How Can I Measure Kinetic Energy in a Non-Inertial Reference Frame?

  • Thread starter Thread starter xkcda
  • Start date Start date
AI Thread Summary
To measure kinetic energy in a non-inertial reference frame, the formula K = (Moment of Inertia about an axis through A * Angular Velocity^2)/2 + (Mass * Velocity^2)/2 is proposed, yielding K = 9.5. However, there is ambiguity regarding the reference frame's rotation and the interpretation of "with respect to point A." It is suggested that if the moment of inertia is calculated about the axis of rotation, the linear kinetic energy term may lead to double counting. Observers in the non-inertial frame must consider the rigid body's instantaneous motion as a combination of linear motion and rotation about its mass center. Clarifying these aspects is essential for accurate kinetic energy measurement.
xkcda
Messages
7
Reaction score
0
TL;DR Summary: I think A is an non inertial reference frame.So how can I measure kinetic energy about it?

I found a solution to the problem which states that Kinetic Energy about A= (Moment of Inertia about an axis passing through A*Angular Velocity^2)/2+(Mass*Velocity^2)/2 .Thus K=9.5.Can anyone please show me the derivation of this formula?
Screenshot from 2023-06-16 00-03-00.png
 
Last edited:
Physics news on Phys.org
No attempt shown.
 
Please show some effort, so we can help you learn.
 
Moved to homework help.
 
The first difficulty is that "with respect to point A" is ambiguous.
It is reasonable to assume, as you have, that it does not mean the fixed point in space where that corner happens to be at some instant; rather, it moves with that corner of the plate. But that still does not answer whether the reference frame is also rotating with the plate. Consider both cases.
In each case, think of what an observer in the frame would see the plate as doing.
xkcda said:
Kinetic Energy about A= (Moment of Inertia about an axis passing through A*Angular Velocity^2)/2+(Mass*Velocity^2)/2
That seems very unlikely to be right. If you take the moment of inertia about the axis of rotation then you should not need to be adding a linear KE term: that would be double counting. Generally speaking, you can consider the instantaneous motion of a rigid body as the sum of the linear motion of its mass centre and its rotation about its mass centre. So if you have an ##mv^2## term for the linear component then the moment of inertia should be about the mass centre.
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top