A particle moves clockwise in a circle of radius `1 m` with centre at `(x, y) = (1m, 0)`. It starts from rest at the origin at time `t = 0`. Its speed increase at the constant rate of `((pi)/(2)) m//s^(2)`. If the net acceleration at `t = 2 sec` is `(pi)/(2) sqrt((1 + Npi^(2))` then what is the value of `N` ?

Explanation

To solve the problem, we need to find the value of \( N \) given the conditions of the particle's motion. Here's a step-by-step breakdown of the solution:

### Step 1: Identify the Given Information
- Radius of the circle, \( R = 1 \, \text{m} \)
- Center of the circle at \( (1, 0) \)
- Initial position of the particle at the origin \( (0, 0) \)
- The particle starts from rest, so initial velocity \( u = 0 \)
- Tangential acceleration \( a_T = \frac{\pi}{2} \, \text{m/s}^2 \)
- Time \( t = 2 \, \text{s} \)

### Step 2: Calculate the Final Velocity at \( t = 2 \, \text{s} \)
Using the first equation of motion:
\[ v = u + a_T t \]
Substituting the values:
\[ v = 0 + \left(\frac{\pi}{2}\right) \cdot 2 = \pi \, \text{m/s} \]

### Step 3: Calculate the Radial Acceleration
The radial acceleration \( a_R \) is given by:
\[ a_R = \frac{v^2}{R} \]
Substituting \( v = \pi \) and \( R = 1 \):
\[ a_R = \frac{\pi^2}{1} = \pi^2 \, \text{m/s}^2 \]

### Step 4: Calculate the Net Acceleration
The net acceleration \( a \) is the vector sum of tangential acceleration \( a_T \) and radial acceleration \( a_R \):
\[ a = \sqrt{a_T^2 + a_R^2} \]
Substituting the values:
\[ a = \sqrt{\left(\frac{\pi}{2}\right)^2 + (\pi^2)^2} \]
Calculating \( a_T^2 \):
\[ a_T^2 = \left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4} \]
Calculating \( a_R^2 \):
\[ a_R^2 = (\pi^2)^2 = \pi^4 \]
Now substituting these into the equation for \( a \):
\[ a = \sqrt{\frac{\pi^2}{4} + \pi^4} \]
Factoring out \( \pi^2 \):
\[ a = \sqrt{\pi^2 \left(\frac{1}{4} + \pi^2\right)} = \pi \sqrt{\frac{1}{4} + \pi^2} \]

### Step 5: Set the Net Acceleration Equal to the Given Expression
According to the problem, the net acceleration at \( t = 2 \, \text{s} \) is given as:
\[ a = \frac{\pi}{2} \sqrt{1 + N\pi^2} \]
Setting the two expressions for \( a \) equal:
\[ \pi \sqrt{\frac{1}{4} + \pi^2} = \frac{\pi}{2} \sqrt{1 + N\pi^2} \]

### Step 6: Simplify and Solve for \( N \)
Dividing both sides by \( \pi \) (assuming \( \pi \neq 0 \)):
\[ \sqrt{\frac{1}{4} + \pi^2} = \frac{1}{2} \sqrt{1 + N\pi^2} \]
Squaring both sides:
\[ \frac{1}{4} + \pi^2 = \frac{1}{4} (1 + N\pi^2) \]
Multiplying through by 4:
\[ 1 + 4\pi^2 = 1 + N\pi^2 \]
Subtracting 1 from both sides:
\[ 4\pi^2 = N\pi^2 \]
Dividing by \( \pi^2 \) (assuming \( \pi \neq 0 \)):
\[ N = 4 \]

### Final Answer
The value of \( N \) is \( 4 \).

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