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Rotational Second Law Physics 
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#1
Mar1412, 06:35 PM

P: 2

1. The problem statement, all variables and given/known data
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? 2. Relevant equations Ugrav=mgh KE=0.5mv^2=0.5Iw^2 angular acceleration alpha=atan/r torquenet=I*alpha 3. 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=(2ghmv^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) yy0=(v0+vy)/2 *t 2.00m0m=(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? 


#2
Mar1412, 07:10 PM

P: 343

I think that you may need to rethink 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. 


#3
Mar1412, 07:27 PM

P: 2

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? 


#4
Mar1412, 08:19 PM

P: 343

Rotational Second Law Physics
Could you show your FBDs, please?



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