Question

A particle of mass m starts moving in a circular path of canstant radiur r , such that iss centripetal accelerationa_(c) is varying with time a=t as (a_(c)=k^(2)r//t) , where K is a contant. What is the power delivered to the particle by the force acting on it ?

Answer 1

Given:

A particle of mass m.

Circular path of constant radius r

Centripetal acceleration, a(t) = k² r / t , where K is a contant

To Find:

The power delivered to the particle by the force acting on it.

Solution:

We know,

• Centripetal acceleration = V²/r = k²r/t
• V² = k²r²/t
• V  = k r / √t

Tangential acceleration is given by dv/dt

Therefore from above V ,

• dv/dt = -kr/2$t^{3/2}$

Power delivered = Force x Velocity

Therefore,

• Power = ma x V
• Power = m x dv/dt x V

• Power = m x  -k r/2$t^{3/2}$ x k r / √t

• Power = $\frac{-mk^{2}r^{2}}{2t^{2}}$

The power delivered to the particle by the force acting on it is   P =  $\frac{-mk^{2}r^{2}}{2t^{2}}$ W

Answer 2

Answer:

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Explanation: