Need a bit more clarification on W=F.ds

In summary, the work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. This applies to both point-sized particles and extended bodies, both rigid and non-rigid. For rigid bodies, the displacement term is the displacement of the center of mass, while for non-rigid bodies it is the displacement of the point of contact. The normal force does not do work because it has no displacement, but potential energy can still be gained through rotation or internal stored energy. Additionally, the concept of "center of mass work" may be used for rigid, non-rotating bodies.
  • #1
GKRM
22
0
I was a bit confused about the work energy theorem. I perfectly understood it's applications for point sized particles but I'm a bit confused about its application on extended bodies both rigid and non rigid.
Case 1 {for rigid bodies} :-
Consider a rod of definite mass hinged at the top. It's initially vertical. Now you apply a horizontal force F to the lowest point in horizontal direction of constant magnitude. The magnitude is low enough so that the rod doesn't gain any significant kinetic energy. However it gains potential energy as it rotates. The hinge doesn't do any work because there is no displacement of the point of contact. The applied force F is always horizontal. So F.ds in vertical direction is always zero. My question is how does it gain potential energy then?
OR
Consider a square block kept on a rough horizontal surface (friction is sufficient enough to prevent slipping/sliding). You apply a horizontal force F on the topmost point ,constant in magnitude and direction. The magnitude is adjusted so that the block doesn't gain significant kinetic energy. Due to rotation about the axis passing through the line of contact of the ground and block, the potential energy of the block increases as it loses contact with the surface. How does it gain potential energy?
Case 2 {non-rigid body}:-
Consider 2 blocks each of mass M, connected by a massless spring. One of the blocks is in contact with a vertical wall. You now compress the spring and let the motion begin after you have left the other block. The normal force doesn't do any work because there is no displacement of the point of contact. So How does the system gain kinetic energy?

On the whole I want to understand what is actually meant by the elemental displacement term in the work energy equation. I assume that in case of a rigid body it's displacement of centre of mass and in case of non rigid body it's displacement of point of contact. Am I correct assuming this?
 
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  • #2
GKRM said:
So F.ds in vertical direction is always zero. My question is how does it gain potential energy then?
Work is a scalar. It doesn't have a direction.

GKRM said:
The normal force doesn't do any work because there is no displacement of the point of contact. So How does the system gain kinetic energy?
From internally stored potential energy.
 
  • #3
Can you please write the corresponding equation for case1 in work energy.. for both sub cases?
 
  • #4
I also need more clarification on the meaning of ds. Is it displacement of centre of mass or point of contact?
 
  • #5
GKRM said:
I also need more clarification on the meaning of ds. Is it displacement of centre of mass or point of contact?
Point of contact.

If you want to get picky, it is the displacement of the material in the object on which the force is being exerted at the point where the force is currently being applied that is relevant. This nit-picky point becomes crucial when discussing, for instance, the work done on a tire by the road. The "point of contact" is moving. But the "material at the point of contact" is not. So the road does zero work on the tire.

Now, before @Doc Al jumps into point it out, there is a concept of "center of mass" work in which it is the displacement of the center of mass that counts. So that the road is seen to do center of mass work on the car.

For a rigid, non-rotating body, "center of mass work" is equal to plain old "work" and there is no distinction to be made.
 
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Likes Dale and Doc Al
  • #6
GKRM said:
Consider a rod of definite mass hinged at the top. It's initially vertical. Now you apply a horizontal force F to the lowest point in horizontal direction of constant magnitude. The magnitude is low enough so that the rod doesn't gain any significant kinetic energy. However it gains potential energy as it rotates. The hinge doesn't do any work because there is no displacement of the point of contact. The applied force F is always horizontal. So F.ds in vertical direction is always zero. My question is how does it gain potential energy then?

I think the answer is F must have a vertical component. If there is no friction then the force acting on the rod must be normal to the rod, and at all angles (other than vertical) that has a vertical component.
 
  • #7
I was surfing on the internet and I came across this theorem
https://en.m.wikipedia.org/wiki/Chasles'_theorem_(kinematics)
If I apply this for a rigid body I think displacement of centre of mass would do( by treating it as a particle). But I'm still confused about non rigid bodies like ropes and bodies with many point of contact.
 
Last edited:
  • #8
CWatters said:
I think the answer is F must have a vertical component. If there is no friction then the force acting on the rod must be normal to the rod, and at all angles (other than vertical) that has a vertical component.
I think it's not necessary. You can even perform this experiment at home using a pencil held in your hands. The vertical component of hinge reaction may exceed mg by infinitesimal amount to lift the centre of mass.
 
  • #9
A.T. said:
Work is a scalar. It doesn't have a direction.
But we can use F.ds in any direction we choose.
 
  • #10
GKRM said:
But we can use F.ds in any direction we choose.
F has a direction. ds has a direction. Their dot product does not.

If F is purely horizontal then clearly it has a non-zero horizontal component and a zero vertical component. If the pendulum is moving in a narrow region around bottom dead center then clearly ds has a non-zero horizontal component. So their dot product will certainly be non-zero.
 
  • #11
What I'm trying to quote is that the horizontal force F doesn't have any vertical components hence the work done by F is zero. But hinge reaction has a vertical component. Hence maybe hinge is doing work by displacing centre of mass. I know energy is scalar.
 
  • #12
IMG-20170623-WA0002.jpeg
 
  • #13
In the above case I have tried to proceed from centre of mass energy to conservation of energy for a rigid body. Can anybody help me out on how does this work for non rigid bodies like ropes.
 
  • #14
For example a rope falling from a desk{with a hole} I can write the centre of mass energy but I can't proceed further.
 
  • #15
GKRM said:
What I'm trying to quote is that the horizontal force F doesn't have any vertical components hence the work done by F is zero.
This is not how work is defined.

GKRM said:
Hence maybe hinge is doing work
A static frictionless hinge does no work.
 
  • #16
A.T. said:
This is not how work is defined.A static frictionless hinge does no work.
So How would you explain the situation
?
 
  • #17
GKRM said:
So How would you explain the situation
?
See post #10.
 
  • #18
I'm talking about centre of mass energy above.
 
  • #19
GKRM said:
I'm talking about centre of mass energy above.
While there is a concept of center-of-mass work, the term "center of mass energy" is undefined (to my knowledge). The concept of center-of-mass work is poorly suited to a conservation of energy problem since it ignores some of the energy transfer due to the force.
 

1. What is the equation for "Need a bit more clarification on W=F.ds"?

The equation is W = F * d * cos(theta), where W represents work, F represents force, d represents distance, and theta represents the angle between the force and the displacement.

2. How is work defined in the equation "Need a bit more clarification on W=F.ds"?

In this equation, work is defined as the product of force and displacement in the direction of the force. It is a measure of the energy transferred to an object by a force.

3. What is the significance of the angle theta in the equation "Need a bit more clarification on W=F.ds"?

The angle theta represents the direction of the force relative to the displacement. If the force is applied in the same direction as the displacement, theta is 0 degrees and the cosine of 0 is 1, resulting in the maximum amount of work being done. If the force is applied perpendicular to the displacement, theta is 90 degrees and the cosine of 90 is 0, resulting in no work being done.

4. How is the equation "Need a bit more clarification on W=F.ds" used in real-life situations?

This equation is used in various fields, such as physics, engineering, and biomechanics, to calculate the amount of work done by a force on an object. It can be used to determine the energy required to move an object a certain distance, or the force needed to move an object a certain distance with a given amount of energy.

5. What are some other important factors to consider when using the equation "Need a bit more clarification on W=F.ds"?

In addition to force, distance, and angle, it is important to consider the units of measurement used for each variable. Force is typically measured in Newtons, distance in meters, and angle in radians. It is also important to take into account any external factors, such as friction, that may affect the amount of work done by a force.

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