JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 14)
A modern grand - prix racing car of mass m is travelling on a flat track in a circular arc of radius R with a speed v. If the coefficient of static friction between the tyres and the track is $$\mu$$s, then the magnitude of negative lift FL acting downwards on the car is : (Assume forces on the four tyres are identical and g = acceleration due to gravity)
_17th_March_Morning_Shift_en_14_1.png)
$$m\left( {g - {{{v^2}} \over {{\mu _s}R}}} \right)$$
$$ - m\left( {g + {{{v^2}} \over {{\mu _s}R}}} \right)$$
$$m\left( {{{{v^2}} \over {{\mu _s}R}} - g} \right)$$
$$m\left( {{{{v^2}} \over {{\mu _s}R}} + g} \right)$$
Explanation
_17th_March_Morning_Shift_en_14_2.png)
$$f = {{m{v^2}} \over R}$$
$$ \Rightarrow $$ $$\mu N = {{m{v^2}} \over R}$$
$$\mu (mg + {F_L}) = {{m{v^2}} \over R}$$
$$ \Rightarrow $$ $$mg + {F_L} = {{m{v^2}} \over {R\mu }}$$
$$ \Rightarrow $$ $${F_L} = {{m{v^2}} \over {\mu R}} - mg$$
$$ = m\left( {{{{v^2}} \over {\mu R}} - g} \right)$$
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