Rotation around which axis

Question

Imagine you have a rigid rod which is free to move and a force is applied at a point away from the center of mass of the rod. This would create torques at multiple points on the rod as the force will have a line of action. Will this mean the rod will rotate around all these points due to the same torque? Won't this create multiple axis and violate the rigid property of rod?

How do we decide around which axis will the rotation start?

How is "a force applied at a point" when at the same time "the force will have a line of action"? — Steeven, Aug 29 '18 at 11:25
Also, when the force is applied at a point, how will it create torques at multiple points? — Shah M Hasan, Aug 29 '18 at 11:30
I don't get what the problem is. The force is acting at some point say B which is x meter above from center of mass. At a point A y meter below center of mass, do you mean to say there will be no torque from this force — Aladdin, Aug 29 '18 at 11:36
According to wikipedia, "All rigid body movements are rotations, translations, or combinations of the two. " So, the problem, as it stands, seems to be a combination of both. There will always be some torque which will be present if x is away from the centre of mass. — Shah M Hasan, Aug 29 '18 at 11:43
That's what I am asking. If the force produces translations and rotstions both the how do we know where it produces rotation. It can very well be any point on the rod — Aladdin, Aug 29 '18 at 11:52
Possible duplicate of What is the proof that a force applied on a rigid body will cause it to rotate around its center of mass? — stafusa, Aug 30 '18 at 12:46

Vincent Thacker · Accepted Answer · 2018-09-15T12:16:04.160

3

The rod accelerates as if all the forces acting on it are acting upon the center of mass. Thus, if the rod is free to move on a frictionless surface, the rod will still obey the equation $$\vec F = m\vec a$$ Again, if the rod is free to move (not constrained to rotate about a certain axis, and no other force acting on it), it will always rotate about its center of mass. That is, if the force is applied at a distance $r$ from the center, we have $$Fr = I\alpha$$ where $I = \frac{1}{12}mL^2$ is the moment of inertia about the center of mass, and $\alpha$ is the angular acceleration.

The translational motion and the rotational motion can be separated. Imagine the rod as a point mass (that is, a dot) accelerating with $\vec F = m\vec a$, and rotating about it.

edited Sep 15 '18 at 12:16

answered Aug 29 '18 at 11:37

Vincent Thacker

7,926

Why does it rotate only around the centre of mass? Aren't there other possible points where it can rotate around. Will it mean any system will always rotate around its centre of mass if left unconstrained – Aladdin Aug 29 '18 at 11:43
I suggest you take a look at this post: https://physics.stackexchange.com/questions/53465/what-is-the-proof-that-a-force-applied-on-a-rigid-body-will-cause-it-to-rotate-a – Vincent Thacker Aug 29 '18 at 11:51
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If the rod could rotate about another point by itself, a centripetal force is required (such as a pivot) to keep the center of mass traveling in a circle around the point. The center of mass is the only point of rotation where no external force is required. If you switch to the center of mass frame (which ignores the translation), the force will simply cause a pure rotation. Separating the movement into translation and rotation, we can see this more clearly. – Vincent Thacker Aug 29 '18 at 12:06
So the motion will be rotation as well as translation about centre of mass frame? – Aladdin Aug 29 '18 at 12:38
Yes. An only possible exception to this is where the applied force is itself centripetal in nature. A more fundamental answer can be found here: https://physics.stackexchange.com/a/151382/174766 – Vincent Thacker Aug 29 '18 at 12:52
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@GENESECT If it rotates about anything other than the COM, then momentum is not conserved. – JEB Aug 29 '18 at 18:20

score 2 · Answer 2 · answered Aug 29 '18 at 17:07

The laws of motion state the following:

The net force acting on a free body describe the motion of the center of mass (only the point).
The net torque about the center of mass describe the motion about the center of mass (rotation).

An eccentric load-line causes the center of mass the move linearly (in line with the force) and the body to rotate about the center of mass due to the moment arm of the force w.r.t. the center of mass.

The combined motion yields an instant center of rotation that is "on the other side" of the center of mass as the applied load and at a distance from the center of mass. The distance of the instant center of rotation to the center of mass is $$ c = \frac{I}{m\, d}$$

_{Here $m$ is the mass, $I$ is the mass moment of inertia and $d$ the moment arm of the load.}

Specifically, with your question, each (small) part of the body has only linear motion and no rotation (a point cannot rotate). The rotation of a rigid body is an emergent property from the combined motion of all the particles in the body. This becomes clear when you can derive the equation of motion of a rigid body from the definition of momentum of each individual particle.

score 2 · Answer 3 · answered Aug 29 '18 at 18:41

Imagine a mass described by a density function $\rho({\vec r})$ rotating about a point ${\vec r}_0$ with angular velocity $\vec{\omega}$. The linear momentum of the mass is:

$${\vec p}=\int{\rho({\vec r})\vec v(\vec r)d^3r} $$

where the velocity of a point is given by:

$$\vec v(\vec r)=\vec\omega\times[\vec r-\vec r_0] $$

So:

$$\vec p =\int{\vec{\omega}\times[\vec r - \vec r_0]\rho({\vec r})d^3r} $$

Since there are no external forces:

$$ \vec p = 0$$

Hence:

$$ \int{[\vec\omega\times\vec r]\rho(\vec r)d^3r}=\int{[\vec\omega\times\vec r_0]\rho(\vec r)d^3r} $$

At this point, you can solve for $\vec r_0$, or plug in the center-of-mass:

$$ \vec r_{com}\equiv \frac{\int{\vec r\rho(\vec r)d^3r}}{\int{\rho(\vec r)d^3r}}$$

and verify that it is the solution.

Rotation around which axis

3 Answers3