A cyclist turns around a curve at 15, miles/h | vedclass

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(B) The centripetal force required for a cyclist to turn around a curve is given by $F = \frac{mv^2}{r}$,where $m$ is the mass,$v$ is the velocity,and

Vedclass · Accepted Answer

Quadrupled

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(B) The centripetal force required for a cyclist to turn around a curve is given by $F = \frac{mv^2}{r}$,where $m$ is the mass,$v$ is the velocity,and $r$ is the radius of the curve.From this relation,we can see that the force $F$ is directly proportional to the square of the velocity,i.e.,$F \propto v^2$.If the speed $v$ is doubled $(v' = 2v)$,the new force $F'$ becomes $F' \propto (2v)^2 = 4v^2$.Therefore,$F' = 4F$.This means the tendency to overturn becomes four times (quadrupled) the original value.

$A$ cyclist turns around a curve at $15\, miles/hour$. If he turns at double the speed,the tendency to overturn is

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