Question 37: Work, Energy and Power & Circular Motion

Question

A car of mass ' $m$ ' moves on a banked road having radius ' $r$ ' and banking angle $	heta$. To avoid slipping from 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

Accepted Answer

$$ N \cos	heta - \mu N \sin	heta = mg $$
$$ N \sin	heta + \mu N \cos	heta = \frac{mv_0^2}{r} $$
Dividing the second equation by the first gives
$$ \frac{v_0^2}{r} = \frac{g\left(\sin	heta + \mu \cos	hetaight)}{\cos	heta - \mu \sin	heta}. $$
Multiplying both sides by $\cos	heta - \mu\sin	heta$ leads to
$$ \frac{v_0^2}{r}(\cos	heta - \mu \sin	heta) = g(\sin	heta + \mu \cos	heta). $$
Expanding and grouping the terms with $\mu$:
$$ \frac{v_0^2}{r}\cos	heta - g\sin	heta = \mu\left(\frac{v_0^2}{r}\sin	heta + g\cos	hetaight). $$
Thus, solving for $\mu$:
$$ \mu = \frac{\frac{v_0^2}{r}\cos	heta - g\sin	heta}{\frac{v_0^2}{r}\sin	heta + g\cos	heta}. $$
Multiplying the numerator and the denominator by $r$ gives
$$ \mu = \frac{v_0^2\cos	heta - rg\sin	heta}{v_0^2\sin	heta + rg\cos	heta}. $$
Dividing both the numerator and denominator by $\cos	heta$ (assuming $\cos	heta 
eq 0$):
$$ \mu = \frac{v_0^2 - rg	an	heta}{v_0^2	an	heta + rg}. $$
This result matches the expression
$$ \mu=\frac{v_0^2 - rg\,	an	heta}{rg + v_0^2\,	an	heta}. $$
Thus, the correct answer is:
Option C

Answer

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

Answer

$\mu=\frac{v_0^2-r g 	an 	heta}{\mathrm{rg}-\mathrm{v}_{\mathrm{o}}^2 	an 	heta}$

Answer

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

Answer

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

Work, Energy and Power & Circular Motion

A car of mass ' m ' moves on a banked road having radius ' r ' and banking angle \theta. To avoid slipping from 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

Choose the correct answer: