# Moment of inertia in a wheel?

## Homework Statement

So basically there is a wheel with a radius of 0.3 m. A light cord wrapped around the wheel supports a 0.75-kg object that accelerates at 3 m/s^-2 downwards. What is the moment of inertia of the wheel? (no friction)

I=mr^2
T=I(alpha)
(alpha)=ra

## The Attempt at a Solution

Seems like it should be an easy question, but the inclusion of the acceleration threw me off. The two approaches I have thought about are:

I=mr^2, I=(0.3*0.75^2) which gives an option in the answers (0.15), but that seems too easy and I did not use the acceleration.

The other approach is using Torque=I (angular acceleration), using the above value for inertia (0.15), and then calculating angular acceleration by r * linear acceleration (0.3*3), but this produced a number not in the solution list (0.135).

Are there other approaches I have missed? Involving forces or the like?

cepheid
Staff Emeritus
Gold Member
I=mr^2, I=(0.3*0.75^2) which gives an option in the answers (0.15), but that seems too easy and I did not use the acceleration.
This is wrong because this equation is for the moment of inertia of a single point mass, not for an extended body. And without knowing more about the shape of the wheel, you cannot come up with the proper equation for its moment of inertia.

The other approach is using Torque=I (angular acceleration), using the above value for inertia (0.15), and then calculating angular acceleration by r * linear acceleration (0.3*3), but this produced a number not in the solution list (0.135).

Are there other approaches I have missed? Involving forces or the like?
This is close, but you did something pretty weird. You're trying to solve this equation for I. So I don't understand why you plugged in a value for I, when that is the thing that you are trying to solve for.

You know the torque around the centre of the wheel that is produced by the weight of the hanging mass. You also know the angular acceleration of the wheel because, as you pointed out, it is related to the linear acceleration of the mass. So, given both torque and angular acceleration, you can solve for the moment of inertia.

EDIT: except it you should have written:

angular acceleration = (linear acceleration)/r

divided by r, not multiplied by r.

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Ohhhh right. Wow, quite a fail on my behalf.

So now I calculated Torque as F*d ((3*0.75)*0.3) and got 0.675.

Then I divided this by 10 (calculated angular acceleration - 3/0.3) and got 0.068.

This is an option...but is it correct? Or should I factor in gravity to my force calculation?

cepheid
Staff Emeritus
Gold Member
Ohhhh right. Wow, quite a fail on my behalf.

So now I calculated Torque as F*d ((3*0.75)*0.3) and got 0.675.
Where does the 3 come from? Side note: always include units in your calculations. What you have written here is actually totally meaningless, because it doesn't have units.

Not only are units am essential part of any meaningful physical calculation, but they also help you catch errors. If the units don't work out, you've made a mistake. Your final answer should be in Nm, since you're computing a torque.

This is an option...but is it correct? Or should I factor in gravity to my force calculation?
The force being exerted here is the weight of the mass that is hanging. So what is that force equal to (I don't mean what is the numerical value, I just mean, how would you compute it)?

cepheid
Staff Emeritus
Gold Member
On second thought...since the mass is not in free fall, the net force on it must be smaller than its weight. (The net force is ma, and a is smaller than the acceleration due to gravity) This means that the rope must be pulling up on the mass to counter its weight partly.

So think maybe the tangential force that is producing the torque is equal to the net force after all, and not to the weight.

Hmm...I'll have to think about it some more.

gneill
Mentor
Hint: The 0.75kg object is not falling at g, so there must be a tension T in the cord that is slowing its fall. This same tension will be responsible for the torque on the wheel.

cepheid
Staff Emeritus
Gold Member
Hint: The 0.75kg object is not falling at g, so there must be a tension T in the cord that is slowing its fall. This same tension will be responsible for the torque on the wheel.
Yeah, my intuition was having a little bit of trouble with it, but I resolved the difficulty by just realizing that if you draw a FBD for the wheel only, then the only thing that can be applying a tangential force at the rim of the wheel is the tension in the rope.

So I guess the OP has to solve for the tension in the rope by figuring out the difference between the downward force on the mass due to gravity, and the net force that ends up acting on it.

cepheid
Staff Emeritus
Gold Member
Actually, hang on. What if the mass WERE in free-fall? Then your analysis would lead us to believe the tension in the rope would be zero. Where, then, would the tangential force that is spinning up the wheel be coming from?

gneill
Mentor
Actually, hang on. What if the mass WERE in free-fall? Then your analysis would lead us to believe the tension in the rope would be zero. Where, then, would the tangential force that is spinning up the wheel be coming from?
If the mass were in free-fall and the tension thus zero, then the moment of inertia of the wheel would have to be zero, too.

cepheid
Staff Emeritus