A particle travels along the arc of a circle of radius r . Its speed depends on the distance travelled l as v = a l , where a is a constant. The angle α between the vectors of total acceleration and the velocity of the particle is

Let, when particle is at angular position θ then distance travelled = l v = a l But a t = d v d t = a 2 l d l d t = a v 2 l = a 2 2 And a c = v 2 r = ( a 2 l r ) Angle between a n e t & a t : α = tan − 1 ( 2 l r )

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Physics - Circular Motion Question with Solution | TestHub

PhysicsCircular MotionMiscellaneousEasy2 minPYQ_2020

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A particle travels along the arc of a circle of radius $r$ . Its speed depends on the distance travelled $l$ as $v = a \sqrt{l}$ , where $a$ is a constant. The angle $α$ between the vectors of total acceleration and the velocity of the particle is

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Answer:

Solution:

Let, when particle is at angular position $θ$ then distance travelled $= l$
$v = a \sqrt{l}$

But $a_{t} = \frac{d v}{d t} = \frac{a}{2 \sqrt{l}} \frac{d l}{d t} = \frac{a v}{2 \sqrt{l}} = \frac{a^{2}}{2}$

And $a_{c} = \frac{v^{2}}{r} = (\frac{a^{2} l}{r})$

Angle between $a_{n e t}$ & $a_{t}$ :

$α = \tan^{- 1} (\frac{2 l}{r})$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Circular MotionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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