Rotational Work versus Linear Work

[alkaspeltzar]

In the link below, there is a force tangent to a wheel that travels distance Ds. There is also a torque that moved thru an angle. Why do they consider them equal? There are two works, shouldn't they be added together?

https://courses.lumenlearning.com/p...nal-kinetic-energy-work-and-energy-revisited/

Or is it because in this simple case, the force travels the distance DS, but the object relative to the pivot doesn't move. This force moved thru this distance traveled because this force creates a torque which moves theta, such that work calculated linearly or rotationally are both the same.

Thank for clarification. Just trying to understand how they are the same in this case since there is both a force and torque.

 

[jbriggs444]

For a rigid, rotating object, one can decompose the work done by a force acting at a point on the object into two parts:

1. The center-of-mass work done by multiplying [or integrating] the dot product of the force and the motion of the center of mass.

2. The rotational work done by mutiplying [or integrating] the dot product of the torque about the center of mass and the rotation of the object about its center of mass.

Or one could simply multiply [or integrate] the dot product of the force and the motion of the [object's material at the] point of application.

 

[russ_watters]

Summary:: In case of work done by a force, such as rotating a wheel in the link provided, why don't we add the rotational work and linear work? Are they the same thing in this example?

In the link below, there is a force tangent to a wheel that travels distance Ds. There is also a torque that moved thru an angle. Why do they consider them equal? There are two works, shouldn't they be added together?

https://courses.lumenlearning.com/p...nal-kinetic-energy-work-and-energy-revisited/

It's not so much that they are equal, but rather it is only one applied force, being expressed in two different ways.

 

[alkaspeltzar]

It's not so much that they are equal, but rather it is only one applied force, being expressed in two different ways.

So in Layman's term's, what you are saying is the force creates a torque. And in doing so, we can look at the work as the distance traveled thru the force(linear work) as being equal to the work done by torque. So really it is same work, just looked at two different ways, because of that relationship with force/torque on the body.

And as jbriggs44 said above, there is also no motion about the center of mass. IF it was an object traveling in a straight line, then hitting a pivoting door, energy has to be conserved. So in that case, we had kinetic energy, So therefore we only have the rotational kinetic energy. Not like we can have more than we started with.

That sound about right? It makes sense.

 

[Dale]

Summary:: In case of work done by a force, such as rotating a wheel in the link provided, why don't we add the rotational work and linear work? Are they the same thing in this example?

Why do they consider them equal? There are two works, shouldn't they be added together?

In this scenario there are two systems of interest and one interaction between them. This interaction can be characterized by several quantities:

1) force characterizes the transfer of momentum
2) torque characterizes the transfer of angular momentum
3) work characterizes the transfer of energy

These three quantities are related, so they can be written in terms of each other. However, there are not two works. There is one interaction and thus only one transfer of energy. That one transfer of energy may be related to either the torque or the force.

 

[alkaspeltzar]

In this scenario there are two systems of interest and one interaction between them. This interaction can be characterized by several quantities:

1) force characterizes the transfer of momentum
2) torque characterizes the transfer of angular momentum
3) work characterizes the transfer of energy

These three quantities are related, so they can be written in terms of each other. However, there are not two works. There is one interaction and thus only one transfer of energy. That one transfer of energy may be related to either the torque or the force.

Dale, can you see my reply to Russ? Isn't what i said essentially the same and correct? There is a force, which creates a torque. But the work done by the force is the same as the work done by the torque, because we are not moving the CofM, but rotating. The force creates torque, and therefore all the work put in becomes converted to rotational work/rotational kinetic energy. As you said there is only one work!

 

[alkaspeltzar]

Dale, can you see my reply to Russ? Isn't what i said essentially the same and correct? There is a force, which creates a torque. But the work done by the force is the same as the work done by the torque, because we are not moving the CofM, but rotating. The force creates torque, and therefore all the work put in becomes converted to rotational work/rotational kinetic energy. As you said there is only one work!

In this scenario there are two systems of interest and one interaction between them. This interaction can be characterized by several quantities:

1) force characterizes the transfer of momentum
2) torque characterizes the transfer of angular momentum
3) work characterizes the transfer of energy

These three quantities are related, so they can be written in terms of each other. However, there are not two works. There is one interaction and thus only one transfer of energy. That one transfer of energy may be related to either the torque or the force.

Thank you

 

[etotheipi]

The answers above are perfect, I'll add some more context (specifically to @jbriggs444's great suggestion). Consider an unconstrained rigid body acted upon by some external force $\vec{F}(t)$ applied at some material point $\mathcal{P}$ on the body, then by definition the work this force does is$$W = \int_{t_1}^{t_2} \vec{F} \cdot \frac{d\vec{x}}{dt} dt$$with $\vec{x} = \vec{x}(t)$ parameterising the trajectory of the point of application of the force through the configuration space. Now let the position of the centre of mass be $\vec{\mathcal{X}}(t)$, and $\vec{x}' = \vec{x} - \vec{\mathcal{X}} = \overrightarrow{\mathcal{C} \mathcal{P}}$ be the position of the point of application, $\mathcal{P}$ with respect to the the centre of mass, $\mathcal{C}$. Clearly then$$W= \int_{t_1}^{t_2} \vec{F} \cdot \frac{d(\vec{\mathcal{X}} + \vec{x}')}{dt} dt = \int_{t_1}^{t_2} \vec{F} \cdot \frac{d\vec{\mathcal{X}}}{dt} dt + \int_{t_1}^{t_2} \vec{F} \cdot \frac{d\vec{x}'}{dt} dt$$Let's label the first term $W_{\mathcal{A}}$, and the second term $W_{\mathcal{B}}$. If the momentum of the entire rigid body, in the lab frame, is $\vec{P}$, then $\vec{P} = m \dot{\vec{\mathcal{X}}} \implies \vec{F} = m \ddot{\vec{\mathcal{X}}}$. Thus the first term becomes$$\begin{align*}
W_{\mathcal{A}} = m\int_{t_1}^{t_2} \ddot{\vec{\mathcal{X}}} \cdot \dot{\vec{\mathcal{X}}}dt = \frac{1}{2} m\int_{t_1}^{t_2} \frac{d}{dt} \left( \dot{\vec{\mathcal{X}}}^2 \right) dt = \frac{1}{2}m \Delta \left(\dot{\vec{\mathcal{X}}}^2 \right)
\end{align*}
$$The second term, $W_{\mathcal{B}}$, is the work done in the centre of mass frame. Considering that the force is always applied to the same point $\mathcal{P}$ on the rigid body, $|\overrightarrow{\mathcal{C} \mathcal{P}}|$ is constant and so $\vec{x}'(t) = R(t) \vec{x}'(0)$ or equivalently $\dot{\vec{x}}' = \vec{\omega} \times \vec{x}'$, then due to the cyclic symmetry of the triple product,$$W_{\mathcal{B}} = \int_{t_1}^{t_2} \vec{F} \cdot (\vec{\omega} \times \vec{x}') dt = \int_{t_1}^{t_2} \vec{\omega} \cdot (\vec{x}' \times \vec{F}) dt = \int_{t_1}^{t_2} \vec{\omega} \cdot \vec{\tau}' dt$$where $\vec{\tau}'$ is the torque of $\vec{F}$ with respect to $\mathcal{C}$, the centre of mass. $W_{\mathcal{B}}$ is then the change in kinetic energy of the rigid body in the centre of mass frame. The sum $W = W_{\mathcal{A}} + W_{\mathcal{B}}$ is the total change of kinetic energy in the lab frame.

Alternatively, you could have chosen to compute the original line integral for $W$, without decomposing anything. That is fine also!

 

[alkaspeltzar]

The answers above are perfect, I'll add some more context (specifically to @jbriggs444's great suggestion). Consider an unconstrained rigid body acted upon by some external force $\vec{F}(t)$ applied at some material point $\mathcal{P}$ on the body, then by definition the work this force does is$$W = \int_{t_1}^{t_2} \vec{F} \cdot \frac{d\vec{x}}{dt} dt$$with $\vec{x} = \vec{x}(t)$ parameterising the trajectory of the point of application of the force through the configuration space. Now let the position of the centre of mass be $\vec{\mathcal{X}}(t)$, and $\vec{x}' = \vec{x} - \vec{\mathcal{X}} = \overrightarrow{\mathcal{C} \mathcal{P}}$ be the position of the point of application, $\mathcal{P}$ with respect to the the centre of mass, $\mathcal{C}$. Clearly then$$W= \int_{t_1}^{t_2} \vec{F} \cdot \frac{d(\vec{\mathcal{X}} + \vec{x}')}{dt} dt = \int_{t_1}^{t_2} \vec{F} \cdot \frac{d\vec{\mathcal{X}}}{dt} dt + \int_{t_1}^{t_2} \vec{F} \cdot \frac{d\vec{x}'}{dt} dt$$Let's label the first term $W_{\mathcal{A}}$, and the second term $W_{\mathcal{B}}$. If the momentum of the entire rigid body, in the lab frame, is $\vec{P}$, then $\vec{P} = m \dot{\vec{\mathcal{X}}} \implies \vec{F} = m \ddot{\vec{\mathcal{X}}}$. Thus the first term becomes$$\begin{align*}
W_{\mathcal{A}} = m\int_{t_1}^{t_2} \ddot{\vec{\mathcal{X}}} \cdot \dot{\vec{\mathcal{X}}}dt = \frac{1}{2} m\int_{t_1}^{t_2} \frac{d}{dt} \left( \dot{\vec{\mathcal{X}}}^2 \right) dt = \frac{1}{2}m \Delta \left(\dot{\vec{\mathcal{X}}}^2 \right)
\end{align*}
$$The second term, $W_{\mathcal{B}}$, is the work done in the centre of mass frame. Considering that the force is always applied to the same point $\mathcal{P}$ on the rigid body, $|\overrightarrow{\mathcal{C} \mathcal{P}}|$ is constant and so $\vec{x}'(t) = R(t) \vec{x}'(0)$ or equivalently $\dot{\vec{x}}' = \vec{\omega} \times \vec{x}'$, then due to the cyclic symmetry of the triple product,$$W_{\mathcal{B}} = \int_{t_1}^{t_2} \vec{F} \cdot (\vec{\omega} \times \vec{x}') dt = \int_{t_1}^{t_2} \vec{\omega} \cdot (\vec{x}' \times \vec{F}) dt = \int_{t_1}^{t_2} \vec{\omega} \cdot \vec{\tau}' dt$$where $\vec{\tau}'$ is the torque of $\vec{F}$ with respect to $\mathcal{C}$, the centre of mass. $W_{\mathcal{B}}$ is then the change in kinetic energy of the rigid body in the centre of mass frame. The sum $W = W_{\mathcal{A}} + W_{\mathcal{B}}$ is the total change of kinetic energy in the lab frame.

Alternatively, you could have chosen to compute the original line integral for $W$, without decomposing anything. That is fine also!

would you also agree with my post #4 and 6? This is alittle beyond me, just trying to get a gut feel for it

 

[etotheipi]

Intuition will only get you so far, it’s best that you develop the mathematics! I think you’d be well served by obtaining a classical mechanics textbook.

 

[alkaspeltzar]

Intuition will only get you so far, it’s best that you develop the mathematics! I think you’d be well served by obtaining a classical mechanics textbook.

I have a physics book from college. I was simply asking if in the situation i attached in the link, what it is showing and saying thru the math is that the work done by the force, ds x force, is equal to the work done by torque, T x theta. Isnt that true? My intuition says it is because in the case the force acts thru the arc length ds due to the torque it created.

 

[Lnewqban]

I have a physics book from college. I was simply asking if in the situation i attached in the link, what it is showing and saying thru the math is that the work done by the force, ds x force, is equal to the work done by torque, T x theta. Isnt that true? My intuition says it is because in the case the force acts thru the arc length ds due to the torque it created.

Your intuition is correct.
The center of rotation (constraint force) drives the hand to describe a non-linear trajectory, but still is the magnitude of that space (the length of an circular arc in this case) what counts.
For an infinite radius, ΔS becomes a linear distance or a displacement s in a straight line in the direction of the force.

We can always establish a parallel between linear and rotational movements.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

The force of the hand only does work or transfers energy to the disc if it is overcoming certain amount of resistive tangential force or torque in oposite direction.
The torque applied by the hand or by any resistive torque are always the product of a linear force and a lever; therefore, we are dealing with only one amount of mechanical work.

 

[jbriggs444]

The force of the hand only does work or transfers energy to the disc if it is overcoming certain amount of resistive tangential force or torque in oposite direction.

That's Newton's third law. The hand only applies a force on the target if there is an opposite force acting on the hand from the target. Forces come in pairs.

The notion of "resistance" is not appropriate to a correct understanding of Newton's third law.

 

[alkaspeltzar]

Your intuition is correct.
The center of rotation (constraint force) drives the hand to describe a non-linear trajectory, but still is the magnitude of that space (the length of an circular arc in this case) what counts.
For an infinite radius, ΔS becomes a linear distance or a displacement s in a straight line in the direction of the force.

We can always establish a parallel between linear and rotational movements.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

The force of the hand only does work or transfers energy to the disc if it is overcoming certain amount of resistive tangential force or torque in oposite direction.
The torque applied by the hand or by any resistive torque are always the product of a linear force and a lever; therefore, we are dealing with only one amount of mechanical work.

Thank you. ANd i get the resistant force idea. No need to clarify

 

[Lnewqban]

Thank you. ANd i get the resistant force idea. No need to clarify

You are welcome.