Conservation of Energy & circular motion

Click For Summary
SUMMARY

The discussion centers on the application of the conservation of energy principle in circular motion, specifically addressing a classical physics problem involving a mass moving in a vertical circular path. The total energy in the system is defined as m*g*L, where m is mass, g is gravitational acceleration, and L is the radius of the circular path. The participant successfully derives the relationship between the velocity at the top of the circle (Vt) and the height (L-x) using energy conservation equations, ultimately confirming that 1/2*m*Vt² + m*g*(2*(L-x)) = m*g*L is valid.

PREREQUISITES
  • Understanding of classical mechanics, particularly circular motion
  • Familiarity with the conservation of energy principle
  • Basic algebra and manipulation of equations
  • Knowledge of gravitational force and its effects on motion
NEXT STEPS
  • Study the derivation of centripetal force in circular motion
  • Explore advanced applications of conservation of energy in non-linear motion
  • Learn about potential and kinetic energy transformations in physics
  • Investigate real-world examples of circular motion, such as roller coasters and satellites
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the principles of energy conservation in circular motion.

nrahim
Messages
2
Reaction score
0
The problem is stated on the sheet i have attached along with my attempt to the possible solution. This is a proof and i am very bad at working with proofs. This seems to be a classical problem so I'm sure some of you guys might have seen it, please guide me!

http://img22.imageshack.us/img22/1405/53811275.th.jpg

I have posted the problem with what i have tried... as you can see i was stuck, please guide me if my logic or math is wrong
 
Last edited by a moderator:
Physics news on Phys.org
Welcome to PF.

Consider the conditions imposed.

Total energy in the system is m*g*L

To complete a circle then with Vt the velocity at the top of the circle about x

m*Vt2/(L-x) = m*g

m*Vt2 = m*g*(L-x)

So ... at the top of that loop

1/2*m*Vt2 + m*g*(2*(L-x)) = m*g*L

Substituting ...

1/2*m*g*(L-x) + m*g*(2*(L-x)) = m*g*L
 
thanks a lot my friend.. you saved me a lot of hassle:D
 

Similar threads

Replies
8
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
5
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
55
Views
3K
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K