Question

A particle of mass 8kg is moving in a circular path of radius 16m its velocity varies with time as v=√t. Calculate the centripetal force on the particle when its net acceleration vector is making an angle of 45° with the velocity vector at that moment.​

Answer 1

Given:

A particle of mass 8kg is moving in a circular path of radius 16m its velocity varies with time as v=√t.

To find:

Centripetal force when net acceleration vector is making an angle of 45° with Velocity vector.

Calculation:

Tangential acceleration

$\therefore \: a_{t} = \dfrac{dv}{dt} = \dfrac{d( \sqrt{t} )}{dt} = \dfrac{1}{2 \sqrt{t} }$

Centripetal acceleration

$\therefore \: a_{c} = \dfrac{ {v}^{2} }{r} = \dfrac{t}{r} = \dfrac{t}{16}$

Now , net acceleration is making 45° angle with Velocity vector , hence it will also make 45° angle with a_(t).

$\therefore \: \tan( {45}^{ \circ} ) = \dfrac{a_{t}}{a_{c}}$

$= > a_{t} = a_{c}$

$= > \dfrac{1}{2 \sqrt{t} } = \dfrac{t}{16}$

$= > \: {t}^{ \frac{3}{2} } = 8$

$= > \: t = {(8)}^{ \frac{2}{3} }$

$= > \: t = {2}^{2}$

$= > \: t = 4 \: sec$

So , centripetal force at t = 4 sec:

$\therefore \: f = \dfrac{m {v}^{2} }{r}$

$= > \: f = \dfrac{m {( \sqrt{t}) }^{2} }{r}$

$= > \: f = \dfrac{m t}{r}$

$= > \: f = \dfrac{8 \times 4}{16}$

$= > \: f = 2 \: newton$

So, final answer is:

$\boxed{\bf{centripetal \: force = 2 \: newton}}$

Answer 2

Explanation:

Centripetal acceleration, ac=k2rt2

where, ac=rv2

⇒rv2=k2rt2

⇒v=krt ........(1)

Tangential acceleration, at=dtdv=kr.........(2)

Tangential force acting on the particle, F=mat=mkr

Power delivered, P=F.v=Fvcosθ

∴   P=Fv=(mkr)×krt  (∵θ=0o)

⟹P=mk2r2t

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