Solving the Spool's Acceleration with a Constant Force

AI Thread Summary
A spool of wire unwinds under a constant force F, and the goal is to determine the acceleration of its center of mass. The initial calculations incorrectly used the torque as FR instead of the correct value of 2FR, which arises from using the point of contact with the floor as the axis of rotation. The moment of inertia is calculated using the parallel axis theorem, yielding I = (3/2)MR^2. By substituting the correct torque into the equations, the acceleration can be derived as a = 4F/3M. Understanding the torque's dependence on the distance from the axis of rotation is crucial for accurate calculations.
thenewbosco
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Diagram: http://snipurl.com/c8h6

A spool of wire is unwound with a constant force F. The spool is a solid cylinder mass M, radius R, and doesn't slip.
Show the acceleration of the centre of mass is 4F/3M.

what i have done:
\sum Torque = I\frac{a}{R}
FR=I\frac{a}{R}
\frac{FR^2}{I}=a

Now, for I, I used the parallel axis theorem and got
I=\frac{1}{2}MR^2+ MR^2 = \frac{3}{2}MR^2

but plugging into my formula above yields a=\frac{2F}{3M}

how do i get a=4F/3M?

thanks for your help
 
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thenewbosco said:
\sum Torque = I\frac{a}{R}
FR=I\frac{a}{R}
Since you are using the point of contact with the floor as your axis of rotation, the torque is 2FR, not FR.
 
Thank you for your help. Can you explain why the torque is 2FR instead of just FR??

thanks
 
thenewbosco said:
Can you explain why the torque is 2FR instead of just FR??
Torque is F times the perpendicular distance to the axis. Since the axis you are using is the point of contact with the floor, the distance is 2R, not R. (FR is the torque about the center of mass.)
 
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