A particle of mass m moves in a circular path of radius r, under the action of force which delivers it constant power p and increases its speed. The angular acceleration of particle at time (t) is proportional

Options

A. `(1)/(sqrtt)`
B. `sqrtt`
C. `t^(0)`
D. `t^(3//2)`

Answer

To solve the problem step by step, we need to analyze the relationship between power, velocity, and angular acceleration of a particle moving in a circular path under constant power.

### Step-by-Step Solution:

1. **Understanding Power and Velocity**:
- The power \( P \) delivered to the particle is given by the formula:
\[ P = F \cdot v \]
where \( F \) is the force acting on the particle and \( v \) is its tangential velocity.

2. **Relating Power to Mass and Acceleration**:
- The force \( F \) can also be expressed in terms of mass \( m \) and tangential acceleration \( a_t \):
\[ F = m \cdot a_t \]
- Therefore, we can write:
\[ P = m \cdot a_t \cdot v \]

3. **Expressing Acceleration**:
- The tangential acceleration \( a_t \) can be expressed as the time derivative of velocity:
\[ a_t = \frac{dv}{dt} \]
- Substituting this into the power equation gives:
\[ P = m \cdot \frac{dv}{dt} \cdot v \]

4. **Rearranging the Equation**:
- Rearranging the equation to isolate \( dv \):
\[ \frac{dv}{dt} = \frac{P}{m \cdot v} \]

5. **Separating Variables**:
- We can separate variables to integrate:
\[ v \, dv = \frac{P}{m} \, dt \]

6. **Integrating Both Sides**:
- Integrating both sides:
\[ \int v \, dv = \int \frac{P}{m} \, dt \]
- This results in:
\[ \frac{v^2}{2} = \frac{P}{m} t + C \]
- Assuming the initial velocity is zero when \( t = 0 \), we find \( C = 0 \):
\[ \frac{v^2}{2} = \frac{P}{m} t \]

7. **Finding Velocity**:
- Rearranging gives:
\[ v^2 = \frac{2P}{m} t \]
- Taking the square root:
\[ v = \sqrt{\frac{2P}{m}} \sqrt{t} \]
- This shows that \( v \) is proportional to \( t^{1/2} \).

8. **Relating Velocity to Angular Speed**:
- The tangential velocity \( v \) is related to angular speed \( \omega \) by:
\[ v = r \cdot \omega \]
- Thus:
\[ r \cdot \omega \propto t^{1/2} \]
- Since \( r \) is constant, we can say:
\[ \omega \propto t^{1/2} \]

9. **Finding Angular Acceleration**:
- Angular acceleration \( \alpha \) is defined as:
\[ \alpha = \frac{d\omega}{dt} \]
- Differentiating \( \omega \propto t^{1/2} \):
\[ \alpha \propto \frac{1}{2} t^{-1/2} \]
- This means:
\[ \alpha \propto \frac{1}{\sqrt{t}} \]

### Final Answer:
The angular acceleration \( \alpha \) of the particle at time \( t \) is proportional to \( \frac{1}{\sqrt{t}} \).

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