Rotational Mechanics:Small Problems

[BobbyBear]

Please explain these points giving necessary examples.

1.The moment induced by two equal and opposite forces is equal to the moment of one force about the point of action of the other. It doesn’t matter which force you use to do this calculation.

That is because the moment produced by a couple (two forces of equal magnitude and opposing direction) is constant through all of space, and its value is F times d, where d is the separation between the lines of action of the two forces. Obviously, if you take the moment with respect to a point on the line of action of one of the forces, that force's contribution will be zero. But regardelss of the point about which you take it, the moment produced by the couple,
\vec{M} = \vec{r_1}\times \vec{F_1} + \vec{r_2}\times \vec{F_2}
is the same w.r.t. every point of space. Try it out and see that it's true! :P

2.If an object translates in space in a circular trajectory, with constant speed,then a centripetal acceleration, and hence a centripetal force, is needed, so the object cannot be in equilibrium.
But if an extended object does not translate, it only rotates
around a fixed axis, in the absence of dissipative forces and with no other forces, it will continue to rotate.

I'd like to simply lay down a couple of ideas first that might help. Thinking of a particle translating in a circular trajectory with constant speed, the centripetal force is always perpendicular to the trajectory (or velocity vector), and therefore is doing no work upon the particle. In the absence of any other force, the (kinetic) energy of the particle must be conserved (so I think this is simply an application of the law of conservation of energy, yes?)

In the case of the solid rotating without translating about an axis, you could apply this reasoning to each individual particle?
Or alternatively, you could apply Newton's second law for rotation to the solid:
\vec{M} = I \frac{d \vec{L}}{d t}}
: since there is no external moment, the angular momentum and thus the angular velocity of the solid is conserved.

Can someone comment on these ideas, please?

The second point speaks first about a mass point and then about a rigid body-which contains a collection of mass points-. A mass point under no external force will, by Newton's second law, continue along a straight line with its original speed; it needs a centripetal force to revolve about an axis in space. But in case of the rigid body mentioned, the individual mass points not lying on the axis of rotation will be subjected to radial forces from other mass points though there is no external force on the body as a whole. So apply Second law for individual mass points and you see why they go along a circle around an axis passing through the COM of the rigid body; apply it for the entire rigid body and you see why the COM doesn't accelerate.

Now I have a slight problem . . . firstly, with respect to sganesh88's comment above, does the rotation of the solid have to be about an axis passing through its COM? Urmi only talked about 'an axis', so why did you assume an axis passing through the COM? If the axis of rotation was not through the COM, would that mean that the solid would be translating as well as rotating (because the COM would be describing some trajectory)? And how do you apply Newton's law of rotation to a solid?? I mean:

what is meant by "net torque on a rod" for example? We always talk about torque of a force wrp a point, so what does the 'torque of a solid' mean? You don't always have a 'couple' acting upon a solid, do you?

Say, if you have a rod with a magnet at one end suspended in a magnetic field, there will be a force upon the rod but it will only be applied at one end of the rod. The rod will rotate until its length is aligned with the magnetic field, but about which point is it rotating? What is the torque on the rod? Since the torque (due to the magnetic force) w.r.t. evert point in space is different, how can we talk of "the torque" on the rod, and if we do, what exactly do we mean by it?

 

[D H]

I'd like to simply lay down a couple of ideas first that might help. Thinking of a particle translating in a circular trajectory with constant speed, the centripetal force is always perpendicular to the trajectory (or velocity vector), and therefore is doing no work upon the particle. In the absence of any other force, the (kinetic) energy of the particle must be conserved (so I think this is simply an application of the law of conservation of energy, yes?)

This is true only for an object rotating around one of its principle axes. In general, it is not true. The motion in general is not circular, even in the absence of external forces / torques.

Or alternatively, you could apply Newton's second law for rotation to the solid:
\vec{M} = I \frac{d \vec{L}}{d t}}
: since there is no external moment, the angular momentum and thus the angular velocity of the solid is conserved.

Can someone comment on these ideas, please?

BobbyBear, what you wrote is the freshman physics version of the rotational analog of Newton's second law. This expression is only true under special circumstances. Freshman physics rotational mechanics problems are carefully constructed so as to make this simple form applicable. Angular momentum is always conserved in the absence of external torques. Angular velocity is not.

For a rigid body, the inertia tensor is constant in a frame fixed with respect to the body. That does not necessarily mean it is constant in an inertial frame. The problem is that the inertia tensor is a second order tensor, not a simple scalar. Suppose no external forces act on a rigid body. Angular momentum is conserved along with linear momentum and energy. The angular momentum of the body is \boldsymbol L = \mathbf I \boldsymbol{\omega}. Because the inertia tensor is not necessarily constant in an inertial frame, angular velocity is not conserved. There is no such thing as a law of conservation of angular velocity.

Because the inertia tensor is constant in a frame fixed with respect to a constant mass rigid body, this is one of those cases where working in a rotating frame is easier than working in an inertial frame. The derivative of any vector quantity \boldsymbol q from the perspectives of a rotating frame versus an inertial frame are related by

\left.\frac{d \boldsymbol q}{dt}\right|_{\text{inertial}} =<br /> \left.\frac{d \boldsymbol q}{dt}\right|_{\text{rotating}} +<br /> \boldsymbol{\omega}\times \boldsymbol q

Applying this to the angular momentum vector \boldsymbol L = \mathbf I \boldsymbol{\omega} and taking advantage of the fact that the inertia tensor is constant in the rotating frame,

\left.\frac{d \boldsymbol L}{dt}\right|_{\text{inertial}} =<br /> \mathbf I \left.\frac{d \boldsymbol \omega}{dt}\right|_{\text{rotating}} +<br /> \boldsymbol{\omega}\times (\mathbf I \boldsymbol{\omega})

Back to Newton's second law. F=ma is also a freshman physics version of Newton's second law. A more general expression is d\boldsymbol p/dt = \boldsymbol F_{\text{ext}}, where \boldsymbol p is the linear momentum of the object in question and F_{\text{ext}} is the external force acting on the object. The rotational analog of this is d\boldsymbol L/dt = \boldsymbol {\tau}_{\text{ext}}, where \boldsymbol {\tau}_{\text{ext}} is the external torque acting on the body. This expression is only true from the perspective of an inertial frame. Combining this with the rotating frame expression yields

\mathbf I \left.\frac{d \boldsymbol \omega}{dt}\right|_{\text{rotating}} +<br /> \boldsymbol{\omega}\times (\mathbf I \boldsymbol{\omega})<br /> = \boldsymbol {\tau}_{\text{ext}}

In the absence of external torques, the angular velocity will be constant only if the angular velocity is zero or if it is parallel to the the angular momentum.

Now I have a slight problem . . . firstly, with respect to sganesh88's comment above, does the rotation of the solid have to be about an axis passing through its COM?

The motion of a rigid body can always be expressed as a translation of the center of mass plus a rotation about an axis through the center of mass. Moreover, when one models the motion in this way, the translational and rotational mechanics are decoupled. (This is not true for a non-rigid body.)

 

[BobbyBear]

*flails*
Okay, D.H. I think I'm a bit of a freshbear as far as solid dynamics goes :P I shall strive to brush up a little on the subject when I can - thanks for your insights, which I shall take some time to process :P

And as far as my other question (about the rod), you said:

The motion of a rigid body can always be expressed as a translation of the center of mass plus a rotation about an axis through the center of mass. Moreover, when one models the motion in this way, the translational and rotational mechanics are decoupled. (This is not true for a non-rigid body.)

Yes, I know that . . . so does that mean that if the solid is rotating about an axis that does not contain its centre of mass, its movement is not a pure rotation: it can be viewed as the superposition of a translation of its centre of mass and a rotation about an axis through the centre of mass. Is that correct then?

 

[D H]

BobbyBear,

What I didn't tell you is that the instantaneous motion of a rigid body can be expressed as the linear translation of any point in the body plus a rotation about an axis passing through that point.

A freely rotating object in space can be viewed as rotating about any axis parallel to the central axis. Asking which axis is the correct one implies that all others are somehow "incorrect". This is not the case. The equations of motion, when done correctly, will yield the same results regardless of which point is chosen as the central point. The reason for choosing the center of mass as the central point is because this choice makes the translational and rotational dynamics decouple. This makes the math easier.

An example of where a different choice might make more sense is a rocket. A rocket that launches something into orbit is oriented vertically at takeoff but is oriented horizontally when it reaches orbital velocity about ten minutes after takeoff. Meanwhile, the rocket has burned off 90% or so of if its mass. The center of mass of a rocket moves as the rocket consumes fuel. Some people prefer to model the rotational dynamics in a frame fixed with respect to a point on the rocket structure; others prefer to model the rotational dynamics with a center of mass frame. Which is correct? Wrong question. Both viewpoints are correct. Which is easier? Both viewpoints are messy. The rocket is not a constant mass rigid body. The translational and rotational equations of motion do not decouple in either a center of mass frame or a structural frame (or any other frame).

 

[sganesh88]

Now I have a slight problem . . . firstly, with respect to sganesh88's comment above, does the rotation of the solid have to be about an axis passing through its COM?

With no external force, rotation must be about an axis passing through the COM.

what is meant by "net torque on a rod" for example? We always talk about torque of a force wrp a point, so what does the 'torque of a solid' mean? You don't always have a 'couple' acting upon a solid, do you?

Yes. It doesn't make sense under the definition of torque. But if someone says so, it has to mean torque about the COM or a pivot point.

 

[sganesh88]

I'd like to simply lay down a couple of ideas first that might help. Thinking of a particle translating in a circular trajectory with constant speed, the centripetal force is always perpendicular to the trajectory (or velocity vector), and therefore is doing no work upon the particle. In the absence of any other force, the (kinetic) energy of the particle must be conserved (so I think this is simply an application of the law of conservation of energy, yes?)

This is true only for an object rotating around one of its principle axes. In general, it is not true. The motion in general is not circular, even in the absence of external forces / torques.

But he was talking about particles only. So i think he's right; Motion of a particle when the force acting on it is continuously perpendicular to its velocity vector is circular.

 

[D H]

But he was talking about particles only. So i think he's right; Motion of a particle when the force acting on it is continuously perpendicular to its velocity vector is circular.

Two points:

1. Urmi Roy in post #95 was not talking about particles. He was talking about rotation, not revolution.
2. Classically, particles have zero angular momentum due to rotation. It is a bit silly to talk about rotational mechanics with respect to particles.

 

[sganesh88]

Two points:

1. Urmi Roy in post #95 was not talking about particles. He was talking about rotation, not revolution.
2. Classically, particles have zero angular momentum due to rotation. It is a bit silly to talk about rotational mechanics with respect to particles.

Sorry. I thought you were addressing the quoted text of Bobbybear which talked about revolution of a particle about some point; not rotation(which of course doesn't make sense as you said). And Urmi Roy is a female i guess; the tone of her messages suggests that. ;-)

 

[iitjee10]

it would be better if you start a new thread for that

 

[BobbyBear]

Heylo :)

okay with regard to post 104, thank you D H for your insights. As far as I know, the central axis is the locus of points with the smallest linear velocity (if this velocity is zero, it is termed the instantaneous axis of rotation). Thus the movement of a rigid body is the superposition of a translation given by the velocity of the points on the central axis, plus a rotation about this axis. I never thought about describing the motion as a rotation around any other axis . . . but I suppose what you say makes sense, because all movement is relative? And I'm not sure how all this ties in with the centre of mass either.. and how the equations of rotation and translation become decoupled. Anyhow, I think it's best for me to actually study the equations in order to see all this more clearly. I don't have time right now, but maybe in a few weeks . . . anyway, may I ask if you could suggest any good textbooks explaining these thigns, as I do not have any at the moment? I'll keep your posts so as to go over them again when I do get down to studying them in more depth : ) Thanks again for your insights xD


About post 106, actually I was argumenting why the energy is conserved for a particle moving in a circular trajectory, which I think was the first consideration that Urmi brought up in point No.2 of post 95 . . . and I was trying to extrapolate this reasoning to a rotating solid by argumenting that each individual particle of that solid could be viewed as a particle describing a circular translational motion, and thus the energy of each particle is conserved, and hence the energy of the solid is conserved. Correct me if I'm wrong, but it seems to me that there are only three fundamental laws of motion (at least as far as Newtonian mechanics goes), and these are the three laws that Newton put forth for particles. Solids are simply a collection of particles stuck together, and the expressions that describe/govern the dynamics of solids are merely derived from the principles put forth for particles, using the adequate maths. That is, they are not new principles, simply operative expressions that can be applied directly to finite bodies - but in the end, they simply say the same thing as applying Newton's principles to each particle of the solid and tying all these movements together would say about the movement of the solid, if you get what I mean :P

Thank you sganesh for clearing my doubt about 'torque of a solid'. But still, if you were to have a solid suspended in space, and apply only one force say at one end of the solid (not through its COM), would it rotate at all? (I'm assuming there are no physical constraints upon the solid at all - no physical pivot or anything). And about which point would it rotate? I can't seem to imagine why, if it should rotate, it should do so about any particular point.

 

[sganesh88]

Thank you sganesh for clearing my doubt about 'torque of a solid'. But still, if you were to have a solid suspended in space, and apply only one force say at one end of the solid (not through its COM), would it rotate at all? (I'm assuming there are no physical constraints upon the solid at all - no physical pivot or anything). And about which point would it rotate? I can't seem to imagine why, if it should rotate, it should do so about any particular point.

I too had that doubt a year before. I thought rotation is possible only in the presence of a pivot constraint or two opposite forces whose lines of actions are separated by a distance. But the body has to rotate if the line of action of the force doesn't pass through the COM (even if its just a single force). Why? Check whether conservation of angular momentum is conserved if the body doesn't rotate but just translates as a whole when a force acts on its tip. (Though i get a doubt. Why should angular momentum be conserved? Going to read about it now.)

Regarding the axis of rotation, draw a line from the COM to the point of application of the force; the perpendicular to this line, passing through the COM will be the axis.

 

[BobbyBear]

Regarding the axis of rotation, draw a line from the COM to the point of application of the force; the perpendicular to this line, passing through the COM will be the axis.

Hmm, if that is the axis of rotation, then there is a net torque with respect to that axis, and as you said, the 'torque' is always with respect to the COM . . .
D H said that in an inertial frame, we have: <br /> d\boldsymbol L/dt = \boldsymbol {\tau}_{\text{ext}}<br />, so the solid must acquire an angular momentum with respect to an inertial frame? And also as D H said, this does not mean that it must acquire an angular velocity, as the inertia tensor is not constant in an inertial frame - though again, if the body does not rotate, then the inertia tensor would be constant, so it must rotate . . . . Hence if the body does not rotate at all, it would mean that angular momentum is conserved, no? So that can't be true if there's an external force acting on the solid with a certain eccentricity with respect to the centre of mass. Anyhow, I'm a bit dodgy with all of this, I think I need to follow some textbook that develops the general equations and explains all this with certain depth xD

 

[sganesh88]

Torque needn't always be w.r.t COM but in this case of a solid suspended in space, this is preferable.
And i know nothing about this inertia tensor thing. Yet to study about it.

 

[BobbyBear]

Torque needn't always be w.r.t COM but in this case of a solid suspended in space, this is preferable.
And i know nothing about this inertia tensor thing. Yet to study about it.

*flails* that's just the point, you can consider the torque produced by a force with respect to any point, but the motion of the solid cannot depend on which point you choose to consider to apply Newton's law of rotation . . . I think I need to study about it too *wink* if you come across any good books let me know their names xD

 

[D H]

BobbyBear:

To answer your first question regarding the decoupling of translational and rotational equations of motion, consider a system comprising a fixed number of constant mass particles. Such a system will have a constant mass and the individual components won't change (e.g., there are no chemical interactions that change the number of particles).

Denote the mass and position with respect to some inertial frame of the i^th particle as m_i and x_i. The mass, center of mass of the system, and system center of mass velocity are

\aligned<br /> m_{\text{tot}} &amp;\equiv \sum_i m_i \\<br /> m_{\text{tot}} \boldsymbol x_{\text{CM}} &amp;\equiv \sum_i m_i \boldsymbol x_i \\<br /> m_{\text{tot}} \boldsymbol v_{\text{CM}} &amp;\equiv \sum_i m_i \dot{\boldsymbol x}_i<br /> \endaligned

The total linear momentum is the sum of the linear momentum for each particle:

\boldsymbol p_{\text{tot}} = \sum_i m_i \dot{\boldsymbol x}_i

Note that this is the total mass times the system center of mass velocity. Differentiating with respect to time,

\dot{\boldsymbol p}_{\text{tot}} = \sum_i m_i \ddot{\boldsymbol x}_i

Assume each particle obeys Newton's laws of motion:

m_i\ddot{\boldsymbol r}_i = \boldsymbol F_{\text{tot},i}

External and internal forces will act on each particle. The external forces come from outside the system; the internal forces are interactions between pairs of particles. The total force on the i^th particle is the sum of the external force on that particle plus the interactions with the other particles in the system:

\boldsymbol F_{\text{tot},i} =<br /> \boldsymbol F_{\text{ext},i} +<br /> \sum_{j\ne i} \boldsymbol F_{\text{int},ij}

Thus

\aligned<br /> \dot{\boldsymbol p}_{\text{tot}} &amp;= \sum \boldsymbol F_{\text{tot},i} \\<br /> &amp;= \sum_i \left(\boldsymbol F_{\text{ext},i} +<br /> \sum_{j\ne i} \boldsymbol F_{\text{int},ij} \right) \\<br /> &amp;= \boldsymbol F_{\text{ext},\text{tot}} +<br /> \sum_{\text{pairs\,}i,j} (\boldsymbol F_{\text{int},ij} + \boldsymbol F_{\text{int},ji}<br />

By the weak form of Newton's third law, interactions between particles are equal but opposite: \boldsymbol F_{\text{int},ji} = - \boldsymbol F_{\text{int},ij} and thus the change in linear momentum of the system as a whole is equal to the sum of the external forces acting on elements of the system:

\dot{\boldsymbol p}_{\text{tot}} = \boldsymbol F_{\text{ext},\text{tot}}

Using the definition of the system center of mass,

m_{\text{tot}} \ddot{\boldsymbol x}_{\text{CM}} = \boldsymbol F_{\text{ext},\text{tot}}

The system as a whole obeys Newton's second law so long the particles obey the Newton's second and third laws. What about the angular momentum? The total angular momentum of the system is the sum of that of the individual particles:

\boldsymbol L_{\text{tot}}<br /> = \sum_i \boldsymbol L_i <br /> = \sum_i m_i \boldsymbol x_i \times \dot{\boldsymbol x}_i

Differentiating with respect to time,

\aligned<br /> \dot{\boldsymbol L}_{\text{tot}}<br /> &amp;= \sum_i m_i \boldsymbol x_i \times \ddot{\boldsymbol x}_i \\<br /> &amp;= \sum_i \boldsymbol x_i \boldsymbol F_{\text{ext},\text{tot}} \\<br /> &amp;= \sum_i \boldsymbol x_i\times \left(\boldsymbol F_{\text{ext},i} +<br /> \sum_{j\ne i} \boldsymbol F_{\text{int},ij} \right) \\<br /> &amp;= \left(\sum_i \boldsymbol x_i\times \left(\boldsymbol F_{\text{ext},i}\right) +<br /> \left(\sum_{\text{pairs\,}i,j}<br /> (\boldsymbol x_i\times \boldsymbol F_{\text{int},ij} +<br /> \boldsymbol x_j\times \boldsymbol F_{\text{int},ji}\right) \\<br /> &amp;= \left(\sum_i \boldsymbol x_i\times \left(\boldsymbol F_{\text{ext},i}\right) +<br /> \left(\sum_{\text{pairs\,}i,j}<br /> \boldsymbol x_i \times (\boldsymbol F_{\text{int},ij} +\boldsymbol F_{\text{int},ji}) +<br /> (\boldsymbol x_j-\boldsymbol x_i)\times \boldsymbol F_{\text{int},ji}\right)<br /> \endaligned

The terms \boldsymbol F_{\text{int},ij} +\boldsymbol F_{\text{int},ji} vanish per the weak form of Newton's third law. The final set of terms, (\boldsymbol x_j-\boldsymbol x_i)\times \boldsymbol F_{\text{int},ji}, vanish per the strong form of Newton's third law. Thus the total angular momentum depends only on the external forces acting on individual particles:

\dot{\boldsymbol L}_{\text{tot}}<br /> = \sum_i \boldsymbol x_i\times \left(\boldsymbol F_{\text{ext},i}


The linear and angular momenta for a rigid body can be expressed as

\aligned<br /> \boldsymbol p &amp;=<br /> m \dot{\boldsymbol x}_{\text{CM}} \\<br /> \boldsymbol L &amp;=<br /> m \boldsymbol x_{\text{CM}}\times \dot{\boldsymbol x}_{\text{CM}} +<br /> \mathbf I \boldsymbol {\omega}<br /> \endaligned

When expressed in this form, the translational and rotational equations of motion decouple. The translational equations of motion take on the form

m \ddot{\boldsymbol x}_{\text{CM}} = \boldsymbol F_{\text{ext,tot}}

The rotational equations of motion were already described in post #102.

 

[BobbyBear]

THANK YOU D H,
that was very helpful xD