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Suppose that a cord is wrapped around the rim a disk of radius $R$. The disk is allowed to rotate around its central axis (the line passing through the center and perpendicular to the disk surface). The force from the cord is $F$. Then I am told that the magnitude of torque on the disk is $RF$.

I could not understand how $RF$ follows from the definition net torque $T= \sum \vec r_i \times \vec F_i$ when the sum is taken over all particle. Things become more confusing as I notice that the force $F_i$ on any single particle of the object must not be zero, because each particle is rotating together with the rigid object.

Any help is appreciated.

Additional Info: The fact that net $T=\sum \vec r_i \times \vec F_i$ is used, for example, in the proof of Newton's second law for rotation $T = I \varepsilon$. The proof (as far as I know) proceeds from the case of a single particle and then generalizes to rigid objects by considering an object as being a combination of many particles.

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  • $\begingroup$ Note the rotational equations of motion are $$\sum \vec{M}_{cm} = I_{cm} \dot{\vec{\omega}} + \vec{\omega} \times I_{cm} \vec{\omega}$$. $\endgroup$ – ja72 Oct 18 '13 at 12:54
  • $\begingroup$ @ja72 Sorry, may you provide some more explanations? Thank you. And I thought that two vectors $\omega$ and $I\omega$ points in the same direction, so their cross product (your second sum) is zero? $\endgroup$ – tom_a2 Oct 18 '13 at 12:59
  • $\begingroup$ the two vectors are parallel in your case, so that term goes to zero in this problem. However, in general angular momentum and angular velocity are not parallel. $\endgroup$ – Brian Moths Oct 18 '13 at 13:59
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    $\begingroup$ In your case yes, but in general no. Only because a pulley is symmetric. Just wanted to correct you since you stated the 2nd law only contains the $I \dot{\omega}$ parts. $\endgroup$ – ja72 Oct 18 '13 at 13:59
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The sum runs over external forces applied to the object. The only external force is the force on the string. $\vec{r}\times \vec{F}$ for this force is $RF$. Thus this is the total torque on the object.

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  • $\begingroup$ By definition the sum is taken over all particles . You can read, for example page 4 and 5 of hep.fsu.edu/~berg/teach/phy2048/1009.pdf which I found by a quick googling. $\endgroup$ – tom_a2 Oct 18 '13 at 12:44
  • $\begingroup$ Then we are in agreement. $\endgroup$ – Brian Moths Oct 18 '13 at 13:33
  • $\begingroup$ Just to clarify what he means, you are summing over all the particles: at each particle, you add $r_i \times F_i$. However, the only point at which $F_i \neq 0$ is at the point with the cord. So explicitly it's like $r_1 \times 0 + r_2 \times 0 + ... + r_{cord} \times F$. $\endgroup$ – YungHummmma Nov 17 '13 at 16:19

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