$$ \underline{\pmb {Assumption}}$$

Diagram of the rod

Assume a place free from any gravitational or other kind of influences (like friction, drag, etc.). Now assume a thin cylindrical rod of mass $M$ and length $l$. Suppose that a force $\mathbf F$ of constant magnitude and direction (though movable, this means that this force has same magnitude and direction and is always applied to the same point on the rod) is applied on the rod at distance $r$ from the $ \pmb {COM} $ for a displacement $d$($ 0 \lt r \leq \frac {l}{2} $). Similarly assume a point particle on which the same force acts for a displacement $d$.

$$ \underline { \pmb { \text{Constructing the problem}}}$$

Since a force $\mathbf F$ is applied on the rod then a work $$W_{translational} = \mathbf F \cdot \mathbf d$$ is done on the centre of mass. Also the rod is rotating as torque $\mathbf {\tau}$ is generated and the axis is the $ \pmb {COM} $, though it changes in magnitude (every instant) and direction (after a rotation of $90°$). For a rotation of $180°$ zero rotational work is done (You may notice that the rod comes into a periodic motion). This means that if the displacement of $ \pmb {COM} $ was taken for the period of $180°$ rotation then the work done on the system would be same as that done on the point particle for the same displacement. But now comes the problem.

$$\pmb{ \underline {\text {Question}}}$$

Suppose the displacement of the centre of mass is taken for the case of rotation $$0 \lt \theta \lt 180°$$ Then there the object would have translational kinetic energy ( $K_{translational}$) as well as rotational kinetic energy ( $K_{rotational}$). Now for the given case $$W_{COM,extended}=K_{translational} +K_{rotational} $$ but $$W_{point} =K_{translational} $$ Clearly $$W_{COM,extended} \neq W_{point}$$

  • So how does the same force for same displacement do a different work?

  • If possible please give an intuitive explanation.

$$\pmb {\underline {\text {My try}}}$$

I think this dilemma can be solved by concluding that $W_{translational}$ aren't the same for both cases (But then how?). If that's not the case then some internal energy might have converted into kinetic energy (might be temperature). Is there any experimental way to verify this?

  • 1
    $\begingroup$ You have to be careful. What do you mean by "same force"? You are right, the in first case there is rotation. So you aren't actually going to have the "same force" in either case if in both cases the application point follows the object $\endgroup$ Nov 29, 2019 at 14:31
  • 3
    $\begingroup$ How is the force being applied to the rotating rod? Does it remain applied to the same position of the rod and always perpendicular to the rod? Does it remain applied to the same position but keeps the same direction the entire time? Does it remain along the same line so that it is applied to a different location of the rod as it moves? And then once you pick one of those, what is the displacement? It is the displacement of the COM? Is it the displacement of the point of application? Is it the same vertical displacement as the point particle case? $\endgroup$ Dec 4, 2019 at 13:40
  • 1
    $\begingroup$ Remember that work is not just dependent on displacement. It depends on how aligned the force is to the displacement along the path. Therefore, the above questions do matter in resolving your conflict. You have said in both cases $F$ and $d$ are the same, but in reality you have not defined $F$ and $d$ enough for the rotational case in order to make that claim valid. So far, you are not looking at scenarios with the "same force and same displacement". $\endgroup$ Dec 4, 2019 at 13:44

3 Answers 3


You can apply a force of the same magnitude in both cases, and the work will be different. The reason is that the displacements are different. The CM will displace the same amount, so the final kinetic energy of the CM will be the same in both cases. But in the case of the extended object the point of application displaces more than the CM, that is why the work is larger. This extra work will end up as rotational kinetic energy.


cylindrical rod of mass M and length l. Suppose that a force $\mathbf{F}$ of constant magnitude and direction (though movable) is applied on the rod at distance $r$ from the COM for a displacement $\boxed{d}$.

I'm going to assume that $d$ is the displacement of the CM. Correct me if I'm wrong.

I'll briefly describe the setup (so you can correct me if it's not the setup you describe) :

Let a uniform rod of mass $M$ and length $L$ be at rest with its CM at the origin and its axis oriented along the $y$-axis [The end points are at $(0,-L/2)$ and $(0,L/2)$]. Now, a constant force $\mathbf{F}=F\hat{i}$ is applied at the point that is at a distance $r$ above the CM of the rod (Let's call that point $P$). Let $\theta$ be the angle that the rod makes with the vertical direction [Initially, $\theta=0$]. Let's say the motion starts at $t=0$ and ends at $t=T$, and evaluate the work done by $\mathbf{F}$ during that period.

The work done by a force $\mathbf{F}$ on a rigid body = $W$ = $\int \mathbf{F} \cdot\mathbf{v}_{\text{of the point of application of $\mathbf{F}$}}\;dt=\int \mathbf{F} \cdot \mathbf{v}_P \; dt$

$$\text{Force equation: }F=M\dot{v}_{CM} \Rightarrow v_{CM}=\frac{F}{M}t \tag{1}$$

As per your assumption,

$$d=\int_0^T v_{CM}=\frac{F}{2M}T^2 \tag{2}$$

$$\text{Torque equation about CM: }Fr\cos \theta = I_{CM} \ddot{\theta} \tag{3}$$

$$W = \int_0^T \mathbf{F} \cdot \mathbf{v}_P \; dt \stackrel{*}= \int_0^T (F v_{CM} + Fr \dot{\theta} \cos\theta) dt$$ $$\Rightarrow W \stackrel{\text{Using $(1)$ and $(3)$}}= \int_0^T (\frac{F^2}{M} t+I_{CM} \ddot{\theta} \dot{\theta}) dt=\int_0^T \left(\frac{F^2}{M} t+I_{CM}\frac{1}{2} \frac{d \left({\dot{\theta}}^2\right)}{dt}\right)dt$$ $$\Rightarrow W = Fd + \frac{1}{2} I_{CM} (\dot{\theta}_{\text{final}})^2=\frac{1}{2}Mv_{\text{CM | final}}^2+\frac{1}{2}I_{CM} (\dot{\theta}_{\text{final}})^2= KE_{\text{final}} = \Delta KE \tag{4}$$

The work done by force $\mathbf{F}$ is greater than $Fd$ and the excess work is responsible for providing the rotational kinetic energy of the rod.

If it had been a point mass under the action of $\mathbf{F}$, then the work done would have been $W=Fd = \frac{1}{2}Mv_{\text{of the point mass | final}}^2$

$^*$ $\mathbf{v}_P = \mathbf{v}_{\text{P wrt CM}} + \mathbf{v}_{CM}$

OP's concern:

Since a force $\mathbf{F}$ is applied on the rod then a work $$W_{translational} = \mathbf F \cdot \mathbf d$$

What you have said here is correct. Here, I believe you're assuming the displacement of point $P$ to be $\mathbf{d}$ and not the displacement of CM to be $\mathbf{d}$. To maintain consistency in my answer, I'm going to call the displacement of the CM to be $\mathbf{d}$ as before and find the displacement of point $P$ given by $\Delta \mathbf{s}_P$.

$$W = \int \mathbf{F} \cdot d\mathbf{s}_P\stackrel{\text{Since $\mathbf{F}$ is constant in this problem}}=\mathbf{F}\cdot \int d\mathbf{s}_P = \mathbf{F} \cdot \Delta \mathbf{s}_P \tag{5}$$

Let's say the final orientation of the rod is $\theta=\theta_{\text{final}}$. Then, we find $\Delta \mathbf{s}_P = (d + r\sin \theta_{\text{final}})\hat{i}+(r-r\cos \theta_{\text{final}})\hat{j}$. Substitute it in $(5)$ to get:

$$W=Fd + Fr\sin \theta_{\text{final}} \stackrel{\text{Using (4)}}\Rightarrow \dot{\theta}_{\text{final}}=\sqrt{\frac{2Fr\sin \theta_{\text{final}}}{I_{CM}}}\tag{6}$$

The result matches with $(3)$ when you substitute $\ddot{\theta}=\dot{\theta}\frac{d\dot{\theta}}{d\theta}$ and integrate it.

So how does the same force for same displacement do a different work?

From here, you can see that $\Delta \mathbf{s}_P \neq \Delta \mathbf{s}_{CM}=\mathbf{d}$. So, the displacements are not the same.


If you apply the force a distance $r$ along the rod and during that time it rotates by a distance $d_{cm}$ and small_ angle $\theta$, then the point where you apply the force moves by $d_{cm} + r \theta$.

This means the work done and energy added is $F d_{cm} + F r \theta$.

The first term is the translational energy increase and the second is the rotational energy increase.

For larger angles, the calculation involves some trigonometry, but the basic result is the same.

  • $\begingroup$ So what are you assuming about how the force changes? $\endgroup$ Dec 4, 2019 at 16:58
  • $\begingroup$ @AaronStevens in the small angle limit, the assumption is that F is constant. More complex cases are just more complex math, but the result will always confirm the work-energy theorem. $\endgroup$ Dec 4, 2019 at 17:24
  • $\begingroup$ I am not questioning the validity of the work energy theorem. But the analysis does depend on how the force is being applied over the entire path. $\endgroup$ Dec 4, 2019 at 17:26
  • 1
    $\begingroup$ A varying force would require a different analysis. I’d probably use calculus to sum up small displacements as described here. But that doesn’t change the answer to this question, which is based on a constant force F. $\endgroup$ Dec 4, 2019 at 17:28
  • $\begingroup$ So you are assuming a force that keeps its constant magnitude and direction, yet still follows the same application point the entire time? So then the force is not perpendicular to the rod at all times? $\endgroup$ Dec 4, 2019 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.