What causes an object to rotate?

[Like Tony Stark]

Homework Statement

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Relevant Equations

$\Sigma \tau_o=I \alpha$

Hi
I've been taught that any force not going through the centre of mass will create torque.
Consider a rod of length $L$ and negligible mass, with two balls of mass $m$ attached to its ends. Its centre of mass is at $\frac{L}{2}$.
I have two questions:

1) If a force $F$ is applied to the centre of mass and we consider the centre of mass as the origin, the torque will be zero. But what if consider the point $O$ as origin. Now, that force will create a torque $\tau =-F \frac{L}{2}$ and so it will rotate because of the angular acceleration. But we're considering the same force at the same point. Shouldn't we get that the rod will not rotate?

2) is there any point of the rod that, if applied a force to it, will cause the rod to rotate but not to translate?

 

[PeroK]

1) If a force $F$ is applied to the centre of mass and we consider the centre of mass as the origin, the torque will be zero. But what if consider the point $O$ as origin. Now, that force will create a torque $\tau =-F \frac{L}{2}$ and so it will rotate because of the angular acceleration. But we're considering the same force at the same point. Shouldn't we get that the rod will not rotate?

2) is there any point of the rod that, if applied a force to it, will cause the rod to rotate but not to translate?

1) Torque and angular momentum are relative to a point (or axis). You must always say the torque/angular momentum about some point. The torque about the centre of mass is zero, but the the torque about the origin is not. So, you have no angular momentum/rotation about the centre of mass, but you do have angular momentum about the origin.

Note: a particle moving in a straight line has zero angular momentum about any point on that line, but non-zero angular momentum about any other point. And, if the motion is at constant speed, then the angular momentum is conserved.

2) No. Newton's second law applies. In this case, for an extended rigid body, $\vec F = m \vec a_{CoM}$.

 

[jbriggs444]

1) If a force $F$ is applied to the centre of mass and we consider the centre of mass as the origin, the torque will be zero. But what if consider the point $O$ as origin. Now, that force will create a torque $\tau =-F \frac{L}{2}$ and so it will rotate because of the angular acceleration.

Not so fast.

If you consider the point $O$ as the origin then a force (not in line with $O$) will create a torque, yes. And that torque will result in a rate of change of angular momentum, yes.

But that angular momentum need not manifest as a rotation. Linear motion of the center of mass relative to a chosen axis not on the line of motion of the center counts toward angular momentum about the chosen axis.

2) is there any point of the rod that, if applied a force to it, will cause the rod to rotate but not to translate?

$\sum F=ma$. If you apply a non-zero net force, you will get a non-zero translation.

However, if you extend the rod out past both masses then the farther out you apply the force, the greater the ratio of torque to force. If you apply the force far enough out, a negligible force will result in the desired rotation.

You can also apply a "couple". A pair of equal and opposite external forces that do not share the same line of action. Then you get a net external torque without a net external force.

 

[hutchphd]

The object will gain angular momentum unless the sum of the external forces act through the center of mass.
In your example where the origin is moved, the object will not rotate about the center of mass but the system will have angular momentum about the new origin. Torque and angular momentum depend upon choice of origin. Their relationship does not.

 

[Like Tony Stark]

Thanks, I've understood what you told me about the effect of the force applied to the centre of mass.
But I came across with another doubt related to this one:
I know that angular momentum and torque depend on the point considered as the origin. But does the conservation of angular momentum depend on it? Recalling the situation of the rod hit in the centre of mass, if the origin is the centre of mass, there's no torque so angular momentum is conserved. But if we consider $O$ as the origin, there is a torque, so it will not be conserved.
Or consider a point mass in free fall (under the force of gravity). Angular momentum is not conserved because there's an external force caused by gravity, but if the origen is in the same line of the free fall, the position and momentum vectors will be parallel and so the angular momentum will always be zero, which implies that it is conserved.

 

[jbriggs444]

Thanks, I've understood what you told me about the effect of the force applied to the centre of mass.
But I came across with another doubt related to this one:
I know that angular momentum and torque depend on the point considered as the origin. But does the conservation of angular momentum depend on it? Recalling the situation of the rod hit in the centre of mass, if the origin is the centre of mass, there's no torque so angular momentum is conserved. But if we consider $O$ as the origin, there is a torque, so it will not be conserved.
Or consider a point mass in free fall (under the force of gravity). Angular momentum is not conserved because there's an external force caused by gravity, but if the origen is in the same line of the free fall, the position and momentum vectors will be parallel and so the angular momentum will always be zero, which implies that it is conserved.

One answer is that angular momentum is conserved in a closed system. In a closed system there are no external forces.

That is somewhat analogous to Newton's first law

Another answer is that if there is an external force, the rate of change in angular momentum is given by the torque associated with that force. Choose one axis and you get no external torque and no change in angular momentum. Choose another axis and you get a non-zero external torque and a non-zero change in angular momentum. Either way, the rate of change of angular momentum matches the net torque.

This is somewhat analogous to Newton's second law.

Yet another answer is that if there is a torque from one object on another, there is an equal and opposite torque from the other on the one. The net rate of change of the total angular momentum of the two objects is zero. This holds regardless of what axis you choose.

This is a direct consequence of Newton's third law and the idea that the lines of action of third law partner forces coincide.