How Do You Calculate the Moment of Inertia for a Rotating Wheel?

[applemilk]

Homework Statement

You are asked to measure the moment of inertia of a large wheel, for rotation about an axis through its center. You measure the diameter of the wheel to be 0.740 m and find that it weighs 280 N (note, that's not the mass!). You mount the wheel, using frictionless bearings, on a horizontal axis through the wheel's center. You wrap a light rope around the wheel and hang an 8.00 kg mass from the free end of the rope. You release the mass from rest; the mass descends and the wheel turns as the rope unwinds. You find that the mass has a speed of 5.00 m/s after it has descended 2.00 m.

(a) What is the magnitude of the linear acceleration of the 8.00 kg mass?
(b) What is the magnitude of the angular acceleration of the wheel?
(c) What is the magnitude of tension in the rope?
(d) What is the magnitude of the torque on the wheel due to the rope?
(e) What is the moment of inertia of the wheel for this axis?

Homework Equations

Ugrav=mgh
KE=0.5mv^2=0.5Iw^2

angular acceleration
alpha=atan/r

torquenet=I*alpha

The Attempt at a Solution

The only part I know for sure how to solve is e.
mgh=0.5v^2+.5Iw^2
w=(v/r)
I=(2gh-mv^2)/(v^2/r^2)=0.622 kgm^2

I get that the wheel's mass would be:
w=mg=280N
m=280N/9.8m/s^2=28.57kg.

For the other parts I'm confused as to whether I should use the 8 kg mass or the wheel or both. Are the only forces acting on both of them just gravity and then tension of the rope?

For part a)
y-y0=(v0+vy)/2 *t
2.00m-0m=(0m/s+5.00m/s)2*t
t=0.80s

v(t)=v0+at
5.00m/s=0m/s+a(0.80s)
a=6.25m/s^2

Is that right?

 

[OldEngr63]

I think that you may need to re-think part e, along with most of the rest of this problem.

To start again, draw two FBDs, one showing the wheel in the bearings with the rope around in and a tension T acting down on it. The second shows the mass that hangs below with the weight of the mass acting down and the tension T acting upwards. Now write all of the equations of motion and the kinematic constraints. See what you can do from there.

 

[applemilk]

I think that you may need to re-think part e, along with most of the rest of this problem.

To start again, draw two FBDs, one showing the wheel in the bearings with the rope around in and a tension T acting down on it. The second shows the mass that hangs below with the weight of the mass acting down and the tension T acting upwards. Now write all of the equations of motion and the kinematic constraints. See what you can do from there.

I know that my answer for part e is correct.

For the FBDs, the wheel would have w and T in the negative y direction and the mass has positive T and negative w.

8.00kg mass:
w=mg=8kg*9.8m/s^2=78.4N, but it's -78.4N since it's in the negative direction
y components:
T+w=ma
T+-78.4N=8kg*a

wheel:
w=280N
T+w=ma
T+280N=28.57kg*a

Is that right?

 

[OldEngr63]

Could you show your FBDs, please?

 

[asteriatic]

For part b)
I'm not sure how to find the angular acceleration without knowing the length of the rope or the radius of the wheel.

For part c)
T=mv^2/r
T=(8kg)(5.00m/s)^2/(0.740m/2)
T=271.6N

For part d)
Torquenet=I*alpha
Torquenet=0.622kgm^2*6.25m/s^2
Torquenet=3.89N

Your attempt at solving the problem is on the right track. However, there are a few things that need to be clarified and corrected.

First, for part a), you are correct in using the 8 kg mass to find the linear acceleration. This is because the problem states that the mass is the only object that is accelerating in a linear manner. The wheel is rotating, but it is not accelerating in a linear manner.

Your calculation for the linear acceleration is correct, but the units are incorrect. The units should be m/s^2, not m/s.

For part b), to find the angular acceleration, you can use the relationship between linear and angular acceleration: a = r*alpha, where a is the linear acceleration, r is the radius of the wheel, and alpha is the angular acceleration. You can rearrange this equation to solve for alpha: alpha = a/r.

In this case, you have already calculated the linear acceleration in part a) and the radius of the wheel is given in the problem as 0.740 m. So, you can plug in these values to find the angular acceleration.

For part c), you are correct in using the tension in the rope to find the magnitude of the tension. However, the equation you used is incorrect. The correct equation is T = m*a, where T is the tension in the rope, m is the mass of the object, and a is the linear acceleration.

For part d), you are correct in using the equation Torquenet = I*alpha. However, you have used the incorrect value for the moment of inertia. The moment of inertia of the wheel is not 0.622 kgm^2. This value is for the 8 kg mass. To find the moment of inertia of the wheel, you can use the equation I = m*r^2, where m is the mass of the wheel and r