Physics Help Forum Determining Mass from Momentum

 Aug 7th 2017, 02:42 PM #1 Junior Member   Join Date: Aug 2017 Posts: 3 Determining Mass from Momentum Is it possible to determine the mass of an object (i.e Bicycle) given the (Coefficient rolling resistance), (Initial Velocity) and the (Deceleration). Given that you are able to change the (Coefficient Rolling Resistance) and from this determine a new (Deceleration). Wind resistance and internal frictions are neglected. If so how? This is for a personal project where for one part I need to determine the weight of the bicycle, by a simple roll test. Last edited by j95; Aug 8th 2017 at 01:03 PM.
 Aug 8th 2017, 10:41 AM #2 Senior Member     Join Date: Jun 2016 Location: England Posts: 338 Yes But How is the coefficient of rolling resistance expressed? Would I be correct in guessing that this is a function of velocity? What you want to get to is something in the "F=ma" line. you have "a" (the deceleration), you need to determine "F" which I suspect should be derivable from the coefficient of rolling resistance. __________________ ~\o/~
Aug 8th 2017, 11:22 AM   #3
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 Originally Posted by j95 Is it possible to determine the mass of an object (i.e Bicycle) given the (Coefficient rolling resistance), (Initial Velocity) and the (Deceleration). Given that you are able to change the (Coefficient Rolling Resistance) and from this determine a new (Deceleration). Wind resistance and internal frictions are neglected. If so how?
Possible? Yes. Easy? No.

 Aug 8th 2017, 12:50 PM #4 Junior Member   Join Date: Aug 2017 Posts: 3 well yes, but the trouble is that you also need the mass to do that. To find the mass we would need to know either the momentum or the friction. But all we have is the "time to stop" which is dependent on both momentum and friction (I would think equally). So how to relate the time or (deceleration) to a force/momentum relation. For some reason I feel a linear approximation is necessary somewhere, because mass and the (tire size, tire pressure etc.) coefficients stay the same.
 Aug 9th 2017, 07:10 AM #5 Senior Member     Join Date: Jun 2016 Location: England Posts: 338 you indicate that you can change the coefficient of rolling resistance. This should allow you to set up a simultaneous equations scenario which will in turn allow you to solve a problem with more than one unknown. The key to this is the precise definition of "coefficient of rolling resistance" without knowing exactly what that means, I can't advise further. __________________ ~\o/~
 Aug 9th 2017, 08:10 AM #6 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,271 Rolling resistance is typically modelled as being equal to the coefficient of the rolling resistance (which is a constant) times the normal force. So mass cancels out: $\displaystyle F = ma = -m \mu g$ So $\displaystyle a = -\mu g$ As for changing the value of mu: I suppose you could deflate the tires to increase mu, but all that will do is shorten the time to decelerate to zero - it still doesn't help with determining the value for m.
Aug 9th 2017, 08:36 PM   #7
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 Originally Posted by Woody you indicate that you can change the coefficient of rolling resistance. This should allow you to set up a simultaneous equations scenario which will in turn allow you to solve a problem with more than one unknown. The key to this is the precise definition of "coefficient of rolling resistance" without knowing exactly what that means, I can't advise further.
Coefficient rolling "friction" (u) i suppose, would be the unit-less number in this case. Which I am able to find out, and change by moving to gravel or another higher coefficient.

The work done during the coasting would be is W=(m)(g)(u)(d)
Then the Impulse Momentum (change in momentum) would be Jn = (Fnet)(t) = m(Vi-Vf)

and obviously W = Fnet(d)
The work and Impulse would have to equal out in some way because they're the only two forces acting on the horizontal.

 Tags determining, mass, momentum

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