Contradiction in work-energy theorem for rigid bodies?

We've a rigid body at rest. A force acts on it which leads it to translate (velocity of center of mass is $$V_{cm}$$ )and rotate about a fixed axis with $$\omega$$

Kleppner and Kolnekow prove that:

Work done by the force equals the change in kinetic energy of the center of mass. So $$W=\Delta K_{cm}=\frac{1}{2} m V_{c m}^{2}$$ and here is their proof

To derive the translational part, we start with the equation of motion for the center of mass \begin{aligned} \mathbf{F} &=M \frac{d^{2} \mathbf{R}}{d t^{2}} \\ &=M \frac{d \mathbf{V}}{d t} \end{aligned} The work done when the center of mass is displaced by $$d \mathbf{R}=\mathbf{V} d t$$ is \begin{aligned} \mathbf{F} \cdot d \mathbf{R} &=M \frac{d \mathbf{V}}{d t} \cdot \mathbf{V} d t \\ &=d\left(\frac{1}{2} M V^{2}\right) \end{aligned} Integrating, we obtain $$\oint_{\mathbb{R}}^{R_{i}} \mathbf{F} \cdot d \mathbf{R}=\frac{1}{2} M V_{b}^{2}-\frac{1}{2} M V_{a}^{2}$$

But we also know from work kinetic energy theorem that the Work done by an external force equals the change in kinetic energy of the body so $$W=\Delta K=\frac{1}{2} m V_{c m}^{2}+ \frac{1}{2} I \omega^{2}$$.

Then these two equations contradict each other!

• The quote starts with "to derive the translational part". Why would you expect the rotation to be included? – nasu Jan 9 at 21:06
• Dear nasu, The work done by external force equals the change in kinetic energy ( rot+trans) – Kashmiri Jan 10 at 3:24
• This question was born due to another one here physics.stackexchange.com/questions/606440/… Id be grateful if you could have a look. – Kashmiri Jan 10 at 3:24

If the force is applied at some $$R$$ above the CM, then $$\mathrm{d}r=\mathrm{d}r_{\text{CM}}+R\omega \mathrm{d}t$$, so you have an extra term. This extra term becomes:

$$W_{\text{rot}}=\int FR\omega \mathrm{d}t=\int FR\omega \mathrm{d}\omega /\alpha=\tfrac{1}{2}I\omega^2$$

(where we replaced $$\alpha=FR/I$$ before integrating)

• Thank you, say we've a spinning body spinning at $\omega_0$ set down on a horizontal plane and it skids and then begins to roll .The work done by friction is $W_{f}=\frac{1}{2} m \cdot V_{c m}^{2} - 0=\frac{1}{2} m \cdot \omega^{2} R^{2}$ But work done by friction will also be change in total kinetic energy which is $W_f=\Delta K\Rightarrow \frac{1}{2} \operatorname{m} \omega^{2} R^{2}=\left(\frac{1}{2} m \omega^{2} R^{2}+\frac{1}{2} I \omega^{2}\right)-\frac{1}{2} I \omega_{0}^{2}$. – Kashmiri Jan 10 at 3:21
• But this is a contradiction, could you please have a look here physics.stackexchange.com/questions/606440/… – Kashmiri Jan 10 at 3:22
• the question I answered is correct, now you are asking me a new and different question. – Wolphram jonny Jan 10 at 4:15
• I have no problem on giving it a look though – Wolphram jonny Jan 10 at 4:15
• Yes dear, I don't have any teacher to go to. I have only prayers to give you – Kashmiri Jan 10 at 4:41

If the line of action for the applied force does not pass through the center of mass, then the point of application of the force experiences a tangential acceleration. This means that for a given (short) time, the dR for the force is not the same as the dR for the center of mass. This difference is not considered in the given derivation. It is not valid if rotation occurs. Basically, the extra distance traveled by the force does the work to supply the rotational energy.