Kinetics of a rigid body (conservation of energy)

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Homework Help Overview

The discussion revolves around the kinetics of a rigid body, specifically focusing on the conservation of energy principles as they relate to spring mechanics and potential energy calculations. Participants are examining the relationship between vertical displacement, spring extension, and energy transformations in a system involving a spring and gravitational forces.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the meaning of vertical displacement in relation to spring extension and are questioning the application of the Pythagorean theorem in their calculations. There are discussions about the setup of conservation of energy equations, with some participants suggesting to express kinetic and potential energy in terms of relevant quantities. Others are exploring the implications of using different unit systems and questioning the assumptions about potential energy contributions from gravity versus the spring.

Discussion Status

The discussion is active, with participants providing insights into the calculations of spring extension and potential energy. Some guidance has been offered regarding the setup of energy equations, and there is an ongoing exploration of the role of gravitational potential energy in the context of the problem. Multiple interpretations of the energy contributions are being considered, particularly regarding the spring and gravitational effects.

Contextual Notes

There are mentions of using English (Imperial) units versus metric units, which may influence the calculations and understanding of mass and energy in the context of the problem. Participants are also addressing the initial conditions of the spring's stretch and its implications for potential energy calculations.

sseebbeekkk
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Homework Statement


eZTS4A6.jpg
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Homework Equations


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(1) (pink) Is it vertical displacement ?
* 4 - 2 (unstretched length of the spring) ?

(2) (blue) What does it mean ?
* I was thinking that maybe they used The Pythagorean theorem (62+42)=7,2 ≅ 7) but I'm not sure. Besides I don't understand why they used it.

3. Attempt

My reasoning: In general I think they are using this formula,
upload_2015-7-17_13-20-26.png

where in that case the left side is the initial state - the right side - the final state of the spring.
 

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I recommend setting it up carefully in terms of conservation of energy, expressing T and V in terms of all the relevant quantities (leaving as symbols), making simplifications, solving for the unknown (with symbols), and plugging in numbers at the end.
 
The 4-2 and 7-2 are both calculations of spring extension. In each case, 2 is the relaxed length. At the horizontal position, the total length is 4, so the extension is 4-2. After rotating 30 degrees, the total length is 4+6 sin(30)=7.
I don't like the way they divide the mass by 32.2 in the KE terms. It seems more natural to me to multiply by 32.2 in the other terms - it comes to the same thing, of course. Maybe their way is standard in this non-metric system. I have no background in it.
 
The English (Imperial) unit of mass is slugs. One obtains mass by dividing weight (lbs) by the acceleration of gravity (32.2 ft/s/s) to obtain the mass in slugs.

I thought ballistics was the last holdout still using English units for real physics.

https://en.wikipedia.org/wiki/English_Engineering_units
 
Dr. Courtney said:
The English (Imperial) unit of mass is slugs. One obtains mass by dividing weight (lbs) by the acceleration of gravity (32.2 ft/s/s) to obtain the mass in slugs.

I thought ballistics was the last holdout still using English units for real physics.

https://en.wikipedia.org/wiki/English_Engineering_units
Ok, thanks.
 
Thank you for help.

Dr. Courtney - is this equation correct ?

X0yeXaC.png
 
Is the only potential energy due to the spring?
 
If a spring is not stretched or compressed, then there is no elastic potential energy stored in it.

However in this case at the beginning the spring was stretched 2 metres (4-2=2), hence 1/2*k*s1^2

At the end, the spring was stretched 5 meters (7-2=5), hence 1/2*k*s2^2

'W' stands for: work of a weight -> WΔy= 50 * (1.5)
 
  • #10
I think the only potential energy is due to the spring. If the spring lies the ground then mgh=0
 
  • #11
sseebbeekkk said:
I think the only potential energy is due to the spring. If the spring lies the ground then mgh=0
The massive bar is rotating in a vertical plane. Are you saying its centre of gravity does not change in height?
 
  • #12
Ok, it does. Thanks.
 

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