A particle of mass 10 g moves along a circle of radius 64 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. `0.02 m//s^(2)`
B. `0.01 m//s^(2)`
C. `0.015 m//s^(2)`
D. `0.018 m//s^(2)`

Answer

To solve the problem step by step, we will follow the principles of circular motion, kinetic energy, and the relationships between linear and angular quantities.

### Step 1: Understand the Problem
We have a particle of mass \( m = 10 \, \text{g} = 0.01 \, \text{kg} \) moving in a circle of radius \( R = 64 \, \text{cm} = 0.64 \, \text{m} \) with a constant tangential acceleration \( a_T \). The kinetic energy after 2 revolutions is given as \( KE = 8 \times 10^{-4} \, \text{J} \).

### Step 2: Calculate the Total Angle Covered
The particle completes 2 revolutions. The total angle \( \theta \) in radians for 2 revolutions is:
\[ \theta = 2 \times 2\pi = 4\pi \, \text{radians} \]

### Step 3: Use Kinetic Energy Formula
The kinetic energy of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
We also know that the linear velocity \( v \) can be expressed in terms of angular velocity \( \omega \) as:
\[ v = \omega R \]
Thus, we can write the kinetic energy in terms of \( \omega \):
\[ KE = \frac{1}{2} m (\omega R)^2 = \frac{1}{2} m \omega^2 R^2 \]

### Step 4: Relate Angular Velocity to Angular Acceleration
Since the particle starts from rest, the final angular velocity \( \omega \) after covering \( \theta \) radians with constant angular acceleration \( \alpha \) is given by:
\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]
Here, \( \omega_0 = 0 \) (initial angular velocity), so:
\[ \omega^2 = 2\alpha\theta \]

### Step 5: Substitute \( \theta \) and Solve for \( \omega^2 \)
Substituting \( \theta = 4\pi \):
\[ \omega^2 = 2\alpha(4\pi) = 8\pi\alpha \]

### Step 6: Substitute \( \omega^2 \) into Kinetic Energy Equation
Now substitute \( \omega^2 \) back into the kinetic energy equation:
\[ KE = \frac{1}{2} m (8\pi\alpha) R^2 \]
Setting this equal to the given kinetic energy:
\[ 8 \times 10^{-4} = \frac{1}{2} \times 0.01 \times (8\pi\alpha) \times (0.64)^2 \]

### Step 7: Simplify and Solve for \( \alpha \)
\[ 8 \times 10^{-4} = 0.005 \times 8\pi\alpha \times 0.4096 \]
\[ 8 \times 10^{-4} = 0.016384\pi\alpha \]
\[ \alpha = \frac{8 \times 10^{-4}}{0.016384\pi} \]

### Step 8: Calculate Tangential Acceleration
The tangential acceleration \( a_T \) is given by:
\[ a_T = \alpha R \]
Substituting \( R = 0.64 \):
\[ a_T = \frac{8 \times 10^{-4}}{0.016384\pi} \times 0.64 \]

### Step 9: Final Calculation
Calculating \( a_T \):
1. Calculate \( \alpha \):
\[ \alpha \approx \frac{8 \times 10^{-4}}{0.051835} \approx 0.0154 \, \text{rad/s}^2 \]
2. Calculate \( a_T \):
\[ a_T \approx 0.0154 \times 0.64 \approx 0.009856 \, \text{m/s}^2 \approx 0.01 \, \text{m/s}^2 \]

### Conclusion
The magnitude of the tangential acceleration is approximately \( 0.01 \, \text{m/s}^2 \).

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