# Finding Moment of Inertia for Wagon Wheel

• PascalPanther
In summary, the conversation discusses the moment of inertia of a wooden wagon wheel and the methods used to find it experimentally. The first approach using the equation I = MR2 is proven to be incorrect due to the mass being more evenly distributed. The second approach, using tension and torque, is shown to be a more accurate method for finding the moment of inertia.
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?

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?

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.

## 1. What is moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is similar to mass in linear motion and is affected by the object's mass and distribution of that mass.

## 2. How is moment of inertia calculated?

The moment of inertia for a wagon wheel can be calculated by integrating the mass of the wheel at different distances from the axis of rotation. This calculation takes into account the shape and distribution of the wheel's mass.

## 3. Why is it important to find the moment of inertia for a wagon wheel?

Knowing the moment of inertia for a wagon wheel allows us to understand how the wheel will respond to different forces and torques. This information is critical for designing the wheel to function properly and safely.

## 4. What factors affect the moment of inertia for a wagon wheel?

The moment of inertia for a wagon wheel is affected by its mass, shape, and distribution of that mass. The more mass an object has, the larger its moment of inertia will be. Objects with a larger radius or concentrated mass further away from the axis of rotation will also have a larger moment of inertia.

## 5. How does the moment of inertia for a wagon wheel affect its performance?

The moment of inertia for a wagon wheel affects its rotational acceleration and deceleration. A larger moment of inertia means that the wheel will require more force or torque to start or stop spinning, while a smaller moment of inertia will allow for easier acceleration and deceleration.

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