A cyclist riding the bicycle at a speed of 14 3 m s - 1 takes a turn around a circular road of radius 20 3 m without skidding. Given g = 9.8 m s - 2 What is his inclination to the vertical ?

θ = t a n - 1 v 2 r g = t a n - 1 ( 14 3 ) 2 20 3 × 9.8 = t a n - 1 [ 3 ] = 60 °

Physics - Circular Motion Question with Solution | TestHub

PhysicsCircular MotionMiscellaneousEasy2 minPYQ_2020

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A cyclist riding the bicycle at a speed of $14 \sqrt{3} m s^{- 1}$ takes a turn around a circular road of radius $20 \sqrt{3} m$ without skidding. Given $g = 9.8 m s^{- 2}$ What is his inclination to the vertical ?

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$θ = t a n^{- 1} (\frac{v^{2}}{r g}) = t a n^{- 1} [\frac{(14 \sqrt{3})^{2}}{20 \sqrt{3} \times 9.8}] = t a n^{- 1} [\sqrt{3}] = 60 °$

Stream:JEESubject:PhysicsTopic:Circular MotionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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Physics - Circular Motion Question with Solution | TestHub

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