A particle of mass m moves along a horizontal circle of radius R such that normal acceleration of particle varies with time as `a_(n)=kt^(2).` where k is a constant.
Total force on particle at time t s is

Options

A. `msqrt(k(R^(2)+kt^(4)))`
B. `msqrt(k(R+kt^(4)))`
C. `2msqrt(k(R+kt^(4)))`
D. `m/2sqrt(k(R+kt^(4)))`

Answer

To solve the problem, we need to find the total force acting on a particle of mass \( m \) moving in a horizontal circle of radius \( R \) with a normal acceleration that varies with time as \( a_n = kt^2 \). Here are the steps to derive the total force:

### Step 1: Identify the Normal Acceleration
The normal acceleration \( a_n \) is given by:
\[ a_n = kt^2 \]
This is the centripetal acceleration that keeps the particle moving in a circular path.

### Step 2: Relate Normal Acceleration to Normal Force
The normal force \( F_n \) acting on the particle can be expressed as:
\[ F_n = m \cdot a_n = m \cdot kt^2 \]
Thus, we have:
\[ F_n = mkt^2 \tag{1} \]

### Step 3: Find the Velocity of the Particle
From the relationship between normal acceleration and velocity, we know:
\[ a_n = \frac{v^2}{R} \]
Setting this equal to the given normal acceleration:
\[ \frac{v^2}{R} = kt^2 \]
Rearranging gives:
\[ v^2 = kRt^2 \tag{2} \]

### Step 4: Differentiate to Find Tangential Acceleration
To find the tangential acceleration \( a_t \), we differentiate equation (2) with respect to time \( t \):
\[ \frac{d(v^2)}{dt} = \frac{d(kRt^2)}{dt} \]
Using the chain rule on the left side:
\[ 2v \frac{dv}{dt} = 2kRt \]
Solving for \( \frac{dv}{dt} \):
\[ \frac{dv}{dt} = \frac{kRt}{v} \]
Substituting \( v \) from equation (2):
\[ \frac{dv}{dt} = \frac{kRt}{\sqrt{kRt^2}} = \frac{kR}{\sqrt{kR}} = \sqrt{kR} \tag{3} \]

### Step 5: Calculate Tangential Force
The tangential force \( F_t \) is given by:
\[ F_t = m \cdot a_t = m \cdot \frac{dv}{dt} \]
Substituting from equation (3):
\[ F_t = m \cdot \sqrt{kR} \]

### Step 6: Find the Total Force
The total force \( F \) acting on the particle is the vector sum of the normal and tangential forces:
\[ F = \sqrt{F_n^2 + F_t^2} \]
Substituting \( F_n \) and \( F_t \):
\[ F = \sqrt{(mkt^2)^2 + (m \sqrt{kR})^2} \]
This simplifies to:
\[ F = m \sqrt{(kt^2)^2 + kR} \]
Thus, the total force on the particle at time \( t \) is:
\[ F = m \sqrt{k^2 t^4 + kR} \tag{4} \]

### Final Answer
The total force on the particle at time \( t \) is:
\[ F = m \sqrt{k^2 t^4 + kR} \]