Question

A particle of mass \(m\) is moving in a circular path of constant radius \(r\) such that its centripetal acceleration \(a_{c}\) is varying with time \(t\) as \(a_{c} = k^{2}rt^{2}\), where \(k\) is a constant. The power delivered to the particle by the forces acting on it is:

• A. \(2\pi mk^{2}r^{2}t\)
• B. \(mk^{2}r^{2}t\)
• C. \(\frac{1}{3} mk^{4}r^{2}t^{5}\)
• D. 0
• E.

Hint:

Remember that centripetal force does zero work because it is always perpendicular to the displacement. Only the tangential component of force contributes to a change in speed and thus to the power delivered.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In circular motion, power is delivered only by the tangential force, as the centripetal force is always perpendicular to the velocity. Power is the rate of change of kinetic energy or the product of tangential force and velocity.

Step 2: Key Formula or Approach
1. Centripetal acceleration \(a_c = \frac{v^2}{r}\).
2. Tangential acceleration \(a_t = \frac{dv}{dt}\).
3. Power \(P = F_t \cdot v = (m a_t) \cdot v\).

Step 3: Detailed Explanation
1. Find velocity (\(v\)): Given \(a_c = k^2 r t^2\). Since \(a_c = \frac{v^2}{r}\): \[ \frac{v^2}{r} = k^2 r t^2 \implies v^2 = k^2 r^2 t^2 \implies v = krt \]
2. Find tangential acceleration (\(a_t\)): Differentiate velocity with respect to time: \[ a_t = \frac{dv}{dt} = \frac{d}{dt}(krt) = kr \]
3. Calculate Power (\(P\)): \[ P = F_t \times v = (m \times a_t) \times v \] \[ P = (m \times kr) \times (krt) = mk^2r^2t \]

Step 4: Final Answer
The power delivered to the particle is \(mk^{2}r^{2}t\).