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Physics - Circular Motion Question with Solution | TestHub

PhysicsCircular MotionKinematicsMedium2 minQB

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A particle rotates in a circular path of radius 32 m with varying speed $v = 2 t^{2}$ . Here $v$ is in m/s and $t$ in second. Find the angle between velocity and acceleration at $t = 2 s$ .

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Answer:

Solution:

$a_{t} = \frac{d v}{d t} = 4 t = 8 m / s^{2} at t = 2 s; a_{r} = \frac{v ^{2}}{R} = \frac{64}{32} = 2 m / s^{2} . tan θ = \frac{a _{r}}{a _{t}} = \frac{2}{8} = \frac{1}{4}, so θ = tan^{- 1} (\frac{1}{4}) .$

Stream:JEESubject:PhysicsTopic:Circular MotionSubtopic:Kinematics

⏱ 2mℹ️ Source: QB

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