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In the classic spool problem, where a spool of string of mass M is unrolled with a force F, like in the diagram, after working through the equations with Newton's laws of rotation and translation, we get a acceleration of 4F/3M, more than F/M, with a process like.

Diagram

Standard solution(f for static friction force):

F+f=MA

(F-f)R = Ia = 1/2 MR^2 * A/R --> F-f= 1/2 MA

or(setting it up with friction pointing other way)

F-f=MA

(F+f)R = Ia = 1/2 MR^2 * A/R --> F-f= 1/2 MA

2F=3/2 MA

F=3/4 MA --> A= 4M / 3F

The textbook explanation is that there is a force of static friction F/3 pointing in the same direction as the applied force that causes this, but if the linear acceleration was 4/3 (F/M) as well as the spool accelerating rotationally, wouldn't this violate conservation of energy as only F force is applied to the spool? Where does this "extra" force come from or how can we explain it? I'm not sure how the spool can accelerate faster and gain rotational energy compared to a sliding frictionless block of the same mass with the same force applied to it. Are there problems with the underlying assumptions in the calculations?

Textbook solution is this:

Solution given by textbook is this:

J Duan
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  • $\vec{f}$ points in the opposite direction of $\vec{F}$. Friction forces always oppose motion. So in scalars your equation of motion is $F-f=ma$. The spool is accelerating more slowly than an equivalent sliding frictionless block. – Gert Nov 12 '15 at 02:26
  • In this problem, you can set it up that way, but then you get a negative value for the frictional force and the same result that it points in the positive direction. The frictional force is opposing the sliding by opposing the rotational motion, thus pointing to the right.

    In the case of the spool being pulled from the center, the applied force doesn't apply torque so the friction does point the other way and the answer of 2F/3M makes sense, being slower than a sliding block, but this case is different which is what confuses me.

     – J Duan Nov 12 '15 at 02:30
  • No, the frictional force is the cause of the rotation because it provides the torque $F_fR$ needed for the angular acceleration (clockwise) of the spool. For that it must point to the left. See e.g.: http://physics.stackexchange.com/questions/217843/understanding-rotational-motion-on-a-rough-surface – Gert Nov 12 '15 at 02:43
  • The difference between this problem and that is where the force is applied. In that case, the applied force doesn't apply a torque and the rotation is provided by the friction, but in this problem, the applied force applies a torque of F*R. – J Duan Nov 12 '15 at 02:44
  • This is the explanation given by my textbook for this problem, but the 4F/3M result just seems off. http://i.imgur.com/jYdrzsV.png – J Duan Nov 12 '15 at 02:46
  • Sorry, the torque is of course $(f+F)R$. – Gert Nov 12 '15 at 02:50

1 Answers1

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Your issue is that a force $F$ applied through a distance $\Delta x$ results in more kinetic energy here than if the same force was applied through the same distance on a sliding frictionless block.

The problem is the equation $W = F \Delta x$ only applies to point particles, or objects that move totally uniformly like point particles. When you have a rotating and moving object, $\Delta x$ is ambiguous; it should be the $\Delta x$ of the point where the force is applied. In this case, the top of a rolling ball moves twice as fast as its center, so $\Delta x$ is actually twice what it appears to be.

Then the final energy should be at most $2W$, not $W$. You can check that it's indeed less than $2W$.

(Another way to see that $\Delta x$ is actually double is to consider what is actually doing the work. Suppose there's a machine pulling in the rope far off to the right. In the rolling situation, the machine pulls in twice as much rope length as in the sliding situation. From its perspective, it might be just lifting a weight twice as far, so the energy output has to be double.)

knzhou
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  • Ah, thanks alot. I thought that you'd need a force of 2F to produce a translational acceleration of 4/3 MA and a rotational acceleration of I/R^2 A = 2/3 MA but it makes sense now why the force of F is sufficient. – J Duan Nov 12 '15 at 04:44
  • So in a sense, is this situation like a moving pulley with a IMA of 2? – J Duan Nov 12 '15 at 04:46
  • Yup! Pulleys work the exact same way, i.e in both cases it's all about the distance over which the force acts. – knzhou Nov 12 '15 at 04:47