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Excelled
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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 b.
Ideal damper: F_b (t) = bv(t)
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 \frac{dv}{dt} by drawing a curve of v(t) and estimate the slopes at points in order to find the damping constant b.
I solved that problem (at least I think I did) by saying that the resulting force on the mass is -bv(t) which is equal in size to m \frac{dv}{dt}, so b = \frac{-m}{v(t)} \frac{dv}{dt}; I got b 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 b 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: m \frac{dv}{dt} = mg_{\text{parallel}} - bv(t) (where i have g_{\text{parallel}} = 9.82 \sin 5.8°). My thinking was that I could just, as before, solve for b and plug in values for v(t) and approximations of \frac{dv}{dt} without even having to solve a differential equation, but I only get wildly varying values of b. I.e. if b(t) = \frac{mg_{\text{parallel}} - m \frac{dv}{dt}}{v(t)}, then b(10)=404, b(20)=267, b(30)=222, b(40)=195, b(50)=183, b(60)=171. Solving the differential equation to v(t) = \frac{mg_{\text{parallel}}}{b} (1-e^{-\frac{bt}{m}}), inputting known values and solving for b (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 b.
Homework Equations
Ideal damper: F_b (t) = bv(t)
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 \frac{dv}{dt} by drawing a curve of v(t) and estimate the slopes at points in order to find the damping constant b.
I solved that problem (at least I think I did) by saying that the resulting force on the mass is -bv(t) which is equal in size to m \frac{dv}{dt}, so b = \frac{-m}{v(t)} \frac{dv}{dt}; I got b 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 b 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: m \frac{dv}{dt} = mg_{\text{parallel}} - bv(t) (where i have g_{\text{parallel}} = 9.82 \sin 5.8°). My thinking was that I could just, as before, solve for b and plug in values for v(t) and approximations of \frac{dv}{dt} without even having to solve a differential equation, but I only get wildly varying values of b. I.e. if b(t) = \frac{mg_{\text{parallel}} - m \frac{dv}{dt}}{v(t)}, then b(10)=404, b(20)=267, b(30)=222, b(40)=195, b(50)=183, b(60)=171. Solving the differential equation to v(t) = \frac{mg_{\text{parallel}}}{b} (1-e^{-\frac{bt}{m}}), inputting known values and solving for b (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.
