How Does Speed Influence Complete Revolutions in Circular Motion?

Click For Summary

Homework Help Overview

The discussion revolves around a problem involving circular motion, specifically analyzing the energy dynamics of a small ring projected along a vertical circular wire. The participants are tasked with determining the conditions under which the ring will complete its revolutions based on its initial speed and the gravitational force acting on it.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the potential and kinetic energy of the ring at different points in its motion, questioning how to apply conservation of energy principles. There are attempts to clarify the expressions for potential and kinetic energy, as well as the relationship between height and radius.

Discussion Status

Some participants have provided guidance on the correct expressions for potential and kinetic energy, while others are exploring how to manipulate these equations to derive the required condition for complete revolutions. There is an acknowledgment of confusion regarding the variables involved, particularly the height in relation to the radius of the circle.

Contextual Notes

Participants express uncertainty about the initial setup and the definitions of variables, particularly the height in relation to the radius of the circular path. There is a recognition of the challenge posed by the problem's requirements and the need for clarity in the energy expressions.

teme92
Messages
185
Reaction score
2

Homework Statement



A smooth circular wire of radius a is fixed with its plane vertical. A small ring
threaded on the wire is projected with speed u from the lowest point of the
circle. Taking gravitational acceleration to be the constant g, calculate the
potential energy and the kinetic energy of the ring. Assuming conservation
of energy, show that the ring will describe complete revolutions if:

u^2 > 4ga

Homework Equations



I know all relevant circular motion and SHM equations but don't know where to begin.

The Attempt at a Solution



I genuinely have no idea how to approach this problem. Any help will be much appreciated.
 
Last edited:
Physics news on Phys.org
Write the potential and kinetic energies of the ring, at the bottom of the wire and top of the wire.
And use conservation of energy.
 
  • Like
Likes   Reactions: 1 person
Hi nasu thanks for the speedy reply.

I'm having trouble visualizing the problem. Would the potential energy be equal to mgh + mg(0) as at the bottom of the wire h=0? And for kinetic energy do I use 0.5(m)(v^2)?
 
You don't add the potential energies.
The potential energy at the bottom may be zero, yes, if we measure it from that level.
At the PE at the top point will be mgh, where h is the height of the top pf the circle.

And yes, this is the formula for KE.
 
  • Like
Likes   Reactions: 1 person
Ok and for the second part of the question where I'm asked to show that the ring will describe complete revolutions. What would show it describes complete revolutions?

Thanks again for the help.
 
Conservation of energy. I told you already.
But first you need the correct expressions for PE and KE energy.
 
Thanks for your patience, I'm new to these type of problems and I'm finding them tricky to understand.

So conservation of energy is PE=KE

PE=mgh

KE= 0.5(m)(v^2)

mgh=0.5(m)(v^2)

Putting in the form the question requires and I get:

u^2=2gh,

which isn't the required answer. Clearly the 'h' isn't part of the answer so how to I go about getting rid of it?
 
What is h in terms of a, the radius of the circle?
 
  • Like
Likes   Reactions: 1 person
I completely forgot 'a' was the radius, I must have been half asleep last night doing this. I was thinking it was acceleration. Thanks a million, I have the solution now.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 14 ·
Replies
14
Views
5K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K