(B) The maximum speed $v_{m}$ at which the cyclist will not skid is given by the formula $v_{m} = \sqrt{\mu r g}$.Substituting the given values: $\mu

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(B) The maximum speed $v_{m}$ at which the cyclist will not skid is given by the formula $v_{m} = \sqrt{\mu r g}$.Substituting the given values: $\mu = 0.2$,$r = 3 \; m$,and $g = 10 \; m \cdot s^{-2}$.$v_{m} = \sqrt{0.2 \times 3 \times 10} = \sqrt{6} \approx 2.45 \; m \cdot s^{-1}$.To convert this speed into $km \cdot h^{-1}$,we multiply by $\frac{18}{5}$:$v_{m} = 2.45 \times 3.6 = 8.82 \; km \cdot h^{-1}$.The cyclist will not skid if their speed is less than or equal to $8.82 \; km \cdot h^{-1}$.Among the given options,$7.2 \; km \cdot h^{-1}$ is the only speed less than $8.82 \; km \cdot h^{-1}$.

Class 11 Physics Newton's Laws of Motion and Friction Circular motion with Friction

Medium

$A$ cyclist on a level road takes a sharp circular turn of radius $3 \; m$ $(g = 10 \; m \cdot s^{-2})$. If the coefficient of static friction between the cycle tyres and the road is $0.2$,at which of the following speeds will the cyclist not skid while taking the turn?

NEET 2017 All NEET

A
$9 \; km \cdot h^{-1}$
B
$7.2 \; km \cdot h^{-1}$
C
$10.8 \; km \cdot h^{-1}$
D
$14.4 \; km \cdot h^{-1}$

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Newton's Laws of Motion and Friction

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Circular motion with Friction

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$A$ cyclist on a level road takes a sharp circular turn of radius $3 \; m$ $(g = 10 \; m \cdot s^{-2})$. If the coefficient of static friction between the cycle tyres and the road is $0.2$,at which of the following speeds will the cyclist not skid while taking the turn?

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$A$ gramophone record is rotating with an angular velocity $\omega$. $A$ coin is placed at a distance $r$ from the center of the record. The coefficient of static friction is $\mu$. The coin will revolve with the record if .......

The coefficient of friction between the tyres and the road is $0.25$. The maximum speed with which a car can be driven round a curve of radius $40 \,m$ without skidding is ........ $ms^{-1}$ (assume $g = 10 \,ms^{-2}$)

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