A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as `a_c = k^2rt^2`, where k is a constant. Calculate the power delivered to the particle by the force acting on it.

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript while elaborating on each step.

### Step 1: Understanding Centripetal Acceleration
We are given that the centripetal acceleration \( a_c \) of a particle moving in a circular path is given by the equation:
\[ a_c = k^2 r t^2 \]
where \( k \) is a constant, \( r \) is the radius of the circular path, and \( t \) is time.

### Step 2: Relating Centripetal Acceleration to Velocity
Centripetal acceleration can also be expressed in terms of velocity \( v \) as:
\[ a_c = \frac{v^2}{r} \]
Setting the two expressions for centripetal acceleration equal gives us:
\[ \frac{v^2}{r} = k^2 r t^2 \]

### Step 3: Solving for Velocity
From the equation above, we can solve for \( v^2 \):
\[ v^2 = k^2 r^2 t^2 \]
Taking the square root of both sides, we find:
\[ v = k r t \]

### Step 4: Finding Tangential Acceleration
Tangential acceleration \( a_t \) is defined as the rate of change of velocity with respect to time:
\[ a_t = \frac{dv}{dt} \]
Substituting our expression for \( v \):
\[ a_t = \frac{d(k r t)}{dt} = k r \]

### Step 5: Calculating Tangential Force
The tangential force \( F_t \) acting on the particle can be calculated using Newton's second law:
\[ F_t = m a_t = m (k r) \]

### Step 6: Calculating Power Delivered to the Particle
Power \( P \) delivered to the particle by the tangential force is given by the product of force and velocity:
\[ P = F_t \cdot v \]
Substituting the expressions for \( F_t \) and \( v \):
\[ P = (m k r) \cdot (k r t) \]
\[ P = m k^2 r^2 t \]

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

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