For a particle in uniform circular motion , the acceleration ` vec(a)` at a point ` p ( R, theta )` on the circle of radiu ` R` is ( Here ` theta` is measured from the ` x- axis` )

Options

A. `(v^(2))/(R)hati+(v^(2))/(R)jhat`
B. `-(v^(2))/(R)cos thetahati+(v^(2))/(R)sin thetajhat`
C. `-(v^(2))/(R)sin thetahati+(v^(2))/(R)cos thetajhat`
D. `-(v^(2))/(R)cos thetahati-(v^(2))/(R)sin thetajhat`

Answer

To solve the problem regarding the acceleration of a particle in uniform circular motion at a point \( P(R, \theta) \) on the circle of radius \( R \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Uniform Circular Motion**:
- In uniform circular motion, a particle moves in a circular path with constant speed. The direction of the velocity vector changes continuously, which results in an acceleration directed towards the center of the circle.

2. **Identifying the Radius and Angle**:
- The particle is located at point \( P \) on the circle with radius \( R \) and angle \( \theta \) measured from the positive x-axis.

3. **Expression for Centripetal Acceleration**:
- The magnitude of the centripetal acceleration \( a \) for uniform circular motion is given by:
\[ a = \frac{v^2}{R} \]
where \( v \) is the constant speed of the particle.

4. **Direction of Acceleration**:
- The acceleration vector points towards the center of the circle. At point \( P \), we can resolve this acceleration into its components along the x-axis and y-axis.

5. **Resolving Acceleration into Components**:
- The acceleration vector \( \vec{a} \) can be expressed in terms of its components:
- The x-component of the acceleration is directed towards the center and can be expressed as:
\[ a_x = -a \cos(\theta) = -\frac{v^2}{R} \cos(\theta) \]
- The y-component of the acceleration is also directed towards the center and can be expressed as:
\[ a_y = -a \sin(\theta) = -\frac{v^2}{R} \sin(\theta) \]

6. **Writing the Acceleration Vector**:
- Combining both components, we can write the acceleration vector \( \vec{a} \) in vector form:
\[ \vec{a} = a_x \hat{i} + a_y \hat{j} = -\frac{v^2}{R} \cos(\theta) \hat{i} - \frac{v^2}{R} \sin(\theta) \hat{j} \]

7. **Final Result**:
- Therefore, the acceleration vector at point \( P(R, \theta) \) is:
\[ \vec{a} = -\frac{v^2}{R} \cos(\theta) \hat{i} - \frac{v^2}{R} \sin(\theta) \hat{j} \]