Finding Moment of Inertia for Wagon Wheel

[PascalPanther]

While working on your latest novel about settlers crossing the Great Plains in a wagon train, you get into an argument with your co-author regarding the moment of inertia of an actual wooden wagon wheel. The 70-kg wheel is 120-cm in diameter and has heavy spokes connecting the rim to the axle. Your co-author claims that you can approximate using I = MR2 (like for a hoop) but you anticipate I will be significantly less than that because of the mass located in the spokes. To find I experimentally, you mount the wheel on a low-friction bearing then wrap a light cord around the outside of the rim to which you attach a 20-kg bag of sand. When the bag is released from rest, it drops 3.77 m in 1.6 s.

Here is what I did initially:

v = 3.77m/1.6s
mgh = (1/2)mv^2 + (1/2)*I*omega^2
omega = v/R
(20kg)(9.8)(3.77m) = (1/2)(20kg)(2.36 m/s)^2 + (1/2)I((2.36)^2/(0.60)^2)
I = 88.3 ... which is more than I = MR^2 = 25.2 ...

So this is entirely wrong, because I know it should be less, since the mass is more evenly distributed.
Second approach, I believe the drops 3.77 m in 1.6 s is still due to gravity, is not constant velocity. There must also be tension then since it is less than gravitational acceleration.

mg - T = ma
y=(1/2)a*t^2
(3.77m) = (1/2)a(1.6)^2
a= 2.95 m/s^2

(20 kg)(9.8) - T = (20kg)(2.95)
T = 137N

torque = F*R
Only T does something in relation to the rotation of the axis? So:
torque = T*R = I * alpha
alpha = a/R = 2.95m/s^2 / 0.60m = 4.92 rad/s^2

137N * 0.60m = I * (4.92 rad/s^2)
I = 16.6 N s

Well I got the less inertia part now. But was that the right way to do it?

 

[OlderDan]

Both ways should give you the same result. You made an error in your first calculation. A hint as to where that error is can be found in your second calculation (a = 2.95m/s²). How fast is the mass moving at the end of 1.6 seconds?

 

[kyle8921]

I would first commend you for taking the initiative to experimentally determine the moment of inertia of the wagon wheel rather than just relying on an approximation. Your approach of using the equation I = MR^2 for a hoop would have been appropriate if the wheel was a solid disk without spokes. However, since the wheel has heavy spokes connecting the rim to the axle, it cannot be treated as a solid disk and a different approach is needed.

Your second approach of using the relationship between tension and acceleration is a better method of determining the moment of inertia for the wagon wheel. However, there are a few things that could be improved upon. Firstly, the equation you used (mg - T = ma) assumes that the wheel is in equilibrium, which is not the case as the bag of sand is being released from rest. Instead, you should use the equation F = ma, where F is the net force acting on the system (in this case, the bag of sand and the wheel) and m is the total mass. This will give you the correct value for acceleration.

Secondly, the equation torque = T*R = I * alpha is only valid if the wheel is rotating at a constant angular velocity. In this case, the wheel is accelerating and therefore, the equation should be rewritten as torque = T*R = I * alpha + I * alpha_0, where alpha_0 is the initial angular acceleration of the wheel. Since the wheel starts from rest, alpha_0 = 0 and the equation simplifies to torque = T*R = I * alpha.

Lastly, you have correctly calculated the torque (137N * 0.60m) and the angular acceleration (4.92 rad/s^2) but you have not taken into account the radius of the wheel in your calculation of moment of inertia. The correct equation should be I = (T*R)/alpha = (137N * 0.60m)/4.92 rad/s^2 = 16.6 kg m^2.

In conclusion, your approach to find the moment of inertia of the wagon wheel experimentally was correct, but there were some minor errors in your calculations. By taking into account the net force, the initial angular acceleration, and the radius of the wheel, you have correctly determined the moment of inertia to be 16.6 kg m^2, which is significantly less than the value obtained from the approximation I = MR^2.