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October 23, 2014

October 23, 2014

Posted by **Chris** on Sunday, May 10, 2009 at 1:15am.

- Physics - THIS IS DUE TMRW PLEASE HELP! -
**Damon**, Sunday, May 10, 2009 at 10:51amFcentripetal = m Ac = m v^2/r = m v^2/26

toppling moment = height Fc = (.6/26) m v^2

Righting moment = m g (1.4/2)

start to tip when toppling moment = righting moment

(.6/26)v^2 = 9.8 (1.4/2)

- Physics - THIS IS DUE TMRW PLEASE HELP! -
**drwls**, Sunday, May 10, 2009 at 11:02amThe car will topple over if the centripetal force, applied at the center of mass, reaches a value such that the net torque about the wheels on the OUTSIDE of the curve is zero. Under these conditions, the centripetal-force torque about a line through the outer wheel's contact points with the ground will be equal and opposite to the gravity-force torque. At that speed, the inner wheels will have no ground force on them, and will start to lift. The car will topple.

The equation to solve(for V) is

(M V^2/R) * y = M g L

where y = 0.6 m , L = 1.40 m and R = 26.0 m. Note that M cancels out. Long wheelbase (high L) and low CG (low value of y) tend to increase the toppling velocity.

- Physics - THIS IS DUE TMRW PLEASE HELP! -
**AAD**, Thursday, November 12, 2009 at 8:34amThe max velocity is

Sqrt(Rgl/2h)

R = 26 m

g = 9.8 m/s^2

l = 1.4 m

h = 0.6 m

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