Rotational Motion & Energy Equation (Intro Physics)

Click For Summary
SUMMARY

The discussion centers on the application of the Energy Equation in the context of rotational motion and kinetic energy in introductory physics. Participants confirm that the initial potential energy (Mgh) of a mass M converts into kinetic energy as the mass falls, necessitating the inclusion of both the linear kinetic energy (½Mv²) of the falling mass and the kinetic energy of the rotating table in the final energy equation. The consensus emphasizes the importance of accounting for all energy forms while assuming no energy loss due to friction.

PREREQUISITES
  • Understanding of the Energy Equation in physics
  • Familiarity with concepts of kinetic and potential energy
  • Basic knowledge of rotational motion dynamics
  • Ability to manipulate algebraic expressions involving mass and velocity
NEXT STEPS
  • Study the conservation of energy principles in rotational systems
  • Learn how to derive and apply the Energy Equation in various scenarios
  • Explore the relationship between linear and rotational kinetic energy
  • Investigate the effects of friction on energy conservation in mechanical systems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of energy conservation in rotational motion and its applications in problem-solving.

carlodelmundo
Messages
133
Reaction score
0

Homework Statement



[PLAIN]http://carlodm.com/phys/1.png

Homework Equations



Energy Equation

The Attempt at a Solution



As shown in the scan.

Is this the correct approach? I have a feeling that the velocity, v, needs to be used. I used the Energy Equation for this problem but wasn't sure if it was the right equation to address kinetic energy, potential energy, and rotational motion.

Please lead me to the right direction if I'm incorrect.

Thanks
 
Last edited by a moderator:
Physics news on Phys.org
This assumes there is no energy lost due to friction, of course.
The initial energy is all potential in the mass M
At the end, the potential energy lost by the falling mass (Mgh) has appeared as kinetic energy in both the mass and the rotating table.
So you do need to consider the velocity v of the mass.
 
Hi StoneBridge,

Thanks for your response. How do I account for the kinetic energy of the linear 'falling of the mass' when I am only given the velocity?

For the right hand side of the equation, do I just include another kinetic energy term (... KE = (1/2)Mv^2)

Would that account for the velocity?

Thanks
 
The question just asks that you express everything in terms of what's given in the question. You have the mass M and the velocity v of the mass as given.
So yes, ½Mv2 is the k.e. of the falling mass, and in your notes I see the expression for the k.e. of the turntable and the p.e. the mass has at the start.
So it's just a case of including all 3 terms and remembering that energy is conserved (no friction here!)
 
Thanks a lot! I appreciate it!
 

Similar threads

Replies
4
Views
598
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 24 ·
Replies
24
Views
1K
Replies
25
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K