A particle is moving on a circular path such that at any instant its position vector, linear velocity. Angular velocity, angular acceleration with respect to centre are `vec(r),vec(v),vec(w),vec(alpha)` respectively. Net acceleration of the particle is `:-`

Options

A. `(vec(omega)xxvec(v))-(vec(r)xxvec(alpha))`
B. `(vec(omega)xxvec(v))+(vec(r)xxvec(alpha))`
C. `(vec(v)xxvec(omega))+(vec(r)xxvec(alpha))`
D. `(vec(v)xxvec(omega))-(vec(r)xxvec(alpha))`

Answer

To find the net acceleration of a particle moving in a circular path, we need to consider both the centripetal (or radial) acceleration and the tangential acceleration.

### Step-by-Step Solution:

1. **Identify the Components of Acceleration**:
- The net acceleration of the particle in circular motion can be expressed as the sum of centripetal acceleration and tangential acceleration.
- Centripetal acceleration (\( \vec{a}_c \)) is directed towards the center of the circular path and is given by the formula:
\[ \vec{a}_c = \frac{v^2}{r} \hat{r} = \omega^2 r \hat{r} \]
where \( v \) is the linear velocity, \( r \) is the radius of the circular path, and \( \hat{r} \) is the unit vector pointing radially inward.

2. **Tangential Acceleration**:
- The tangential acceleration (\( \vec{a}_t \)) is due to the change in the speed of the particle along the circular path and is given by:
\[ \vec{a}_t = \alpha \times r \]
where \( \alpha \) is the angular acceleration.

3. **Net Acceleration**:
- The net acceleration (\( \vec{a} \)) of the particle can be expressed as the vector sum of the centripetal and tangential accelerations:
\[ \vec{a} = \vec{a}_c + \vec{a}_t \]
- Substituting the expressions for \( \vec{a}_c \) and \( \vec{a}_t \):
\[ \vec{a} = \frac{v^2}{r} \hat{r} + \alpha \times r \]

4. **Final Expression**:
- Therefore, the net acceleration of the particle moving in a circular path can be expressed as:
\[ \vec{a} = \frac{v^2}{r} \hat{r} + \alpha \times \vec{r} \]