A car of mass m moves on a banked road having | vedclass

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(C) For a car moving at maximum speed $v_0$ on a banked road,the forces acting on it are the normal force $N$,gravity $mg$,and the maximum static fric

Vedclass · Accepted Answer

$\mu=\frac{v_0^2-r g 	an 	heta}{r g+v_0^2 	an 	heta}$

Vedclass · Answer

(C) For a car moving at maximum speed $v_0$ on a banked road,the forces acting on it are the normal force $N$,gravity $mg$,and the maximum static friction $f = \mu N$ acting down the incline to prevent slipping outwards.Resolving forces horizontally and vertically:$N \sin 	heta + f \cos 	heta = \frac{m v_0^2}{r}$$N \cos 	heta - f \sin 	heta = m g$Substituting $f = \mu N$ into the equations:$N(\sin 	heta + \mu \cos 	heta) = \frac{m v_0^2}{r}$$N(\cos 	heta - \mu \sin 	heta) = m g$Dividing the two equations:$\frac{\sin 	heta + \mu \cos 	heta}{\cos 	heta - \mu \sin 	heta} = \frac{v_0^2}{r g}$Cross-multiplying:$r g \sin 	heta + \mu r g \cos 	heta = v_0^2 \cos 	heta - \mu v_0^2 \sin 	heta$Rearranging terms to solve for $\mu$:$\mu(r g \cos 	heta + v_0^2 \sin 	heta) = v_0^2 \cos 	heta - r g \sin 	heta$Dividing by $\cos 	heta$:$\mu(r g + v_0^2 	an 	heta) = v_0^2 - r g 	an 	heta$$\mu = \frac{v_0^2 - r g 	an 	heta}{r g + v_0^2 	an 	heta}$

$A$ car of mass $m$ moves on a banked road having radius $r$ and banking angle $\theta$. To avoid slipping from the banked road,the maximum permissible speed of the car is $v_0$. The coefficient of friction $\mu$ between the wheels of the car and the banked road is:

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