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Say we have a rod with center of mass at its geometric center. Rigid bodies rotate around center of mass. If you apply force $\mathbf{F}$ at distance $\mathbf{r}$ from center of mass you generate torque $\boldsymbol{\tau} = \mathbf{r} × \mathbf{F}$, resulting in angular acceleration. If you apply force at center of mass, this results only in linear acceleration. However what happens if we constrain (hang) the rigid body from one of its ends? Why does applying force at the geometric center now generates torque? Does center of mass shift?

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  • $\begingroup$ Hey! You've asked some questions in comments to the answers below. It'd be better if you put those queries in the body of the question above. $\endgroup$ Commented Mar 5, 2020 at 7:46
  • $\begingroup$ '...If you apply force at center of mass, this results only in linear acceleration...'. What happens when the body is already rotating when you apply force at the centre of mass? $\endgroup$ Commented Mar 5, 2020 at 12:19

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Does center of mass shift?

Absolutely not! Centre of mass can shift only if there is a change in distribution of mass of the body.

That means we are still applying force on the centre of mass of the body, and the value of $\vec{r}$ in $\vec{\tau}=\vec{r}×\vec{F}$ is still zero. So we can be sure that the force being applied by us is not producing any torque.

Then what could be the explanation of the torque we witness? Think about it again, what is required for producing a torque about a point? A force whose line of action does not pass through that point!

This implies that there must be at least one such force in the above condition that we have been ignoring till now. Where that force could be? To find it, try comparing the motion of the body in the two different scenarios that you have presented and think about what is causing the motion to differ in the second case. When we say that the rigid body is constrained, what exactly is causing this constraint?

enter image description here

As we can see in the above illustration, in the first case, the topmost point moves with the same acceleration as the rest of the body. But when we hinge this point (constrain the body), its motion stops. From here, we can imply that the hinge applies a force on the point opposing its original motion.

enter image description here

(Please note that I have only shown a component of the hinge force. It can have another componet along the rod too.)

Since the line of action of this force is not passing through the centre of mass of the rod, it is perfectly capable to produce a torque and cause rotation of the rod (which it indeed does).

Why does applying force at the geometric center now generates torque?

The torque is generated because your force at the geometric centre is causing a hinge reaction force (which is not at the geometric centre), and this reaction force is generating the torque.

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  • $\begingroup$ Thanks for the detailed answer! Can I ask what happens to the linear acceleration? Applying force to the center of mass should generate linear acceleration, but we get none. $\endgroup$ Commented Mar 5, 2020 at 5:45
  • $\begingroup$ Also in an example with a cart, if we apply a force to it (cart) the rigid body gets angular acceleration because of torque, but we can say the body also gets linear acceleration right? $\endgroup$ Commented Mar 5, 2020 at 5:55
  • $\begingroup$ Linear acceleration is the acceleration of the cenre of mass of the body, and since the centre of mass is in an accelerated motion, there IS a linear acceleration present. $\endgroup$ Commented Mar 5, 2020 at 7:08
  • $\begingroup$ And yes, the body gets linear acceleration too as, again, the centre of mass is accelerated. $\endgroup$ Commented Mar 5, 2020 at 7:09
  • $\begingroup$ In fact, you can see the motion you are talking about as a composition of a rotation around the center of mass, generated by the hinge reaction, and a linear motion, generated by the force acting on the center of mass. By the way, congratulations to Aumkaar for the nice answer. $\endgroup$ Commented Mar 5, 2020 at 15:04
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If you hang the rigid body from one of the ends, then when you apply your force, there is a normal force applied at the end that ensures the constrained end does not move. This normal force can generate torque. The center of mass doesn't shift.

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  • $\begingroup$ I understand, thanks! But say we constrain the body to a cart and then apply force to the cart which causes it to translate with the rigid body. If we ignore the forces generated by air drag and friction. Would this force cause any torque on the body? The body would get some torque in real life but is it solely caused by air drag? $\endgroup$ Commented Mar 5, 2020 at 4:47
  • $\begingroup$ Depends upon how is the body consrained to the cart and in what manner the cart is being moved. The bottom line is that if there is any force on the body not passing through its centre of mass then it is capable of producing a torque. $\endgroup$ Commented Mar 5, 2020 at 5:22
  • $\begingroup$ If you are saying something like this, then yes there will be a torque on the body as the hinge is applying force in forward direction and this force does not pass through the centre of mass of the body. $\endgroup$ Commented Mar 5, 2020 at 5:29
  • $\begingroup$ @AumkaarPranav Thanks for the reply! I was thinking something like the body at its end being glued onto the cart, not constrained with a hinge. $\endgroup$ Commented Mar 5, 2020 at 5:35
  • $\begingroup$ It's much more convenient to offer such explanations in the answer itself, so please add these follow-up queries of yours in the body of your question. $\endgroup$ Commented Mar 5, 2020 at 7:57
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If you carefully look at your question you have the answer $\boldsymbol{\tau} = \mathbf{r} × \mathbf{F}$, as you said this, now thing is if the force passes through the axis of rotation the cross product becomes zero, so when you pivot the rod using its one of end, weight of mg is always passes through the centre of mass, and hence torque produced by it becomes zero.

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Consider a thin uniform rod of length, L, and mass, m, hanging vertically at rest from a hinge or axle at the top. We will strike the rod at its center with an impulse, Ft, which is perpendicular to the rod (and the axle). Assume, t, is a very short period of time. Relative to the axle, the impulse is an angular impulse which produces an angular momentum: Ft(L/2) = Iw = (mLL/3)w Then: w = (3Ft)/(2mL) This also the angular velocity about the center of mass. We can say that the associated angular momentum was caused by an angular impulse produced by the horizontal component of the force from the axle: ft(L/2) = (mLL/12)w giving: f = [mL/(6t)][3Ft/(2mL)] = F/4. From the angular velocity, the linear (tangential) velocity of the center of mass, v, is (L/2)w = (3/4)Ft/m. We can check this against the net liner impulse: (Ft – ft) = mv, again giving: v = (3/4)Ft/m If the original impulse is applied at some other point along the rod, the analysis is more complicated (two torques acting about the center of mass), but the results are similar.

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You have made principal error here =>

Rigid bodies rotate around center of mass

Nope. They rotate around a reference point, which usually is a fulcrum where rigid body is attached to. If fulcrum will be attached to center of mass - body will rotate around center of mass. Otherwise it will rotate about different point. For example - take a look at wheel. Usually it is rotated around axis which goes through a geometric center of wheel. But nobody said that it must always be so. You can cut a hole in ANY wheel/disk place required, put some rod in there, push the wheel and voila - its starts to rotate around that reference point. Of course such design isn't optimal, because in such way rotating wheel will induce high stress to rotational axis, due to centrifugal force of rotating wheel. In any-case you need to stop thinking that a rigid body rotation has something to do with COM. It can of course, but this is not required.

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