# Where does a torque-invoking force belong in work energy theorem?

First of all, sorry for the poorly phrased question, simply couldn't figure out a better way to put it.

So, the work energy theorem states that work done on an object is equal to the change in its kinetic energy.

$$W=\Delta K$$

Now consider a yo-yo that is dropped from rest. Say it fell a distance h in t seconds. For the correct result we need to take the effect of T into account while calculating h, but not while calculating F.

if:

$$\sum F = mg$$

$$h = \frac{(g - \frac{T}{m})t^2}{2}$$

Since $$W = \Delta K = mgh = K_{rot} + K_{trans} = \frac{3}{4} m v^2$$

$$v=\frac{mg-T}{m}t$$

We find $$T = mg/3$$

But if:

We say $$\sum F = mg - T$$;

We find $$2=3$$ after cancellations.

My actual question is why don't we take T into account while calculating the total force doing the work. Tension and gravity both do work on the object but when I take tension into account it all comes crashing down. What is it that seperates tension?

• Please edit your post using MathJax. Jan 8 at 22:20
• I cant follow what you did, but T is part of the forces to sum
– user65081
Jan 8 at 22:25

The force of tension does 2 'types' of work:

• It acts on the yoyo opposing its motion, affects translational KE, and;
• Provides a torque about the center of rotation (presumably the CoM), affecting the rotational KE.

The work (affecting translational motion) by the tension force is $$W_\mathrm{tension,\ trans}=-Th$$

whereas the work by the tension (affecting rotational motion) is $$W_{\rm tension,\ rot}=Tr\Delta\theta$$

where $$r$$ is the distance between the center of rotation and the point where tension is applied. Assuming rolling without slipping, $$h=r\Delta\theta$$ and thus, $$W_{\rm tension, \ rot}=Th$$

Therefore, it is obvious that $$W_\mathrm{tension,\ trans}+W_{\rm tension,\ rot}=0$$. So, if you decide to include the work by tension, you have to account for the work done on the translation of the CoM AND the torque about the CoM, which both sum to zero, hence can be excluded from calculations.

Also, do not forget to include the rotational kinetic energy in your work-KE theorem.

$$\vec F_{ext} = M \vec a$$ where $$\vec F_{ext}$$ is the net external force, $$M$$ is the total mass, and $$\vec a$$ is the acceleration of the center of mass (CM). $$F_{ext} = mg - T$$ downward where T is the tension in the yo yo string. This is for the motion of the CM. $$T$$ is not constant.

The yo yo has rotational motion about the CM as well as translational motion of the CM. $$\vec \tau_{ext} = d\vec L/dt$$ where $$\vec \tau _{ext}$$ is the total torque with respect to the CM and $$\vec L$$ is the angular momentum with respect to the center of mass. This torque is due to $$T$$.

Regarding your question about T and work, the force T does no work assuming there is no slip at the string/yo yo surface. The force T is a constraint that provides a torque but T does no work on the yo yo as a whole. This is similar to an object rolling down a plane without slipping; here, the force of friction provides a torque but does no work since there is no relative motion between the friction force and the point of contact of the object on the plane. See one of the physics texts by Halliday and Resnick. This allows us to use the conservation of mechanical energy to evaluate the problem as follows. [If the string slips, then $$T$$ does work on the yo yo.]

Assuming no friction loss and no slipping of the yo-yo string, for a yo yo unwinding down starting from rest, from the conservation of energy, after falling a distance $$h$$: $$mgh = mv^2/2 + I \omega ^2/2$$ where $$I$$ is the moment of inertia of the yo yo, $$v$$ is the velocity of the CM, and $$\omega$$ is its angular velocity with respect to the CM, equal to $$v/r$$ where $$r$$ is the distance from the CM to where the string is wound on the yo yo. [Note: this is an evaluation of $$Work = \Delta Kinetic \enspace Energy$$ for the yo yo where $$Work = mg$$ (the work from gravity is considered as a change in potential energy), and $$Kinetic \enspace Energy$$ includes translational energy of the CM and rotational energy about the CM. $$\int_{a}^{b}(mg - T)dx = \Delta KE_{CM} = 1/2Mv^2$$ where x is the downward distance travelled by the CM, but this is just for the kinetic energy of the CM not for the entire yo yo. In this sense T does work on the CM, but not on the yo yo as a whole; the earlier answer by @User256872 discusses this.]

This exchange has numerous discussions of this problem; for example see Simple yo-yo work problem. Also see https://physics.princeton.edu/~mcdonald/examples/yoyo.pdf on the web.

My actual question is why don't we take T into account while calculating the total force doing the work. Tension and gravity both do work on the object but when I take tension into account it all comes crashing down. What is it that seperates tension?

We do take into account $$T$$ (Torque) when calculating the total force doing the work. It takes it into account because the $$\Delta KE$$ in the work energy theorem is the sum of the change in rotational and translational kinetic energy. The torque $$T$$ provide the rotational kinetic energy component of the total change in kinetic energy due to net work (rotational + translational) done.

Hope this helps.