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 tangential speed (at any instant) 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. `1:4`
C. `2:1`
D. `1:2`

Answer

To solve the problem, we need to find the ratio of the tangential speeds of two points on a rotating rod: one at the end of the rod and another at a distance \( L/2 \) from the end.

### Step-by-Step Solution:

1. **Identify the Lengths**:
- The length of the rod is \( 2L \).
- The distance from the center to the end of the rod is \( L \).
- The distance from the center to the point \( L/2 \) from the end is \( L - L/2 = L/2 \).

2. **Angular Acceleration**:
- The angular acceleration \( \alpha \) is given as \( 10 \, \text{rad/s}^2 \).

3. **Angular Velocity**:
- The angular velocity \( \omega \) at any time \( t \) can be calculated using the formula:
\[ \omega = \alpha t = 10t \, \text{rad/s} \]

4. **Tangential Speed Calculation**:
- The tangential speed \( V \) of a point on a rotating body is given by:
\[ V = \omega \times r \]
- For the point at the end of the rod (distance \( L \) from the center):
\[ V_1 = \omega \times L = (10t) \times L = 10tL \]
- For the point at a distance \( L/2 \) from the center:
\[ V_2 = \omega \times \frac{L}{2} = (10t) \times \frac{L}{2} = 5tL \]

5. **Ratio of Tangential Speeds**:
- Now, we find the ratio of the tangential speeds \( V_1 \) and \( V_2 \):
\[ \text{Ratio} = \frac{V_1}{V_2} = \frac{10tL}{5tL} \]
- Simplifying this gives:
\[ \text{Ratio} = \frac{10}{5} = 2 \]

### Final Answer:
The ratio of the tangential speed 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 \( 2 \).