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The car shown above with mass M is turning to the left with an uniform angular speed W on a circular path with radius R. When the angular speed is increased to a critical value C, one of the normal forces vanishes. If W is increased beyond C, the car will roll over, explain why.

I did some calculations and found C = sqrt(gl/hR). However I ran into some troubles explaining why it will roll over. By the free body diagram shown below

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When N1 =0, Fs1 =0 and taking moment about center of mass, the only forces are Fs2 and N2, however unless l is significantly smaller than h, by the free body diagram, the car will actually lean in since N2 > Fs2?

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I believe the following may be useful to consider.

As you point out, once $N_{1}=0$ and $F_{S1}=0$ the only forces remaining which can produce a torque about the center of mass are $N_{2}$ and $F_{S2}$.

However, since we assume the car is still in uniform circular motion around the curve, $F_{S2}$ must adjust as needed in order to maintain the uniform circular motion condition: $\frac{m v^{2}}{R}$. I believe this is exactly balancing the torque by $N_{2}$ at the critical velocity, so any further increase in speed will produce a larger torque than that produced by $N_{2}$ and cause a roll towards the outside of the curve.

I hope this helps.

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  • $\begingroup$ Fs2 has to adjust to maintain the condition, however Fs2 = kN where k is the coefficient of kinetic friction, then shouldn't Fs2*h < N2*l regardless? $\endgroup$
    – tangolin
    Nov 20 '20 at 2:23

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