A particle of mass 10 g moves along a circle of radius `6.4` cm with a constant tangential acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to `8xx10^(-4)`J by the end of the second revolution after the beginning of the motion?

Options

A. `0.15" ms"^(-2)`
B. `0.18" ms"^(-2)`
C. `0.2" ms"^(-2)`
D. `0.1" ms"^(-2)`

Explanation

To solve the problem step by step, we will follow the reasoning presented in the video transcript and apply the relevant physics concepts.

### Step 1: Convert the mass and radius to standard units
- The mass of the particle is given as 10 g. We need to convert this to kilograms:
\[ m = \frac{10 \, \text{g}}{1000} = 0.01 \, \text{kg} \]

- The radius of the circle is given as 6.4 cm. We convert this to meters:
\[ r = 6.4 \, \text{cm} = 6.4 \times 10^{-2} \, \text{m} \]

### Step 2: Write down the kinetic energy formula
The kinetic energy (KE) of the particle is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
We know that the kinetic energy at the end of the second revolution is \(8 \times 10^{-4} \, \text{J}\).

### Step 3: Solve for \(v^2\)
From the kinetic energy formula, we can rearrange it to find \(v^2\):
\[ 8 \times 10^{-4} = \frac{1}{2} \times 0.01 \times v^2 \]
Multiplying both sides by 2:
\[ 16 \times 10^{-4} = 0.01 \times v^2 \]
Now, divide both sides by 0.01:
\[ v^2 = \frac{16 \times 10^{-4}}{0.01} = 16 \times 10^{-2} = 0.16 \, \text{m}^2/\text{s}^2 \]

### Step 4: Calculate the distance traveled in two revolutions
The distance \(s\) traveled after two revolutions can be calculated as:
\[ s = 2 \times (2\pi r) = 4\pi r \]
Substituting the value of \(r\):
\[ s = 4 \pi (6.4 \times 10^{-2}) \approx 4 \times 3.14 \times 6.4 \times 10^{-2} \]
Calculating this gives:
\[ s \approx 4 \times 3.14 \times 0.064 \approx 0.80384 \, \text{m} \]

### Step 5: Use the equation of motion to find tangential acceleration
Using the equation of motion:
\[ v^2 = u^2 + 2a_t s \]
Since the initial velocity \(u = 0\):
\[ v^2 = 2a_t s \]
Substituting \(v^2\) and \(s\):
\[ 0.16 = 2a_t (0.80384) \]
Now, solving for \(a_t\):
\[ a_t = \frac{0.16}{2 \times 0.80384} \approx \frac{0.16}{1.60768} \approx 0.0993 \, \text{m/s}^2 \]
Rounding this gives:
\[ a_t \approx 0.1 \, \text{m/s}^2 \]

### Final Answer
The magnitude of the tangential acceleration is approximately:
\[ \boxed{0.1 \, \text{m/s}^2} \]