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Tuesday, October 12, 2010
Ch8 2 #3
Use equation 8.12. What is the moment of inertia for a cylinder?
4 comments:
Anonymous
said...
I subtracted t1 from t2 using t=Fr. Is this the correct approach?
You do not need any calculus to do this problem. The basic method is to use F=ma for the block, and it's equivalent torgue = I*alpha for the wheel, but realize that the torque = (tension in string)*(radius or wheel) and that the linear acceleration of the wheel is the same as that for the block, so a = alpha*r.
This method always works: add up the forces on a given block (for linear motion) and set it equal to ma, and add up the torques for any wheels (for rotational motion) and set it equal to I*alpha. Then use F*r for each of the forces producing the torques. The last step lets you relate torques to the forces
4 comments:
I subtracted t1 from t2 using t=Fr. Is this the correct approach?
The correct equation to use is T=I*(alpha) =1/2 (MR^2)*alpha
but how do you translate that to r and g?
The answer is 2g/3r. I am only giving it because the way to get the answer, at least for me, involved doing some calculus.
You do not need any calculus to do this problem. The basic method is to use F=ma for the block, and it's equivalent torgue = I*alpha for the wheel, but realize that the torque = (tension in string)*(radius or wheel) and that the linear acceleration of the wheel is the same as that for the block, so a = alpha*r.
This method always works: add up the forces on a given block (for linear motion) and set it equal to ma, and add up the torques for any wheels (for rotational motion) and set it equal to I*alpha. Then use F*r for each of the forces producing the torques. The last step lets you relate torques to the forces
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