I understand that the small wheel will turn more quickly, but I really wanted to know about the torque alone.

I created a diagram so it could more easily be understood. I just need to know if a smaller weighted wheel will have less torque than a bigger weighted wheel.

The diagram can be seen here: http://bizidia.com/wheel.jpg

Thank you so very much for your time!

Sincerely,

JC

Jacob - in the future please do not post a follow up question by start8ing a new thread - it sis much easier to follow if you ask your follow-up questions on the original thread. But to answer your question - in your diagram case 1 uses 4 pounds of weight at 5 inches from the center on the left side and 6 inches on the right. However, given the distributed nature of the weight along the rim you have to take into account the average moment of the force of gravity for the weights. If we define \rho to be the density of the weight in pound/inch, then the torque applied by the weight on one side of the wheel is :


T_1 = \int_0 ^{\pi} \rho r_1^2 sin \theta d \theta = \frac 2 \pi W r_1


This is counteracted by the torque applied on the other side:


T_2 = -\frac 2 {\pi} W r_2


Thus the total torque applied to the wheel for W = 4 pounds, r_1 = 5 inches and r_2 = 6 inches is \frac 8 \pi inch-pounds. For the second case you have 12 pounds on the left at 10 inches and 12 pounds on the right at 11 inches, for a total torque of \frac {24} {\pi} inch-pounds. Thus the wheel with the greater amount of weight distributed along its rim experiences the greater torque. However, a further detail - this calculation is only correct at the instant that the wheel is oriented as you showed. Once the wheel rotates the torque becomes less.

One other point - your diagram seems to indicate that torque is measured at the edge of the wheel. This is incorrect - torque is the turning force experienced at the center of the wheel.