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 :

Options

A. `2pimk^(2)r^(2)t`
B. m`mk^(2)r^(2)t`
C. `1//3mk^(4)r^(2)t_(5)`
D. zero

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the given centripetal acceleration
The centripetal acceleration \( a_c \) is given by the equation:
\[ a_c = k^2 r t^2 \]
where \( k \) is a constant, \( r \) is the radius, and \( t \) is time.

### Step 2: Relate centripetal acceleration to velocity
We know that centripetal acceleration is also defined as:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the linear velocity of the particle.

### Step 3: Solve for velocity squared
From the two equations for centripetal acceleration, we can equate them:
\[ \frac{v^2}{r} = k^2 r t^2 \]
Multiplying both sides by \( r \) gives:
\[ v^2 = k^2 r^2 t^2 \]

### Step 4: Calculate kinetic energy
The kinetic energy \( KE \) of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the expression for \( v^2 \):
\[ KE = \frac{1}{2} m (k^2 r^2 t^2) \]
This simplifies to:
\[ KE = \frac{1}{2} m k^2 r^2 t^2 \]

### Step 5: Calculate power
Power \( P \) is defined as the rate of change of energy with respect to time:
\[ P = \frac{d(KE)}{dt} \]
Substituting the expression for kinetic energy:
\[ P = \frac{d}{dt} \left( \frac{1}{2} m k^2 r^2 t^2 \right) \]
Since \( \frac{1}{2} m k^2 r^2 \) is a constant, we can differentiate \( t^2 \):
\[ P = \frac{1}{2} m k^2 r^2 \cdot \frac{d(t^2)}{dt} \]
The derivative of \( t^2 \) with respect to \( t \) is \( 2t \):
\[ P = \frac{1}{2} m k^2 r^2 \cdot 2t \]
The \( 2 \) cancels out:
\[ P = m k^2 r^2 t \]

### Final Answer
Thus, the power delivered to the particle by the forces acting on it is:
\[ P = m k^2 r^2 t \]

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