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A particle of mass m moves in a circular path of radius r, under the action of force which delivers it constant power p and increases it's speed. The angular acceleration of particle at time (t) is proportional to A) 1/√t B) √t C) t^0 D) t^3/2?
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A particle of mass m moves in a circular path of radius r, under the a...
Answer:

The given conditions can be expressed mathematically as:


  • Force = p

  • Speed = v(t)

  • Angular velocity = ω(t) = v(t)/r

  • Angular acceleration = α(t) = dv(t)/dt * 1/r

  • Power = p = F.v(t) = m.a(t).v(t) = m.v(t).dv/dt = m.v(t).α(t).r


Deriving the expression for angular acceleration:

By substituting the value of power in the equation for power, we get:


  • m.v(t).α(t).r = p

  • α(t) = p/(m.v(t).r)


Since the force is constant, the acceleration is inversely proportional to the speed.


  • v(t).α(t) = p/(m.r)

  • dv/dt * α(t) + v(t).dα/dt = 0

  • v(t).dα/dt = -α^2(t)

  • dα/α^2 = -dv/v

  • 1/α(t) = kt + C (where k and C are constants)

  • α(t) = 1/(kt + C)


Therefore, the angular acceleration of the particle at time (t) is proportional to 1/√t, which means the correct option is A) 1/√t.
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A particle of mass m moves in a circular path of radius r, under the a...
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A particle of mass m moves in a circular path of radius r, under the action of force which delivers it constant power p and increases it's speed. The angular acceleration of particle at time (t) is proportional to A) 1/√t B) √t C) t^0 D) t^3/2?
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