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

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SUMMARY

The discussion focuses on calculating the moment of inertia for a rotating wheel using a practical physics problem involving a 0.740 m diameter wheel weighing 280 N and an 8.00 kg mass. The linear acceleration of the mass is determined to be 6.25 m/s², while the moment of inertia is calculated as 0.622 kg·m². Key equations utilized include Ugrav = mgh and torque_net = I * alpha. The analysis emphasizes the importance of free body diagrams (FBDs) to clarify forces acting on both the wheel and the mass.

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  • Understanding of rotational dynamics and moment of inertia
  • Familiarity with Newton's laws of motion
  • Knowledge of kinematic equations
  • Ability to draw and interpret free body diagrams (FBDs)
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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?
 
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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.
 
OldEngr63 said:
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?
 
Could you show your FBDs, please?
 

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