Angular velocity of a rod and what formula to use while solving.

[brochesspro]

TL;DR Summary

For obtaining the velocity of a rod slipping on a frictionless floor, why should we use the principle of conservation of energy and not the SUVAT equations, and is the normal reaction a conservative force?

The question is:
A uniform rod of length $L$ stands vertically upright on a smooth floor in a position of unstable equilibrium. The rod is then given a small displacement at the top and tips over. What is the rod's angular velocity when it makes an angle of 30 degrees with the floor, assuming the rod does not slip?

I do know what steps to take to solve this problem, and I got the correct answer: $$\omega = {\left(\frac{24g}{13L}\right)}^{1/2}$$

The method is to apply the law of conservation of energy and use a constraint relation of $v = \frac{\sqrt 3}4 L\omega$, to obtain the given answer. You can ask me if you need my help getting the constraint relation.

My question is, why are we applying the law of conservation of energy? It is applied when only conservative forces act on a system, right? But here, along with the earth's gravitational force, normal force by the floor is being applied too.

My other question is whether the acceleration of the rod's centre of mass is constant or not, which would mean that the normal force is constant. Thus, if that is so, we can find the angular velocity of the rod using the SUVAT equations.

That is it, I guess. Thank you for going through my question.

 

[jbriggs444]

A uniform rod of length $L$ stands vertically upright on a smooth floor in a position of unstable equilibrium. The rod is then given a small displacement at the top and tips over. What is the rod's angular velocity when it makes an angle of 30 degrees with the floor, assuming the rod does not slip?

I have an issue with the question. We are told that the floor is smooth (i.e. frictionless). But we are also told to assume that the rod does not slip. Which is it?

My question is, why are we applying the law of conservation of energy? It is applied when only conservative forces act on a system, right? But here, along with the earth's gravitational force, normal force by the floor is being applied too.

The normal force on the rod by the floor acts over what displacement?

 

[A.T.]

My question is, why are we applying the law of conservation of energy? It is applied when only conservative forces act on a system, right?

You can apply conservation of energy to anything that has no net energy input/output.

But here, along with the earth's gravitational force, normal force by the floor is being applied too.

The normal force does no work in the rest frame of the floor, so it's not an energy input/output.

 

[brochesspro]

I have an issue with the question. We are told that the floor is smooth (i.e. frictionless). But we are also told to assume that the rod does not slip. Which is it?
The normal force on the rod by the floor acts over what displacement?

Sorry, the rod is slipping, the not slipping part was a mistake made by me while typing the question, probably because of the pure rolling questions I have solved.
Does the normal force not act throughout the motion? Since the rod never once leaves contact with the ground.

You can apply conservation of energy to anything that has no net energy input/output.The normal force does no work in the rest frame of the floor, so it's not an energy input/output.

Could you give an example for your first statement? Is my statement wrong by any chance?
Also, I do not understand why contact forces like the frictional force(not present in this case) and the normal reaction do zero work in many cases. Why does that occur? For example, the work done by the frictional force on a pure rolling object is zero.

 

[jbriggs444]

Does the normal force not act throughout the motion? Since the rod never once leaves contact with the ground.

Again, what is the displacement through which the normal force acts?

Could you give an example for your first statement? Is my statement wrong by any chance?
Also, I do not understand why contact forces like the frictional force(not present in this case) and the normal reaction do zero work in many cases. Why does that occur? For example, the work done by the frictional force on a pure rolling object is zero.

The normal force is a contact force. If the surface is not moving in the direction of the normal force, then the target object's mating surface must also not be moving in the direction of the normal force. So the dot product of the displacement of the mating surface times the normal force must be zero. That dot product defines the work done by the normal force.

Are you, perhaps, considering the dot product of the normal force and the motion of the center of mass of the target body? That dot product does not, in general, equate to the work done by the normal force. It will match for point-like bodies, for rigid non-rotating bodies and can match, on occasion, by accident.

The work energy theorem has two forms. One relates to real work. The other relates to what I like to call "center of mass work". The work terms in the version about real work involve displacement of the surfaces involved. The work terms in the version about center of mass work involve displacement of the center of mass. The resulting kinetic energy in the real work version is the sum of the kinetic energies of the parts of the object -- it includes rotation and internal motion. The resulting kinetic energy in the center of mass version is the bulk motion of the object as a whole -- it excludes rotation and internal motion.

If you want to invoke conservation of energy, you should be using real work and not center of mass work. Or you will lose track of the kinetic energy associated with rotation or of other internal motion.

 

[brochesspro]

Again, what is the displacement through which the normal force acts?

Oh, if you mean the value, it is half the length of the rod, or $frac L 2$, since the centre of mass moves vertically down, there is no horizontal displacement and thus the bottommost point only moves half the rod's length.

The work energy theorem has two forms. One relates to real work. The other relates to what I like to call "center of mass work". The work terms in the version about real work involve displacement of the surfaces involved. The work terms in the version about center of mass work involve displacement of the center of mass. The resulting kinetic energy in the real work version is the sum of the kinetic energies of the parts of the object -- it includes rotation and internal motion. The resulting kinetic energy in the center of mass version is the bulk motion of the object as a whole -- it excludes rotation and internal motion.

I assume that this is because the centre of mass of a body only experiences translational motion, but why is rotational motion not considered? What if the body is revolving around a point? I think rotation is only neglected when the centre of mass rotates about its own axis, which is negligible. I do not understand why the real work-energy theorem encompasses the rotational and vibratory motions though.

If you want to invoke conservation of energy, you should be using real work and not center of mass work. Or you will lose track of the kinetic energy associated with rotation or of other internal motion.

I understood what you are saying, but I fail to understand why that is the case, I guess you could try elaborating a bit more on the real work-energy theorem and the centre of mass work-energy theorem. Thanks.

 

[jbriggs444]

Oh, if you mean the value, it is half the length of the rod, or $frac L 2$, since the centre of mass moves vertically down, there is no horizontal displacement and thus the bottommost point only moves half the rod's length.

The normal force does not act on the center of mass. It acts on the lower tip of the rod. That lower tip has zero displacement in the direction of the force.

I assume that this is because the centre of mass of a body only experiences translational motion, but why is rotational motion not considered?

Because you are using the wrong notion of work.

What if the body is revolving around a point?

Let us not go there. It is a quagmire.

I think rotation is only neglected when the centre of mass rotates about its own axis, which is negligible.

I do not know what it would even mean for a center of mass to rotate about "its own axis".

I do not understand why the real work-energy theorem encompasses the rotational and vibratory motions though.

I understood what you are saying, but I fail to understand why that is the case, I guess you could try elaborating a bit more on the real work-energy theorem and the centre of mass work-energy theorem. Thanks.

I will try to elaborate. In my experience, it is something that is taught poorly. There is good reason for teaching this poorly. The idea is to avoid confusing the student by teaching too much, too fast.

There are a number of definitions for "work". They do not (in my experience) have standard names that everyone agrees on and uses. I will use two definitions. [The other definitions that I've seen are not relevant -- they are generalizations like thermodynamic work]

1. "Real work"

This is the work done by one force acting on one part of a body over the course of some action. For example, we might consider the work done by the sidewalk on the sole of your shoe.

We multiply the force of the sidewalk on your shoe by the displacement of the sole of your shoe while you take one stride with that shoe on the ground.

We've multipled a force times a [parallel] displacement to yield an amount of work.

[In the case of a stride of a walker, the ground has done zero work on the sole of the shoe because the displacement of the sole was zero].

2. "Net work" or "Center of mass work"

This is the work done by the sum of all forces acting on all parts of a body over the course of some action. For example, we might consider the work done on your body over the course of one stride.

We multiply the force of the sidewalk on your shoe (minus air resistance, if any) by the motion of your body's center of mass during the course of that stride.

[In the case of a stride of a walker, non-zero work has been done on the walkers body since the force was non-zero and the [parallel] displacement of the walker's center of mass was also non-zero]I do not want to leave you with the impression that one of these definitions is "right" and that the other is "wrong". Both notions are useful. They are useful in different situations when computing different things.As I mentioned previously, there is a version of the work energy theorem for each of these definitions. I like to think that the version that goes with "real work" is about conservation of energy and that the version that goes with "center of mass work" is about conservation of momentum.

The work energy theorem for "real work"...

Let us consider "real work". It is a statement about the mechanical energy passing through an interface with a system by virtue of an applied force. This version of the work energy theorem says that:

"The energy transferred by a force to a system over an interface is equal to the product of the displacement [of the target] at the interface and the force across the interface".$$W=\vec{F}\cdot \vec{\Delta S}$$You may be thinking to yourself, "wait just a minute here -- we sped up when we took that step. Surely there was an energy increase. How can there be zero work done!!"

Yes indeed. Work was done. But not by the ground on the soles of your feet. Work was done when your thigh muscles contracted causing your knee to bend. Work was done when your calf muscle contracted causing your ankle to extend.

When we are talking about "real work", internal forces can increase the mechanical energy of a system.

Gun powder can do real work on a system consisting of a gun plus cartridge plus bullet.

Your car's engine can do real work on a system consisting of the car plus engine plus transmission plus wheels plus tires.

Your muscles can do real work on the system consisting of your body.The work energy theorem for "center of mass work"...

Let us consider "center of mass work" this is a statement on how Newton's second law applies to exteral forces on a system considered as a black box.

"The bulk kinetic energy imparted by a set of forces on a system is equal to the product of the displacement of the center of mass and the total force across all system interfaces".$$W = ( \sum_i \vec{F_i} ) \cdot \Delta S_\text{COM}$$Note that the "bulk kinetic energy" of the system is taken as $KE = \frac{1}{2}m_\text{tot}v_\text{COM}^2$

The proof for this version of the theorem is just a simple algebraic manipulation of Newton's second law.

When we are talking about "center of mass" work, no notice is taken when different parts of the system have differing velocities.As a rule of thumb, if you have a rotating system or a system with moving parts and you want to account for the energy tied up in the rotation or relative motion of the internal parts, you need "real work".

But if you are talking about a non-rotating blob and do not care about any internal motion then "center of mass work" is your friend.

If you are talking about a rigid, non-rotating object or about a point-like object then it does not matter since both definitions of work are equivalent.

 

[brochesspro]

I do not know what it would even mean for a center of mass to rotate about "its own axis".

Like, the earth rotates on its axis, but a point like the centre of mass has no dimensions, so rotation about an axis is meaningless.

[In the case of a stride of a walker, the ground has done zero work on the sole of the shoe because the displacement of the sole was zero].

I do not understand this part, does the contact force include frictional force too? And why is the displacement of the sole zero? Reminds me of friction in pure rolling, to be honest, I never understood why the work done there is zero.
Does the work-energy theorem not say that "The work done by all the forces acting on the body is equal to the change in kinetic energy of the body?" I think you say that this is the centre of mass definition of work, and should not be used on a system which has parts moving in multiple ways(rotational and vibrational motion).

"The energy transferred by a force to a system over an interface is equal to the product of the displacement [of the target] at the interface and the force across the interface".

Why energy, and not kinetic energy? Is there any significance?

And in my original problem, the conservation of energy can be used as only conservative forces(gravity) and normal force (whose real work is zero), but why not SUVAT equations? If they can not be used, then that would mean that the acceleration of the rod is not constant, and hence, the normal force is not constant(since weight is constant), could you explain this point?
Also, thanks a lot for making me aware of the nuances of the work-energy theorem and the differences between the two types of work. I am sure it took a lot of time to type. Also, is there any material I can refer to for further clarity on this topic?

 

[malawi_glenn]

Why not assume the rod is supported via a smooth pivot around its end point? E.g. like a physical pendulum? Makes it much easier to grasp.

Anyway you can use external torque = I times alpha. But you can not use suvat because the torque is not constant in time you have to integrate.

 

[jbriggs444]

Like, the earth rotates on its axis, but a point like the centre of mass has no dimensions, so rotation about an axis is meaningless.

I agree that rotation of the center of mass about an axis passing through the center of mass is meaningless.

I do not understand this part, does the contact force include frictional force too? And why is the displacement of the sole zero? Reminds me of friction in pure rolling, to be honest, I never understood why the work done there is zero.

Yes, a contact force usually has both a frictional component and a normal component. The sole of your shoe is motionless on the pavement below. There is no motion in the normal direction. So the normal component does no work. There is no motion in the tangential direction either. So the frictional component does no work.

Yes, this is the same situation in pure rolling. The material at the contact patch is motionless.

[Assuming that we are using the road frame anyway. If we changed to a frame of reference in which the vehicle is stationary and in which the road is moving then the frictional force of road on rolling tire can do non-zero work. In the case of the treads moving rearward under a forward force from tha road, the work "done on" the tire by the road would be negative. Rather than absorbing energy into the system, we are delivering energy from the system out into the [moving] road. Naturally, that energy comes from the car's engine.

It can be jarring to think that the same force can deliver different amounts of energy or carry power at different rates or even in the opposite direction depending only on the frame of reference that one adopts. But you need to remember that kinetic energy depends on velocity and velocity depends on a frame of reference.

Work, Energy and Power are all frame-relative concepts. You have to do your accounting carefully if you plan to use them as "invariant" measures. The invariant work done by a third law force pair accounts for the motion of both participating bodies at the interface between them. You take the real work done by the one force plus the real work done by the third law partner. The total will be zero for the normal force. It can be non-zero for forces between objects in relative motion such as with friction, springs, pistons in cylinders, electrostatic forces and gravity. The total work for the pair will be invariant in any case.

You might want to file this whole section dealing with frames of reference and invariant measures to deal with in the future. It is not important for the problem at hand. I recall a nice little conversation with @Dale years back on this.

Oh, and try to get this material down pat before trying to critique arguments about DDWFTTW]

Does the work-energy theorem not say that "The work done by all the forces acting on the body is equal to the change in kinetic energy of the body?"

No. It does not say that.

The work energy theorem for "net work" or "center of mass work" says that the bulk kinetic energy of the body as a whole ($KE = \frac{1}{2}mv^2$) has increased by that much. The $m$ there being total body mass and the $v$ there being the velocity of the body's center of mass.

This misses part of the kinetic energy of a rotating body.

I think you say that this is the centre of mass definition of work, and should not be used on a system which has parts moving in multiple ways(rotational and vibrational motion).

Your system is a rotating rod. That rotation has mechanical energy. If you want to use a conservation of energy argument, you need to account for that mechanical energy.

Yes, I am saying that "center of mass work" will miss that energy and cannot be properly used in this conservation of energy argument.

Why energy, and not kinetic energy? Is there any significance?

You can do work across an interface without changing the kinetic energy of anything in the system. For instance, a drive belt acting on a wheel. Work is being done, even if the wheel is turning at a steady pace. It might, for instance, be grinding flour.

And in my original problem, the conservation of energy can be used as only conservative forces(gravity) and normal force (whose real work is zero), but why not SUVAT equations?

You do not need to have conservative forces to do an energy accounting. You just have to account for all of the energy transfers. Change in energy = energy in minus energy out.

It is just that with conservation, you have that energy in minus energy out = zero.

If they can not be used, then that would mean that the acceleration of the rod is not constant, and hence, the normal force is not constant(since weight is constant), could you explain this point?

Yes, the acceleration of the rod is not constant and the normal force is not constant. Which makes the SUVAT equations difficult to apply. You'll need to integrate, differentiate or, perhaps, solve a differential equation.

Which is why an energy conservation approach is attractive.

Also, thanks a lot for making me aware of the nuances of the work-energy theorem and the differences between the two types of work. I am sure it took a lot of time to type. Also, is there any material I can refer to for further clarity on this topic?

For additional material, I am not sure. The nuances took me a long time to acquire, after I'd already graduated from university. Doing exercises is relatively easy. Cataloguing the knowledge and fitting it into a coherent world-view that handles the apparent inconsistencies took longer in my case. @Doc Al here (who may have also been Uncle Al on Usenet) gave me the instruction I needed on this.

 

[brochesspro]

Yes, a contact force usually has both a frictional component and a normal component. The sole of your shoe is motionless on the pavement below. There is no motion in the normal direction. So the normal component does no work. There is no motion in the tangential direction either. So the frictional component does no work.

Okay, so here we evaluate the real work(work due to friction on the bottommost point of a rolling object since there is no relative motion), and not the centre of mass work, am I right?

No. It does not say that.

The work energy theorem for "net work" says that the bulk kinetic energy of the body as a whole (KE=12mv2) has increased by that much. The m there being total body mass and the v there being the velocity of the body's center of mass.

https://en.wikipedia.org/wiki/Work_(physics)#Work–energy_principle
Then what about this? Is there a mistake?

You can do work across an interface without changing the kinetic energy of anything in the system. For instance, a drive belt acting on a wheel. Work is being done, even if the wheel is turning at a steady pace.

Does this not contradict the work-energy theorem as stated in Wikipedia? Although I do agree with you.

Yes, the acceleration of the rod is not constant and the normal force is not constant. Which makes the SUVAT equations difficult to apply. You'll need to integrate, differentiate or, perhaps, solve a differential equation.

I do not understand why it is not constant though. Can you prove it mathematically?

For additional material, I am not sure. The nuances took me a long time to acquire, after I'd already graduated from university. Doing exercises is relatively easy. Cataloguing the knowledge and fitting it into a coherent world-view that handles the apparent inconsistencies took longer in my case. @Doc Al here (who may have also been Uncle Al on Usenet) gave me the instruction I needed on this

Okay, thanks for the info. I'll browse for the info first and if I am still not confident, I will ask him for advice.

 

[malawi_glenn]

Consider this first, for the "center of mass work" case.

A massless rod with length 1 m, is allowed to rotate without friction around a pivot at one of its end points. At the distance 1/3 m from the pivot there is a point-mass m fixed and at the distance 3/4 m from the pivot there is another point-mass m fixed. The system is dropped from rest at horisontal position. What is the total work performed on the system by the force of gravity when the rod is making the angle $\theta$ from the horisontal?

Method 1) Compute the work performed on each individual point-mass.

The force of gravity is directed vertically downwards, and the displacement in the vertical direction for the point-mass to the left is $\tfrac{1}{3} \sin \theta$ meters and for the point-mass to the right the vertical displacement is $\tfrac{3}{4} \sin \theta$ meters. The total work performed on the system is therefore $W = mg\tfrac{1}{3} \sin \theta + mg\tfrac{3}{4} \sin \theta= \tfrac{13}{12} mg \sin \theta$ Nm.

Method 2) Work performed by gravity on the CoM.
The CoM is located at the distance $\dfrac{m/3 + 3m/4}{m+m} = \dfrac{13}{24}$ meters from the pivot.

The vertical displacement of the CoM is $\dfrac{13}{24} \sin \theta$ meters.

The mass of CoM is $2m$.

The work performed by the force of gravity on the CoM is $W = 2m \cdot g \cdot \dfrac{13}{24} \sin \theta = \tfrac{13}{12} mg \sin \theta$ Nm, same as we got before.

Try to do these "toy-examples" yourself when in doubt.

SUVAT: https://www.ncl.ac.uk/webtemplate/a...mechanics/kinematics/equations-of-motion.html
are used when acceleration, a, is constant.

It is just not like its "hard" to use SUVAT eqs here, it is impossible.

I do not understand why it is not constant though. Can you prove it mathematically?

For the simpler case where the rod is attached to a pivot:

External torque w.r.t. P is $\mathcal{M_P} = mg \dfrac{L}{2} \sin \theta$ where $L$ is the length of the rod and $m$ is its mass.
Moment of inertia around P is $\mathcal{I}_P = \dfrac{mL^2}{3}$.
Equation of motion $\mathcal{M_P} = \mathcal{I}_P \alpha = \mathcal{I}_P \dfrac{\text{d}^2 \theta}{\text{d}t^2} = mg \dfrac{L}{2} \sin \theta$
Thus, the (angular) acceleration is not constant, the larger the angle $\theta$ is the larger the angular acceleration $\alpha$.

You need to solve the differential equation $\dfrac{\text{d}^2 \theta}{\text{d}t^2} = \dfrac{3g}{2L} \sin \theta$ subject to the constraints $\theta (0) = 0$ (or "close" to 0) and $\dfrac{\text{d} \theta}{\text{d}t} = 0$.
This is not easy.

For your original problem, there is still only gravity performing work on the rod, because i) no friction $F_\text{friction} = 0$ and ii) the normal force from the ground on the rod is not performing any work on the rod because there is no displacement of the contact point in the vertical direction.

 

[hutchphd]

Then what about this? Is there a mistake?

No

Does this not contradict the work-energy theorem as stated in Wikipedia? Although I do agree with you.

Read the system discription in Wikipedia article

I do not understand why it is not constant though. Can you prove it mathematically?

Yes

Okay, thanks for the info. I'll browse for the info first and if I am still not confident, I will ask him for advice.

There you go.

 

[brochesspro]

For the simpler case where the rod is attached to a pivot:
View attachment 324673
External torque w.r.t. P is $\mathcal{M_P} = mg \dfrac{L}{2} \sin \theta$ where $L$ is the length of the rod and $m$ is its mass.
Moment of inertia around P is $\mathcal{I}_P = \dfrac{mL^2}{3}$.
Equation of motion $\mathcal{M_P} = \mathcal{I}_P \alpha = \mathcal{I}_P \dfrac{\text{d}^2 \theta}{\text{d}t^2} = mg \dfrac{L}{2} \sin \theta$
Thus, the (angular) acceleration is not constant, the larger the angle $\theta$ is the larger the angular acceleration $\alpha$.

You need to solve the differential equation $\dfrac{\text{d}^2 \theta}{\text{d}t^2} = \dfrac{3g}{2L} \sin \theta$ subject to the constraints $\theta (0) = 0$ (or "close" to 0) and $\dfrac{\text{d} \theta}{\text{d}t} = 0$.
This is not easy.

For your original problem, there is still only gravity performing work on the rod, because i) no friction $F_\text{friction} = 0$ and ii) the normal force from the ground on the rod is not performing any work on the rod because there is no displacement of the contact point in the vertical direction.

I thought the angular acceleration was about the centre of mass of the rod, I guess I should have just calculated it bout a different point, but just for confirmation, both the values of the angular acceleration will be the same, right?

 

[jbriggs444]

I thought the angular acceleration was about the centre of mass of the rod, I guess I should have just calculated it bout a different point, but just for confirmation, both the values of the angular acceleration will be the same, right?

The orientation of a rod does not depend on what point we choose to regard as the center. The rate of change of orientation is invariant in this sense.

 

[malawi_glenn]

I thought the angular acceleration was about the centre of mass of the rod, I guess I should have just calculated it bout a different point, but just for confirmation, both the values of the angular acceleration will be the same, right?

Note that I gave you a simpler problem, the rod I described was around a fixed pivot and the CoM does not travel a straight line down. Anyway, you can choose any point you want.

For your problem, why CoM does not move at constant acceleration. For that to happen the normal force must be smaller than mg and constant. Now, why should the normal force be constant?