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 January 8th, 2017, 10:20 AM #1 Member   Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 physics hi everyone, i am currently doing a degree in mechanical engineering and i am currently struggling with an assignment. Basically we did an experiment where a roller disc was rolled down 2 rails and we had to calculate how long it takes to get from one end to the other, it was raised at 10cm increments each time and obviously the higher the gradient the quicker it got to the bottom. we then have to calculate things such as radius of gyration, kinetic energy, potential energy and kinetic energy of rotation. now in one of the tables there is values given as S (m) - this is the distance of the rails t (s) - Time in seconds v (m/s) - velocity v^2 (m/s^2) - ? h (m) - height i am also given a statement that says the final linear velocity v is found from s = 1/2(u+v)t now i have transposed this to 2s/t - u = v but how do i find v^2?
 January 8th, 2017, 10:44 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 $v^2 = \left(\dfrac{2s}{t}-u\right)^2$ should be rather simple, espescially if initial velocity, $u=0$
 January 8th, 2017, 10:54 AM #3 Member   Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 so thats all i have to do? maybe i looked a bit too in depth into it. also once plotting a graph with with height on the vertical axis and v^2 on the horizontal how would i find the value of (h/v^2)?
 January 8th, 2017, 11:04 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 $\dfrac{h}{v^2}$ would be the slope of the graph ...
 January 8th, 2017, 11:08 AM #5 Member   Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 and how do i find that? this is the first physics topic we have done and its certainly not my strongest subject
 January 8th, 2017, 11:12 AM #6 Math Team   Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 I would start by plotting a data graph of $h$ vs $v^2$ ... Thanks from Joppy
 January 8th, 2017, 11:17 AM #7 Member   Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 yes i have plotted the graph, i just dont know how to find the gradient
 January 8th, 2017, 11:51 AM #8 Math Team   Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 Have you constructed a line of best fit to your data? The data should follow a linear pattern.
 January 8th, 2017, 11:56 AM #9 Member   Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 yes the line of best fit is a linear pattern
January 8th, 2017, 12:00 PM   #10
Math Team

Joined: Jul 2011
From: Texas

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Quote:
 Originally Posted by Jimkeller1993 yes the line of best fit is a linear pattern
Then measure the slope ... $\dfrac{rise}{run}$, remember?

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