• Support PF! Buy your school textbooks, materials and every day products Here!

Finding the coefficient of kinetic friction using impulse/momentum

  • Thread starter cjdavis7
  • Start date
  • #1
4
0
Hi everyone,

I am trying to calculate the coefficient of kinetic friction in two ways. One using energy and one using the principal of linear impulse and momentum. Not getting the same answer!

The system is a wheeled cart oscillating on a flat metal track between two springs. The motion of the cart was recorded using a motion detector into LoggerPro which calculates the velocity automatically. The mass of the cart and the spring constants are known.

Method 1: using energy/work concepts. Two points on the position graph where the velocity was zero are chosen. All the energy at those points is in the potential energy of the springs. That is calculated using 1/2kx^2 where x is the displacement from the equilibrium point (centered between the two springs). So, then I took subtracted the initial energy from the final energy to determine the amount of work done on the cart. The work in this case we are attributing solely to friction for simplicities sake. Force due to friction acting perpendicular to the motion: mu(normal force)(distance traveled). Normal force is mg. Distance traveled was found by determining the distance between each end of all the oscillations that occurred during the time in question and summing those up.

Method 2: using linear impulse/momentum. Two points 1/4 of an oscillation away from the two points previously chosen were selected. At these two new points, the velocity is at a local max and the potential energy is zero (ie the cart is at the equilibrium point). mv1 + impulse = mv2. Took the final momentum and subtracted the initial momentum from that. The impulse is the integral from time 1 to time 2 of (muN)dt. Again attributing all dampening to friction.

So, the problem is that the result I get from Method 2 is almost exactly twice that of Method 1. Is there something about the oscillatory motion that causes this? Why would I need to divide Method 2 by 2?

Thanks for any insight.
Chris
 

Answers and Replies

  • #2
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Welcome to Physics Forums.

I'd just like to clarify a few things before we get into a technical discussion.
The system is a wheeled cart oscillating on a flat metal track between two springs.
The two springs are attached to the cart, yes?
So, then I took subtracted the initial energy from the final energy to determine the amount of work done on the cart.
What are you taking and the initial and final energies? How did you calculate them?
Force due to friction acting perpendicular to the motion
Surely you mean the frictional force was parallel to the motion?
Distance traveled was found by determining the distance between each end of all the oscillations that occurred during the time in question and summing those up.
Note that for one complete cycle the cart will have travelled (approximately) twice the distance between the stationary points.

The impulse is the integral from time 1 to time 2 of (muN)dt. Again attributing all dampening to friction.
What about the force of the springs acting on the cart?
 
  • #3
4
0
Hi, thanks for the response. Here's some answers that should help clarify things.

1)Yes, the springs are attached to the cart.

2)The initial and final energy of the cart was calculated as being the potential energy in the springs. 1/2kx^2 where x is the displacement from the equilibrium point. The points chosen for the initial and final points were also when the velocity was zero (no kinetic energy).

3)Yes the friction is parallel to the motion, sorry about that.

4)note taken

5)the springs will have an internal dampening, but we are attributing all dampening to friction in this case. I realize that's not precisely accurate, but I'm making the same assumption for both Methods. As far as the force of the springs goes, without the dampening the oscillations would continue forever. I'm looking for the force that causes that dampening.

Thanks!
Chris
 
  • #4
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
5)the springs will have an internal dampening, but we are attributing all dampening to friction in this case. I realize that's not precisely accurate, but I'm making the same assumption for both Methods. As far as the force of the springs goes, without the dampening the oscillations would continue forever. I'm looking for the force that causes that dampening.
I appreciate that. However, when calculating the impulse:
The impulse is the integral from time 1 to time 2 of (muN)dt.
You assume that the only force acting on the cart is that of friction. This is simply not the case, there springs also exert a force on the cart. You must account for the impulse due to the springs as well as the friction. Do you follow?
 
  • #5
4
0
Hi again,

Here's how I'm thinking of it.

If there was no dampening force there would be no change in momentum as measured every time the cart passes through equilibrium because the spring force is a conservative force. The only forces I'm looking for are the nonconservative forces which I'm combining all into friction for simplicities sake.
 
  • #6
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Hi again,

Here's how I'm thinking of it.

If there was no dampening force there would be no change in momentum as measured every time the cart passes through equilibrium because the spring force is a conservative force. The only forces I'm looking for are the nonconservative forces which I'm combining all into friction for simplicities sake.
Ahh right, I see what you're doing now, I had misinterpreted your experimental set up.

Can you show me exactly how you have calculated the work and impulse?
 
  • #7
174
0
Correct me if I'm wrong.
You wrote that the initial energy of the body is k*x^2\2 while X is the displacement from the eq point.
now did You do that for only 1 spring, cause the body has both elastic potential energy from both strings( correct me if I'm wrong again).
good luck.
EDIT:
By reading again I understood that, it is what You did.
What were the results of Your experiment?
 
  • #8
4
0
Dweirdo,
For the value of k, I summed the two springs together because they are acting in parallel (think how it would look if you took the two outside ends of the springs and brought them up and together-the cart would be suspended from both in parallel)

Hootenanny,
The motion was tracked for 30 seconds using a motion detector and LoggerPro with data collected every 0.1 second.

Method one for finding the force causing the dampening: picked two points >20 seconds apart on the position graph where all the energy was potential ie at the extreme ends of the motion when the displacement from equilibrium was at a max and the velocity was zero. Calculated the potential energy of the system at those two points using the k as described above and the displacement from equilibrium. Friction is what I'm attributing the dampening too and rolling friction acts constantly throughout the oscillations. Next figured out the total distance travelled between the two chosen point using the extreme displacements (2nd displacement minus the first...plus...third displacement minus the 2nd...plus...). Set the starting position to be zero so that the integral becomes (F)dx from zero to "total distance". The work done ends up being friction*distance travelled in this case. So, took: (delta potential energy/distance travelled = friction).

Method two for finding the force causing the dampening-this came out to be half the value of the first method and that is my question. What am I missing in this 2nd method? Again took two points separated by >20 sec but this time used points where all the energy was kinetic ie when the cart was at equilibrium and velocity was at a max and potential was zero. These points where chosen as close as I could to the previous two. Found the time elapsed between the two points chosen and set the first time to be zero--these will be the limits of integration. mv1 + integral from 0 to total time of (F)dt = mv2 Solved for F. Got a value that's half of method 1.

Thanks for your help!
Chris
 
  • #9
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
For the value of k, I summed the two springs together because they are acting in parallel (think how it would look if you took the two outside ends of the springs and brought them up and together-the cart would be suspended from both in parallel)
This is your problem, your springs are not in parallel - they are in series. Your cart is connected between two springs, which for all effective purposes is the same as this image,
simg479.gif

Springs in series are not simply additive. Instead they add like resistors in parallel, i.e. the reciprocal of the effective spring constant is equal to the sum of the reciprocals of the individual spring constants. See http://scienceworld.wolfram.com/physics/SpringsTwoSpringsinSeries.html" [Broken] for more information.
 
Last edited by a moderator:

Related Threads for: Finding the coefficient of kinetic friction using impulse/momentum

Replies
1
Views
4K
Replies
4
Views
2K
Replies
2
Views
4K
Replies
3
Views
2K
Replies
8
Views
6K
Replies
3
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
8
Views
2K
Top