How does Sherwood's rotational work equation differ?

In summary, the conversation discusses the derivation of a rotational analogue of the centre of mass work equation, and the difference between two equations presented in the paper. The first equation is only valid for rigid bodies, while the second equation relies on the assumption that the material being acted upon by the force undergoes angular displacements. The question is raised as to why these two equations are presented as different when they seem to be equivalent for rigid bodies.
  • #1
etotheipi
Near the end of this paper, Sherwood presents a rotational analogue of the centre of mass work equation. The derivation is as follows (##\tau_{i, \text{CM}}## is the torque of the ith force about the centre of mass): $$\sum_i \tau_{i, \text{CM}} = I_{\text{CM}} \alpha$$ $$\int \left( \sum_i \tau_{i, \text{CM}} d\theta \right) = \int I_{\text{CM}} \frac{d\omega}{dt} d\theta = \Delta \left(\frac{1}{2}I_{\text{CM}}\omega^2 \right)$$ I have no issues with this derivation. However, he then goes on to say that the actual work done in the centre of mass frame is $$W = \sum_i \left( \int \tau_{i, \text{CM}} d\theta_i \right)$$ The first derivation is only valid in the case that the body is rigid, since otherwise ##\omega## and ##\alpha## are not defined. This means that the power of any given torque is ##P = \mathbf{\tau}_i \cdot \mathbf{\omega}##, where ##\omega## is the angular velocity of the rigid body (regardless of the angular velocity of the force). And if the body is no longer rigid, then the first derivation doesn't apply anyway.

The second equation also relies on the assumption that the material being acted upon by the force also undergo angular displacements of ##d\theta_i##, since that is how work is defined. In the case of a rigid body, then all the angular displacements ##d\theta_i## must be equal.

So I struggle to see why these two equations would ever be different? Would you agree with my interpretation above? Thank you!
 
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  • #2
Can you establish an analogy with equations 1a, 1b and 1c of same paper?
 
  • #3
Lnewqban said:
Can you establish an analogy with equations 1a, 1b and 1c of same paper?

That's part of the problem, he is implying that there is a direct analogy but it doesn't appear to me to be that way. The equations in part 1 say that ##\int \sum_i \mathbf{F}_i \cdot d\mathbf{r}_{CM} = \Delta T_{translational}##, which is the centre of mass work equation. In contrast to ##\int \sum_i \mathbf{F}_i \cdot d\vec{r}_i = \Delta T##, which is the standard work energy theorem dealing with total kinetic energy. These are fine :smile:.

As far as I can tell, there is no such correspondence to the rotational equations he gives, since both when applied to rigid bodies give the increase in kinetic energy w.r.t. the centre of mass.
 
  • #4
Sorry, I can't understand the details of his explanations.
I may be wrong, but it seems to describe the rotational movement as circular layers that slide some angular distance respect to each other while the torque that induces their rotations is applied to a central shaft.
 
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  • #5
etotheipi said:
Near the end of this paper, Sherwood presents a rotational analogue of the centre of mass work equation. The derivation is as follows (##\tau_{i, \text{CM}}## is the torque of the ith force about the centre of mass): $$\sum_i \tau_{i, \text{CM}} = I_{\text{CM}} \alpha$$ $$\int \left( \sum_i \tau_{i, \text{CM}} d\theta \right) = \int I_{\text{CM}} \frac{d\omega}{dt} d\theta = \Delta \left(\frac{1}{2}I_{\text{CM}}\omega^2 \right)$$ I have no issues with this derivation. However, he then goes on to say that the actual work done in the centre of mass frame is $$W = \sum_i \left( \int \tau_{i, \text{CM}} d\theta_i \right)$$ The first derivation is only valid in the case that the body is rigid, since otherwise ##\omega## and ##\alpha## are not defined. This means that the power of any given torque is ##P = \mathbf{\tau}_i \cdot \mathbf{\omega}##, where ##\omega## is the angular velocity of the rigid body (regardless of the angular velocity of the force). And if the body is no longer rigid, then the first derivation doesn't apply anyway.

The second equation also relies on the assumption that the material being acted upon by the force also undergo angular displacements of ##d\theta_i##, since that is how work is defined. In the case of a rigid body, then all the angular displacements ##d\theta_i## must be equal.

So I struggle to see why these two equations would ever be different? Would you agree with my interpretation above? Thank you!
I read this post many times and every time I missed what "you struggle to see." Which two equations are you referring to? The post has three lines with equations. The first and third lines have one equation each and the second line has two. Also "would ever be different" is ambiguous. "Different" from what? Each other or from something else? What are we comparing here and why? Thanks.
 
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  • #6
kuruman said:
I read this post many times and every time I missed what "you struggle to see." Which two equations are you referring to? The post has three lines with equations. The first and third lines have one equation each and the second line has two. Also "would ever be different" is ambiguous. "Different" from what? Each other or from something else? What are we comparing here and why? Thanks.

Sorry for not making myself clear! I refer to the second and third equations I have written. He says this about the former,

1590251289667.png


and then presents the latter equation. I would argue that for a rigid body, the former formula is equivalent to the latter formula in that both will compute the work done by external torques. The reasoning being that the angular velocity is common to all points on the rigid body so the power of all forces will end up only depending on the magnitudes of the torques.

If we're no longer dealing with rigid body dynamics but instead are dealing with something like a system of particles, then there is a difference between them since only the latter equation will apply.

I just wonder why he presents them as different formulae. The first appears to just be a special case of the latter in the case of rigid body motion?
 
  • #7
etotheipi said:
If we're no longer dealing with rigid body dynamics but instead are dealing with something like a system of particles, then there is a difference between them since only the latter equation will apply.
We are no longer dealing with rigid body dynamics. The last equation has ##d\theta_i## in it which allows for the ##i##th and the ##(i+1)##th particles to have different angular separations w.r.t. the CM. Index ##i## does not appear in the previous equations that are, as you noted, appropriate for rigid bodies.
 
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  • #8
kuruman said:
We are no longer dealing with rigid body dynamics. The last equation has ##d\theta_i## in it which allows for the ##i##th and the ##(i+1)##th particles to have different angular separations w.r.t. the CM. Index ##i## does not appear in the previous equations that are, as you noted, appropriate for rigid bodies.

Right, yes. Thanks for clarifying. The latter equation applies to both rigid body dynamics and more general systems of particles, whilst the former is a special case of the latter only applicable if rigid body dynamics applies. Both give the actual work done in the CM frame.

I just found it odd why he noted that the former equation was not an energy equation. It most definitely is equivalent to the actual energy equation for rigid bodies.
 
  • #9
etotheipi said:
I just found it odd why he noted that the former equation was not an energy equation. It most definitely is equivalent to the actual energy equation for rigid bodies.
He explains that in the next sentence, "It leaves no room for thermal energy terms". Such terms could be generated if the ##i##th and the ##(i+1)##th particles rubbed against each other as they moved relative to each other. The relative motion necessitates different angular displacements. The friction between the two would generate heat which would increase the internal energy of the system. So what he's saying is that the previous expressions are too restrictive to allow for total energy conservation in cases where internal friction between components of the system generates heat.
 
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  • #10
kuruman said:
He explains that in the next sentence, "It leaves no room for thermal energy terms". Such terms could be generated if the ##i##th and the ##(i+1)##th particles rubbed against each other as they moved relative to each other. The relative motion necessitates different angular displacements. The friction between the two would generate heat which would increase the internal energy of the system. So what he's saying is that the previous expressions are too restrictive to allow for total energy conservation in cases where internal friction between components of the system generates heat.

Interesting, thanks for clarifying! Internal forces can indeed do work on a non-rigid system.

I guess the distinction is still a little arbitrary, since we consider rigid bodies to have no thermal energy so the former equation is still a valid energy equation for rigid bodies.

But you give me more insight into his thought process here; thanks a bunch for your help! 😁
 

Related to How does Sherwood's rotational work equation differ?

1. What is Sherwood's rotational work equation?

Sherwood's rotational work equation is a mathematical formula used to calculate the work done by a rotating object. It takes into account the rotational speed, mass, and radius of the object to determine the amount of work being done.

2. How is Sherwood's rotational work equation different from other work equations?

Sherwood's rotational work equation is specific to rotating objects and takes into account the rotational speed, which other work equations may not consider. It also accounts for the direction of rotation, making it more accurate for calculating work in rotational systems.

3. Can Sherwood's rotational work equation be used for non-uniform rotational motion?

Yes, Sherwood's rotational work equation can be used for both uniform and non-uniform rotational motion. It takes into account the rotational speed at every point along the object's path, making it applicable for all types of rotational motion.

4. How is Sherwood's rotational work equation derived?

Sherwood's rotational work equation is derived from the principles of rotational dynamics, specifically the relationship between torque and angular acceleration. It is a combination of the work-energy theorem and Newton's second law for rotational motion.

5. What are the units for Sherwood's rotational work equation?

The units for Sherwood's rotational work equation are Joules (J) or Newton-meters (N·m), which are the standard units for work. The rotational speed is typically measured in radians per second (rad/s) and the mass and radius in kilograms (kg) and meters (m), respectively.

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