rolling resistance force ( constant)

1 person