A motorcyclist has to rotate in horizontal ci | vedclass

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(B) For a motorcyclist moving in a horizontal circle inside a cylindrical wall,the forces acting on the motorcyclist are: $1$. The weight $(mg)$ actin

Vedclass · Accepted Answer

$\sqrt{\frac{Rg}{\mu_{s}}}$

Vedclass · Answer

(B) For a motorcyclist moving in a horizontal circle inside a cylindrical wall,the forces acting on the motorcyclist are: $1$. The weight $(mg)$ acting downwards. $2$. The normal reaction $(N)$ from the wall acting towards the center of the circle,which provides the necessary centripetal force: $N = \frac{mv^2}{R}$. $3$. The frictional force $(f)$ acting upwards to balance the weight: $f = mg$. For the motorcyclist not to slip downwards,the frictional force must be less than or equal to the limiting friction: $f \leq \mu_{s} N$. Substituting the values: $mg \leq \mu_{s} \left( \frac{mv^2}{R} \right)$. $g \leq \frac{\mu_{s} v^2}{R}$. $v^2 \geq \frac{Rg}{\mu_{s}}$. Therefore,the minimum speed required is $v_{min} = \sqrt{\frac{Rg}{\mu_{s}}}$.

$A$ motorcyclist has to rotate in horizontal circles inside the cylindrical wall of inner radius '$R$' metre. If the coefficient of friction between the wall and the tyres is '$\mu_{s}$',then the minimum speed required is ($g=$ acceleration due to gravity).

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