About 2 vehicles going down hill

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Hi everyone, I have a simple question. 2 identical vehicles(tyres, shape, design) are set free from rest, going downhill. One vehicle is carrying a large weighted object while the other one is not.Hence, one vehicle has a larger mass than the other.

Which vehicle will reach downhill first? or they reach at the same time? in condition both start at the same place and same time in the first place.
 

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  • #2
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Hi everyone, I have a simple question. 2 identical vehicles(tyres, shape, design) are set free from rest, going downhill. One vehicle is carrying a large weighted object while the other one is not.Hence, one vehicle has a larger mass than the other.

Which vehicle will reach downhill first? or they reach at the same time? in condition both start at the same place and same time in the first place.
It really depends on a number of assumptions that could be made. However, practially speaking, and considering typical conditions, the heavier object will reach the bottom first.

Note that, in the absense of air resistance and friction and other non-ideal behavior, they would arrive at the same time, since gravitational mass and inertial mass are the same.
 
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Why the heavier reaches first then?

Yes, i did mean in the absense of air resistance and friction and other non-ideal behavior.
Is the inertial mass really is the same, considering the larger mass on the other?
 
  • #4
tiny-tim
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Hi Pagedown! :smile:

In the absence of air resistance etc, the heavier vehicle will accelerate slightly more, because of the "rolling mass" of the wheels (which you say are the same on each vehicle).

The force on each car is proportional to the mass, but to find the acceleration we divide by the "effective mass", which is the actual mass plus I/r2 (I is the moment of inertia ) for each wheel (you can check this using conservation of energy and the "rolling constraint" v = ωr) …

for the heavier car, this is slightly less than proportional to mass. :wink:

(If the cars were sliding instead of rolling the accelerations would be the same,)
 
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Note that if you consider perfect ideal case with zero friction, the wheels will not even spin. They will just slide.
 
  • #6
Ranger Mike
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T- Tim..excellent summary...makes perfect sense on a lot of things..this year we are switching flywheels in the race car..rules dictate that the flywheel can only weigh minimum of 10 pounds..but..for the big dollars there is a legal flywheel that has the ten pound mass concentrated in a 4 inch diameter as opposed to the standard 10 inch diameter...rumor has it this combination really has a quick ramp up to hi RPM..but the driver will stall the car off the grid until he gets used to the low torque during launch and has to spin up the revs until he drops the clutch when starting out...
 
  • #7
tiny-tim
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Hi Ranger Mike! :smile:

I'm confused … what is the flywheel attached to? :confused:

(and did you really mean minimum?)
 
  • #8
Ranger Mike
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Sorry T-tim i should have been more specific
I took the moment of inertia discussion into the realm of rotating weight of components in a race car. The flywheel is the traditional design connecting the crankshaft to the manual transmission. Sanctioning rules limit the weight of rotating and reciprocating components of the race car and there is a minimum weight limit of the flywheel. Lighter weight flywheels launch quicker than heavier flywheels..but heavier flywheel will carry a low horsepower , light weight car deeper into the turn than the light flywheel and this combination will " carry the car thru the turn" faster than the light weigh combo.
The moment of inertia topics is a very good subject for my particular application..thank you
 
  • #9
GT1
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Hi Pagedown! :smile:

In the absence of air resistance etc, the heavier vehicle will accelerate slightly more, because of the "rolling mass" of the wheels (which you say are the same on each vehicle).

The force on each car is proportional to the mass, but to find the acceleration we divide by the "effective mass", which is the actual mass plus I/r2 (I is the moment of inertia ) for each wheel (you can check this using conservation of energy and the "rolling constraint" v = ωr) …

for the heavier car, this is slightly less than proportional to mass. :wink:

(If the cars were sliding instead of rolling the accelerations would be the same,)
Why the effective mass is the actual mass+I/r^2 and why it's slightly less than proportional to mass?
 
  • #10
tiny-tim
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Hi GT1! :smile:

Because mgsinθ - friction = ma,

and friction times r = Iα = Ia/r,

so mgsinθ - Ia/r2 = ma,

ie mgsinθ = (m + I/r2)a …

that bracket is therefore the effective mass. :wink:

(and if you increase m but not I, then the bracket doesn't go up proportionally as much as m)
 

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