Question 122: Work, Energy and Power & Circular Motion

Question

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 F
L
acting downwards on the car is : (Assume forces on the four tyres are identical and g = acceleration due to gravity)

Accepted Answer

$$f = {{m{v^2}} \over R}$$
$$ \Rightarrow $$ $$\mu N = {{m{v^2}} \over R}$$
$$\mu (mg + {F_L}) = {{m{v^2}} \over R}$$
$$ \Rightarrow $$ $$mg + {F_L} = {{m{v^2}} \over {R\mu }}$$
$$ \Rightarrow $$ $${F_L} = {{m{v^2}} \over {\mu R}} - mg$$
$$ = m\left( {{{{v^2}} \over {\mu R}} - g} \right)$$

Answer

$$m\left( {g - {{{v^2}} \over {{\mu _s}R}}} \right)$$

Answer

$$ - m\left( {g + {{{v^2}} \over {{\mu _s}R}}} \right)$$

Answer

$$m\left( {{{{v^2}} \over {{\mu _s}R}} - g} \right)$$

Answer

$$m\left( {{{{v^2}} \over {{\mu _s}R}} + g} \right)$$

Work, Energy and Power & Circular Motion

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