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United States24782 Posts
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Okay I'm gonna have a crack at it since I'm bored and there aren't any games on tonight.
I'll start off by using conservation of energy, since the wheels roll without slipping, no energy is lost due to work done by non-conservative forces:
(m1 + 4m2)gh = 1/2 (m1) v^2 + 1/2 (4) I * w^2 (1)
, where I is the inertia of the wheel, w is the angular velocity of the wheel and v is the linear velocity of the wheel once it reaches the bottom of the hill.
Assuming the wheels of the car are thin solid disks, I = 1/2 m2 * R^2. Also, since the car's wheels roll without slipping, we can say w = v/R.
Thus, equation (1) becomes
(m1 + 4m2)gh = 1/2 (m1) v^2 + 2*(1/2) m2 * R^2 * m2^2/R^2 => (m1 + 4m2)gh = v^2 (1/2 m1 + m2).
Therefore,
v = [(m1 + 4m2)gh / (1/2 m1 + m2)]^(1/2).
Since the car is elevated a height h, at an angle theta above the horizontal, the total distance the car travels before it reaches the bottom of the hill is h*sin(theta).
The cart moves under constant acceleration so we can use the equations for 1D kinematics.
vf^2 = v0^2 + 2a(delta s)
=> a = [(m1 + 4m2)gh / (h/sin(theta)(1/2 m1 + m2))].
Thus, h/sin(theta) = 1/2 at^2 =>
t = h/sin(theta)[(1/2m1 + m2)/((m1+4m2)gh)]^(1/2).
Since we have the acceleration now, the angular acceleration of the wheels.
alpha = a/R => alpha = [(m1 + 4m2)gh / (h*R/sin(theta)(1/2 m1 + m2))].
The friction force on each wheel produces the torque on the wheels that cause the wheels to rotate.
Torque = I(alpha) => f(static) * R = I (alpha),
thus the amount of friction force required to rotate each wheel is
f(static) = I(alpha)/R =>
f(static) = 1/2 m2 *[(m1 + 4m2)gh / (h/sin(theta)(1/2 m1 + m2))].
*Edit 1* wow, I need to learn to read. I'm gna go back and figure out the rest of this stuff. Haha.
*Edit 2* okay finished. Hope I didn't just make a fool out of myself.
Cheers :D
*Edit 3* fixed for stupid mistakes. Confirmed that I've succeeded in making a fool out of myself.
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so I guess you didn't go along with what yubee suggested  
edit: I'll give it a shot when I have more time, gtg now
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United States24782 Posts
On July 11 2008 17:36 popnfreshspk wrote:
Since the car is elevated a height h, at an angle theta above the horizontal, the total distance the car travels before it reaches the bottom of the hill is h*sin(theta).
I believe this is erroneous. Sin(theta)=h/hypotenuse
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On July 11 2008 18:21 micronesia wrote:Show nested quote +On July 11 2008 17:36 popnfreshspk wrote:
Since the car is elevated a height h, at an angle theta above the horizontal, the total distance the car travels before it reaches the bottom of the hill is h*sin(theta).
I believe this is erroneous. Sin(theta)=h/hypotenuse
Ah whoops. Went through it too quick. Guess all the * sin(theta) becomes / sin(theta) then? Fixed in my original post.
Sorry about that. It'd be nice if the hypotenuse of the right triangle is longer then one of the legs yeah, yeah? = =.
Were there any other errors?
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used lagrange formalism:
L= T-U = 0,5 m1*v^2+ 2 *I *V^2/R^2 + m1*g*sin()*x
proper potential energy with U(h=0) = 0 would be U=(s*sin-x*sin)*m1*g not just -x*sin*m1*g but it doesnt matter anyway when derivated.
solving the lagrange equation to get the differential equiation for x/v:
d/dt(d/dv L) - d/dx L = 0
=> (m1+ 4* I /R^2) a - m1*g*sin() = 0 with (m1+4 *I / R^2) := b
=> a= m1* g* sin()/b = const.
(reasonable result, for m1>>m2 a becomes g*sin as one would expect)
=> 0,5 a*t^2 = s = h/ sin() => t=sqrt( 2 s/ a) = sqrt( 2h*(m1+4 *I/ R^2) / (m1*sin^2()*g))
ez^^
(if you dont want to ignore the rolling resistance you dont need to add a dissipative force. should work to just decrease the potential energy by Cr*Fn(s-x) where Cr is the coefficent for the roling resistence and Fn the normal component of G. this way the acceleration would be a= m1* g* (sin()- cos()*Cr)/b hence slighty smaller.
(I = 0,5 m2*R^2 if the tires are cylinder.)
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what the hell T_T i attempted it with my little physics I high school knowledge
but..
>.<
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this is way above my level of physics. im like whut?
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On July 11 2008 19:58 aqui wrote:
used lagrange formalism:
L= T-U = 0,5 m1*v^2+ 2 *I *V^2/R^2 + m1*g*sin()*x
proper potential energy with U(h=0) = 0 would be U=(s*sin-x*sin)*m1*g not just -x*sin*m1*g but it doesnt matter anyway when derivated.
solving the lagrange equation to get the differential equiation for x/v:
d/dt(d/dv L) - d/dx L = 0
=> (m1+ 4* I /R^2) a - m1*g*sin() = 0 with (m1+4 *I / R^2) := b
=> a= m1* g* sin()/b = const.
(reasonable result, for m1>>m2 a becomes g*sin as one would expect)
=> 0,5 a*t^2 = s = h/ sin() => t=sqrt( 2 s/ a) = sqrt( 2h*(m1+4 *I/ R^2) / (m1*sin^2()*g))
ez^^
(I = 0,5 m*R^2 if the tires are cylinder.)
Haha aqui, that´s like one of the first things one does in theoretical physics I: mechanics.
Lagrange formalism is ez, that´s true. Anyways nicely done, good old memories 
just for some nitpicking: I=0.5*m2*R^2 if the tires are cylinders 
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United States24782 Posts
Neat. I applied Newton's law to the displacement formula where t and frictional force were unknown, and then did the same thing for rotation to get a second equation, and solved the system.
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EPT
