- #1

Pagedown

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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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- Thread starter Pagedown
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- #1

Pagedown

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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.

- #2

stevenb

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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.

- #3

Pagedown

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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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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/r

for the heavier car, this is slightly

(If the cars were sliding instead of rolling the accelerations

- #5

stevenb

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- #6

Ranger Mike

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- #7

tiny-tim

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Hi Ranger Mike!

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

(and did you really mean minimum?)

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

(and did you really mean minimum?)

- #8

Ranger Mike

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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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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/r^{2}(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 slightlylessthan proportional to mass.

(If the cars were sliding instead of rolling the accelerationswouldbe 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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Because mgsinθ - friction = ma,

and friction times r = Iα = Ia/r,

so mgsinθ - Ia/r

ie mgsinθ = (m + I/r

that bracket is therefore the effective mass.

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

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