# Net torque on an object

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.

• Note the rotational equations of motion are $$\sum \vec{M}_{cm} = I_{cm} \dot{\vec{\omega}} + \vec{\omega} \times I_{cm} \vec{\omega}$$. Commented Oct 18, 2013 at 12:54
• @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? Commented Oct 18, 2013 at 12:59
• 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. Commented Oct 18, 2013 at 13:59
• 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. Commented Oct 18, 2013 at 13:59

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.
• 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 $r_1 \times 0 + r_2 \times 0 + ... + r_{cord} \times F$. Commented Nov 17, 2013 at 16:19