What is the minimum height for a ball to complete a loop and rough surface?

  • Thread starter Thread starter stanford1463
  • Start date Start date
  • Tags Tags
    Loop Loop the loop
Click For Summary

Homework Help Overview

This problem involves energy considerations related to a ball navigating a loop and a rough surface. The question focuses on determining the minimum height required for the ball to successfully complete the loop and traverse the rough surface.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the energy balance involving gravitational potential energy, kinetic energy, and work done against friction. There are attempts to derive expressions for the minimum height based on energy conservation principles.
  • Some participants question whether the constraints for both the loop and the rough surface have been adequately addressed, suggesting a separation of these considerations.
  • There is exploration of the necessary conditions for the ball to maintain motion through the loop and the implications of friction on the required height.

Discussion Status

The discussion is ongoing, with participants presenting different aspects of the problem and questioning the assumptions made. Some guidance has been provided regarding the constraints that must be satisfied for the ball to complete the loop and traverse the rough surface, but no consensus has been reached on the final height calculation.

Contextual Notes

Participants note the importance of considering both the energy required to maintain motion through the loop and the work done against friction on the rough surface. There is an emphasis on ensuring that the potential energy at the starting height exceeds the work done against friction.

stanford1463
Messages
44
Reaction score
0

Homework Statement



So, this problem is about energy. The question is: what must the minimum height (in R) of a height be, for a ball to go down it, through a loop the loop (with radius R), and then through a rough surface 6R's long with u= 0.5.

Homework Equations


gravitational potential energy= mgh
kinetic energy= 1/2mv^2
work of friction= umgs

The Attempt at a Solution


Hopefully this is right: I did initial potential energy= mgh= 1/2mv^2 (kinetic needed to get past the top of the loop) + umgs (work of friction of the surface) + mgh (potential energy at top of loop). Also, I got v^2=Rg, since using the centripetal force equation mv^2/R, I made it equal to the normal force, mg, and v^2=Rg.
The mg's cancel out, so you're left with h=1/2R + (0.5)6R + 2R, which equals 5.5R. Is this right??
 
Physics news on Phys.org
bump?
 
stanford1463 said:

Homework Statement



So, this problem is about energy. The question is: what must the minimum height (in R) of a height be, for a ball to go down it, through a loop the loop (with radius R), and then through a rough surface 6R's long with u= 0.5.

Homework Equations


gravitational potential energy= mgh
kinetic energy= 1/2mv^2
work of friction= umgs

The Attempt at a Solution


Hopefully this is right: I did initial potential energy= mgh= 1/2mv^2 (kinetic needed to get past the top of the loop) + umgs (work of friction of the surface) + mgh (potential energy at top of loop). Also, I got v^2=Rg, since using the centripetal force equation mv^2/R, I made it equal to the normal force, mg, and v^2=Rg.
The mg's cancel out, so you're left with h=1/2R + (0.5)6R + 2R, which equals 5.5R. Is this right??

[tex]m\vec{g}NR-\vec{W}=0[/tex]. Solve. So long as [tex]N>\frac{5}{2}[/tex], you're fine (because that's the min. height for an object to go through any loop-the-loop).
 
There are 2 constraints that need to be met aren't there?

One is the height must be sufficient for the golf ball to stay in the loop the loop. The other is that you must start with enough potential energy to make it through the rough patch. Consider them separately.

Figure first then the work to get through the rough patch.
W = f * d = u*mg*d = .5*6R*mg = 3R*mg
So long as PE > W it will make it right? So Constraint 2 is mg*h > mg*3R or h > 3R

What about the loop the loop?
3R*mg = mV2/2 means V2 will be at least 6*R*g.

At the top of the loop the outward force of the mv2/R must be greater than the weight mg.
mV2/R > mg ? Well substitute the necessary V2 for the rough patch of constraint 2.

m*6*R*g/R = 6*mg > mg ?

If constraint 2 is met then is that sufficient?
 

Similar threads

Replies
8
Views
4K
  • · Replies 12 ·
Replies
12
Views
13K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
Replies
5
Views
2K