Rollover Torque Calculation for a Vehicle in a Circular Path

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This shows a vehicle whose floor is at angle $\theta$ to the road bed. The angle between the floor of the vehicle and the vector from the pivot point at the wheel to the center of gravity is given by $\tan {\theta }_c=\frac{h}{\frac{t}{3}}$ . The radius of the rollover rotation is the hypotenuse of the $(h,\frac{t}{2})$ right triangle, which is $\sqrt{h^2+{\left(\frac{t}{2}\right)}^2}$ . At the rollover point, the center of gravity is over the pivot point. Then the angle $\theta$ is given by $\tan {\theta }_p=\frac{\frac{t}{2}}{h}$ . The centrifugal force $f_c=m{a}_c$ , where $a_c=$ the centripetal acceleration.

In the equation above it appears that the centripetal acceleration $a=\frac{v^2}{r}\mathrm{ }\mathrm{ }$ is constant, but that is not true. Energy is assumed conserved, so $v^2\mathrm{ }\mathrm{ }$ must decrease to $v^2-2g(\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}-h)$ as the center of mass rises from initially $h\mathrm{ }\mathrm{ }$ to $\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}$ at the critical point of rollover. Also, since the wheels are locked to movein a circle of radius $r$ , the radius of the center of mass changes from initially $r\mathrm{ }\mathrm{ }$ to $r+\frac{t}{2}\mathrm{ }\mathrm{ }$ at the critical point of rollover. As functions of $\theta$ , the angle between the vehicle floor and the road bed, the equations are:

${\left[v\left(\theta \right)\right]}^2=v^2-2g[\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}\sin (\theta +{\theta }_c)-h]=v^2-2g[h(\cos \theta -1)+\frac{t}{2}\sin \theta ]$ and $$

$r\left(\theta \right)=r+\frac{t}{2}-\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}\cos (\theta +{\theta }_c)=r+\frac{t}{2}(1-\cos \theta )+h\sin \theta$ ,

where $r\mathrm{ }\mathrm{ }\text{and}\mathrm{ }\mathrm{ }v\mathrm{ }\mathrm{ }\text{are the initial values and}\mathrm{ }\mathrm{ }{\theta }_c=\arctan \left(\frac{h}{\frac{t}{2}}\right)$ is the initial angle between the vector to the center of mass from the pivotpoint and the vehicle floor. Then the torque/weight as a function of $\theta \mathrm{ }\mathrm{ }$ is

$\frac{\tau \left(\theta \right)}{w}=\frac{v^2-2g[\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}\sin (\theta +{\theta }_c)-h]}{g[r+\frac{t}{2}-\sqrt{{\left(\frac{t}{2}\right)}^2+h^2}\cos (\theta +{\theta }_c\left)\right]}(h\cos \theta +\frac{t}{2}\sin \theta )+h\sin \theta -\frac{t}{2}\cos \theta$ ,

$\frac{\tau \left(\theta \right)}{w}=\frac{v^2-2g[h(\cos \theta -1)+\frac{t}{2}\sin \theta ]}{g[r+\frac{t}{2}(1-\cos \theta )+h\sin \theta ]}(h\cos \theta +\frac{t}{2}\sin \theta )+h\sin \theta -\frac{t}{2}\cos \theta$ , where we have used the trigonometric identities $\sin (a+b)=\sin a\cos b+\cos a\sin b\mathrm{ }\mathrm{ }$ and $\cos (a+b)=\cos a\cos b-\sin a\sin b$ .