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This is actually for a engineering course in modelling, and not in physics per se, but it seems to me to be fairly basic physics. Apologies in advance if it's out of place.
A car with a mass of 1000 kg is held still on a slope with an inclination of 5.8 °, and then let go. Its speed v(t) is measured at intervals (t seconds): v(0) = 0 m/s, v(10) = 2.05 m/s, v(20) = 3.30 m/s, v(30) = 4.15 m/s, v(40) = 4.85 m/s, v(50) = 5.20 m/s, v(60) = 5.55 m/s. Find the effective damping constant [itex]b[/itex].
Ideal damper: [itex]F_b (t) = bv(t)[/itex]
Okay, let me preface by saying that the problem before this one was similar: a car with mass 1600 kg is sped up, and the gas pedal is let go at t=0; speed is measured once every ten seconds from t=0 to t=60 s. (Measure data is v(0) = 4.6 m/s, v(10) = 3.1 m/s, v(20) = 2.0 m/s, v(30) = 1.37 m/s, v(40) = 0.88 m/s, v(50) = 0.64 m/s, v(60) = 0.38 m/s.) That problem asked me to model the car as a mass with a damper, and to just approximate [itex]\frac{dv}{dt}[/itex] by drawing a curve of [itex]v(t)[/itex] and estimate the slopes at points in order to find the damping constant [itex]b[/itex].
I solved that problem (at least I think I did) by saying that the resulting force on the mass is [itex]-bv(t)[/itex] which is equal in size to [itex]m \frac{dv}{dt}[/itex], so [itex]b = \frac{-m}{v(t)} \frac{dv}{dt}[/itex]; I got [itex]b[/itex] to between 60-68 kg/s, except for t=60s where it went up to 106 kg/s. So my answer there was that the damping constant [itex]b[/itex] was around 65 kg/s.
Now! This previous problem implies that the problem which is the topic of this thread is to be solved in a similar manner. My thinking is that, if friction and drag can be rolled into this damper model, gravity will make the car accelerate until the damper force cancels it out: [itex]m \frac{dv}{dt} = mg_{\text{parallel}} - bv(t)[/itex] (where i have [itex]g_{\text{parallel}} = 9.82 \sin 5.8°[/itex]). My thinking was that I could just, as before, solve for [itex]b[/itex] and plug in values for [itex]v(t)[/itex] and approximations of [itex]\frac{dv}{dt}[/itex] without even having to solve a differential equation, but I only get wildly varying values of [itex]b[/itex]. I.e. if [itex]b(t) = \frac{mg_{\text{parallel}} - m \frac{dv}{dt}}{v(t)}[/itex], then b(10)=404, b(20)=267, b(30)=222, b(40)=195, b(50)=183, b(60)=171. Solving the differential equation to [itex]v(t) = \frac{mg_{\text{parallel}}}{b} (1-e^{-\frac{bt}{m}})[/itex], inputting known values and solving for [itex]b[/itex] (thank you, Wolfram Alpha) gave me values b=480, b=300, b=239, b=205, b=191, b=179. This damping constant doesn't look very constant to me.
And that's where I hit the wall. Can someone tell me, what am I doing wrong here? Are my assumptions wrong? Am I misunderstanding what is requested? I suspect that the differential equation is lacking something, but I don't know what, and I can't seem to get any clues from the accompanying course text.
Homework Statement
A car with a mass of 1000 kg is held still on a slope with an inclination of 5.8 °, and then let go. Its speed v(t) is measured at intervals (t seconds): v(0) = 0 m/s, v(10) = 2.05 m/s, v(20) = 3.30 m/s, v(30) = 4.15 m/s, v(40) = 4.85 m/s, v(50) = 5.20 m/s, v(60) = 5.55 m/s. Find the effective damping constant [itex]b[/itex].
Homework Equations
Ideal damper: [itex]F_b (t) = bv(t)[/itex]
The Attempt at a Solution
Okay, let me preface by saying that the problem before this one was similar: a car with mass 1600 kg is sped up, and the gas pedal is let go at t=0; speed is measured once every ten seconds from t=0 to t=60 s. (Measure data is v(0) = 4.6 m/s, v(10) = 3.1 m/s, v(20) = 2.0 m/s, v(30) = 1.37 m/s, v(40) = 0.88 m/s, v(50) = 0.64 m/s, v(60) = 0.38 m/s.) That problem asked me to model the car as a mass with a damper, and to just approximate [itex]\frac{dv}{dt}[/itex] by drawing a curve of [itex]v(t)[/itex] and estimate the slopes at points in order to find the damping constant [itex]b[/itex].
I solved that problem (at least I think I did) by saying that the resulting force on the mass is [itex]-bv(t)[/itex] which is equal in size to [itex]m \frac{dv}{dt}[/itex], so [itex]b = \frac{-m}{v(t)} \frac{dv}{dt}[/itex]; I got [itex]b[/itex] to between 60-68 kg/s, except for t=60s where it went up to 106 kg/s. So my answer there was that the damping constant [itex]b[/itex] was around 65 kg/s.
Now! This previous problem implies that the problem which is the topic of this thread is to be solved in a similar manner. My thinking is that, if friction and drag can be rolled into this damper model, gravity will make the car accelerate until the damper force cancels it out: [itex]m \frac{dv}{dt} = mg_{\text{parallel}} - bv(t)[/itex] (where i have [itex]g_{\text{parallel}} = 9.82 \sin 5.8°[/itex]). My thinking was that I could just, as before, solve for [itex]b[/itex] and plug in values for [itex]v(t)[/itex] and approximations of [itex]\frac{dv}{dt}[/itex] without even having to solve a differential equation, but I only get wildly varying values of [itex]b[/itex]. I.e. if [itex]b(t) = \frac{mg_{\text{parallel}} - m \frac{dv}{dt}}{v(t)}[/itex], then b(10)=404, b(20)=267, b(30)=222, b(40)=195, b(50)=183, b(60)=171. Solving the differential equation to [itex]v(t) = \frac{mg_{\text{parallel}}}{b} (1-e^{-\frac{bt}{m}})[/itex], inputting known values and solving for [itex]b[/itex] (thank you, Wolfram Alpha) gave me values b=480, b=300, b=239, b=205, b=191, b=179. This damping constant doesn't look very constant to me.
And that's where I hit the wall. Can someone tell me, what am I doing wrong here? Are my assumptions wrong? Am I misunderstanding what is requested? I suspect that the differential equation is lacking something, but I don't know what, and I can't seem to get any clues from the accompanying course text.
