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Finding g from 2 graphs

  • Thread starter haxtor21
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  • #1
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Homework Statement


I am doing a lab report in which I am supposed to plot 2 graphs from which to approximate g. One is Delta X vs t and one is delta X vs t^2. The data represents the time it takes for an object to travel a certain distance interval (in my case a cart going down an inclined plane, and the time being measured by 2 photo-gates) I have SciDAVIs opened and have the data inside 2 tables, but I have no clue how to graph these two tables so that I get the necessary equations for approximating g. All i have right now is scattered data points.


Homework Equations


a=.5gsin(theta) -> g=(2a)/sin(theta) (for graph x-t)

m=.5gsin(theta); g=(2m)/sin(theta) (for graph x-t^2)


The Attempt at a Solution


Im not sure how these equations and graphs will help me get the experimental value of g. Does the computer just spit out some models and I plug them into each other? The lab just says to find the local value of g from both graphs, extremely helpful -_-. Someone please explain me how this is supposed to work. Thank you.
 

Answers and Replies

  • #2
gneill
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What are the expressions for the theoretical curves that you expect? In other words, given an inclined plane at angle Θ to the horizontal, what is the function expressing x versus t?
 
  • #3
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What are the expressions for the theoretical curves that you expect? In other words, given an inclined plane at angle Θ to the horizontal, what is the function expressing x versus t?
that would be y= a+ ax + ax^2 and y= bx+c

I figured out the fit for my data points, with the x-t being a polynomial and x-tt being a linear fit. Now what am I supposed to do with them? How do these and my initial equations relate?
 
  • #4
gneill
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A cart with frictionless wheel bearings is released from rest on an inclined plane that makes an angle Θ to the horizontal. Derive an expression for the distance x that the cart travels down slope with respect to time t.

Your expression should contain the constant g, the acceleration due to gravity.
 
  • #5
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A cart with frictionless wheel bearings is released from rest on an inclined plane that makes an angle Θ to the horizontal. Derive an expression for the distance x that the cart travels down slope with respect to time t.

Your expression should contain the constant g, the acceleration due to gravity.
general form:
x= x-node + v-node (t) + .5(a-x)t^2

in which case a-x is -g
 
  • #6
gneill
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What is "node"? How can I calculate x(t) from "node"?

What is the specific equation for x versus t in this particular case? Have you not studied blocks sliding down frictionless slopes? There is a function x(t) = ??? which gives the distance that the cart has covered in time t.
 
  • #7
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What is "node"? How can I calculate x(t) from "node"?

What is the specific equation for x versus t in this particular case? Have you not studied blocks sliding down frictionless slopes? There is a function x(t) = ??? which gives the distance that the cart has covered in time t.
well i guess that would be Δx=.5 (g sin(θ)) t^2
 
  • #8
gneill
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Okay! So you would expect your x versus t graph of your data to follow the form

[tex]x(t) = \frac{1}{2}a t^2[/tex]

where a = g sin(θ). If you can find a from your graph, you can find g, right?

What is the slope of the function at some time t = t1? (hint: take the derivative). So pick suitable points along your plotted data and determine the local slope. Use the slope information at point t1 to find g using what you've derived.

For your second graph, where you're plotting x versus t2, essentially what you are doing is replacing t2 with a new variable τ. That is, τ = t2.

Your function becomes:

[tex]x(\tau) = \frac{1}{2}a \tau[/tex]

What's the slope of that graph? It should be easy to determine a and hence g from that one!
 

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