A particle of mass 10 g moves along a circle of the radius `(1)/(pi)` cm with a constant tangential acceleration. What is the magnitude of this acceleration `("in ms"^(-1))` if the kinetic nergy of the particle becomes equal to `8xx10^(-4)J` by the end of the second revolution after the beginning of the motion? (Particle starts from rest)

Answer

To solve the problem step by step, we need to find the magnitude of the tangential acceleration of a particle moving in a circular path. Let's break down the solution:

### Given Data:
- Mass of the particle, \( m = 10 \, \text{g} = 0.01 \, \text{kg} \) (since \( 1 \, \text{g} = 0.001 \, \text{kg} \))
- Radius of the circle, \( r = \frac{1}{\pi} \, \text{cm} = \frac{1}{\pi} \times 0.01 \, \text{m} = \frac{0.01}{\pi} \, \text{m} \)
- Kinetic energy after 2 revolutions, \( K = 8 \times 10^{-4} \, \text{J} \)

### Step 1: Find the angular displacement for 2 revolutions
The angular displacement \( \theta \) for 2 revolutions is:
\[ \theta = 2 \times 2\pi = 4\pi \, \text{radians} \]

### Step 2: Use the relation between angular velocity, angular acceleration, and angular displacement
Since the particle starts from rest, the initial angular velocity \( \omega_0 = 0 \). We can use the equation:
\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]
Substituting the known values:
\[ \omega^2 = 0 + 2\alpha(4\pi) \implies \omega^2 = 8\pi\alpha \]

### Step 3: Relate tangential acceleration to angular acceleration
The tangential acceleration \( a \) is related to angular acceleration \( \alpha \) by:
\[ a = r\alpha \]
Substituting \( r = \frac{0.01}{\pi} \):
\[ \alpha = \frac{a}{r} = \frac{a \pi}{0.01} \]

### Step 4: Substitute \( \alpha \) in the equation for \( \omega^2 \)
Substituting \( \alpha \) into the equation for \( \omega^2 \):
\[ \omega^2 = 8\pi\left(\frac{a \pi}{0.01}\right) = \frac{8\pi^2 a}{0.01} \]

### Step 5: Relate kinetic energy to velocity
The kinetic energy \( K \) is given by:
\[ K = \frac{1}{2}mv^2 \]
Also, we know that \( v = r\omega \):
\[ v^2 = r^2\omega^2 \]
Substituting this into the kinetic energy equation:
\[ K = \frac{1}{2}m(r^2\omega^2) \]
Substituting \( r = \frac{0.01}{\pi} \):
\[ K = \frac{1}{2}m\left(\frac{0.01}{\pi}\right)^2\omega^2 \]

### Step 6: Substitute \( \omega^2 \) into the kinetic energy equation
Substituting \( \omega^2 \) from earlier:
\[ K = \frac{1}{2}m\left(\frac{0.01}{\pi}\right)^2\left(\frac{8\pi^2 a}{0.01}\right) \]
Simplifying:
\[ K = \frac{1}{2}m\left(\frac{0.01 \cdot 8a}{2}\right) = \frac{4ma}{\pi} \]

### Step 7: Solve for \( a \)
Now, substituting the known values:
\[ 8 \times 10^{-4} = \frac{4 \cdot 0.01 \cdot a}{\pi} \]
\[ a = \frac{8 \times 10^{-4} \cdot \pi}{0.04} = 0.02\pi \, \text{m/s}^2 \]

### Step 8: Calculate the numerical value of \( a \)
Using \( \pi \approx 3.14 \):
\[ a \approx 0.02 \cdot 3.14 \approx 0.0628 \, \text{m/s}^2 \]

### Final Answer
The magnitude of the tangential acceleration is approximately \( 0.0628 \, \text{m/s}^2 \).