A hemispherical open bowl of radius R is placed on a smooth floor. A massive ball falls from some height and collides on the circumference of the bowl. Considering contact point of the floor as origin, position vector of the initial contact point of the bowl when the bowl turns `90^(@)` will be :

Options

A. `-(R )/(2)hat(i)+R hat(j)`
B. `-(R )/(3)hat(i)+R hat(j)`
C. `-(R )/(4)hat(i)+R hat(j)`
D. `-R hat(i)+R hat(j)`

Answer

To solve the problem of determining the position vector of the initial contact point of a hemispherical open bowl when it turns 90 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a hemispherical bowl of radius \( R \) placed on a smooth floor.
- The bowl is initially oriented such that its flat surface is on the floor.

2. **Identifying the Initial Contact Point**:
- The initial contact point of the bowl with the floor is at the bottom of the hemisphere, which we can denote as point \( A \) with coordinates \( (0, 0) \).

3. **Analyzing the Rotation**:
- When the bowl turns 90 degrees, it rotates around the edge of the contact point with the floor.
- The center of the bowl moves in a circular arc, and the new position of the bowl will have its circumference touching the floor.

4. **Finding the New Position of the Contact Point**:
- After a 90-degree rotation, the original contact point \( A \) will now be at the position of the edge of the bowl.
- The new contact point will be at the coordinates \( (R, 0) \) since it has moved to the right along the x-axis by the radius \( R \).

5. **Position Vector Representation**:
- The position vector of the new contact point can be represented as \( \vec{r} = R \hat{i} + 0 \hat{j} \).
- Therefore, the position vector of the initial contact point of the bowl when it turns 90 degrees is \( \vec{r} = R \hat{i} \).

### Final Answer:
The position vector of the initial contact point of the bowl when it turns 90 degrees is \( \vec{r} = R \hat{i} \).

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