A point on the periphery of a rotating disc has its acceleration vector making angle of `30^(@)` with the velocity . The ratio `(a_(c)//a_(t) (a_(c)` is centripetal acceleration and `a_(t)` is tangential acceleration) equals

Options

A. `sin 30^@`
B. `cos 30^@`
C. `tan 30^@`
D. none of these

Answer

To solve the problem, we need to find the ratio of centripetal acceleration \( a_c \) to tangential acceleration \( a_t \) given that the acceleration vector makes an angle of \( 30^\circ \) with the velocity vector.

### Step-by-Step Solution:

1. **Understanding the Acceleration Components**:
- The total acceleration \( \vec{a} \) of a point on the rotating disc can be resolved into two components:
- Tangential acceleration \( a_t \) (acts along the direction of velocity).
- Centripetal acceleration \( a_c \) (acts towards the center of the circular path).

2. **Visualizing the Problem**:
- Draw a diagram where the velocity vector \( \vec{v} \) is directed tangentially to the circular path.
- The acceleration vector \( \vec{a} \) makes an angle of \( 30^\circ \) with the velocity vector \( \vec{v} \).

3. **Using Trigonometric Relationships**:
- From the triangle formed by the vectors \( \vec{a} \), \( \vec{a_t} \), and \( \vec{a_c} \), we can use the cosine of the angle to relate the components:
\[ \cos(30^\circ) = \frac{a_c}{a} \]
- And the sine of the angle to relate the tangential acceleration:
\[ \sin(30^\circ) = \frac{a_t}{a} \]

4. **Calculating the Ratios**:
- We know that:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin(30^\circ) = \frac{1}{2} \]
- Therefore, we can express \( a_c \) and \( a_t \) in terms of \( a \):
\[ a_c = a \cdot \cos(30^\circ) = a \cdot \frac{\sqrt{3}}{2} \]
\[ a_t = a \cdot \sin(30^\circ) = a \cdot \frac{1}{2} \]

5. **Finding the Ratio**:
- Now, we can find the ratio \( \frac{a_c}{a_t} \):
\[ \frac{a_c}{a_t} = \frac{a \cdot \frac{\sqrt{3}}{2}}{a \cdot \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \]

6. **Final Result**:
- Thus, the ratio of centripetal acceleration to tangential acceleration is:
\[ \frac{a_c}{a_t} = \sqrt{3} \]