A long thin rod of length 2L rotates with a constant angular acceleration of `10rad//s^(2)` about an axis that is perpendicular to the rod and passes through its center.
What is the ratio of the centripetal acceleration of a point on the end of the rod to that of a point a distance L/2 from the end of the rod?

Options

A. `1:1`
B. `2:1`
C. `1:4`
D. `1:2`

Answer

To solve the problem, we need to find the centripetal acceleration of two points on a rotating rod and then calculate the ratio of these accelerations.

### Step-by-Step Solution:

1. **Identify the Points on the Rod**:
- Point A (end of the rod): Distance from the center = L
- Point B (L/2 from the end): Distance from the center = L/2

2. **Centripetal Acceleration Formula**:
The centripetal acceleration \( a_c \) of a point on a rotating body is given by:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the tangential velocity and \( r \) is the distance from the axis of rotation.

3. **Determine the Angular Velocity**:
The angular velocity \( \omega \) at any time \( t \) is given by:
\[ \omega = \alpha t \]
where \( \alpha = 10 \, \text{rad/s}^2 \). Thus,
\[ \omega = 10t \]

4. **Calculate the Tangential Velocity**:
The tangential velocity \( v \) at a distance \( r \) from the axis of rotation is given by:
\[ v = \omega r \]
- For Point A (end of the rod, \( r = L \)):
\[ v_A = \omega \cdot L = (10t) \cdot L = 10Lt \]
- For Point B (distance \( L/2 \) from the end, \( r = L/2 \)):
\[ v_B = \omega \cdot \frac{L}{2} = (10t) \cdot \frac{L}{2} = 5Lt \]

5. **Calculate the Centripetal Accelerations**:
- For Point A:
\[ a_{cA} = \frac{v_A^2}{L} = \frac{(10Lt)^2}{L} = \frac{100L^2t^2}{L} = 100Lt^2 \]
- For Point B:
\[ a_{cB} = \frac{v_B^2}{\frac{L}{2}} = \frac{(5Lt)^2}{\frac{L}{2}} = \frac{25L^2t^2}{\frac{L}{2}} = 50L^2t^2 \]

6. **Calculate the Ratio of Centripetal Accelerations**:
Now, we find the ratio of the centripetal accelerations:
\[ \frac{a_{cA}}{a_{cB}} = \frac{100Lt^2}{50Lt^2} = \frac{100}{50} = 2 \]

### Final Answer:
The ratio of the centripetal acceleration of a point on the end of the rod to that of a point a distance \( L/2 \) from the end of the rod is:
\[ \text{Ratio} = 2:1 \]