A block of mass m is kept on a horizontal ruler. The friction of coefficient between the ruler and the block is `mu`. The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. a. What can the maximum angular speed of the ruler b) if the angular speed of ruler is uniformly increased from zero to an angular acceleration `alpha` , at what angular speed will the block slip?

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (a): Maximum Angular Speed of the Ruler

1. **Identify Forces Acting on the Block**:
- The block of mass \( m \) experiences gravitational force \( mg \) downward and normal force \( N \) upward. The frictional force \( f \) provides the necessary centripetal force for circular motion.

2. **Centripetal Force Requirement**:
- For circular motion, the required centripetal force \( F_c \) is given by:
\[ F_c = m \omega^2 L \]
- Here, \( \omega \) is the angular speed and \( L \) is the distance from the fixed end to the block.

3. **Maximum Frictional Force**:
- The maximum frictional force \( f_{\text{max}} \) that can act on the block is:
\[ f_{\text{max}} = \mu N = \mu mg \]
- Where \( \mu \) is the coefficient of friction.

4. **Setting Up the Inequality**:
- For the block to not slip, the centripetal force must be less than or equal to the maximum frictional force:
\[ m \omega^2 L \leq \mu mg \]

5. **Simplifying the Inequality**:
- Canceling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \omega^2 L \leq \mu g \]
- Rearranging gives:
\[ \omega^2 \leq \frac{\mu g}{L} \]

6. **Finding Maximum Angular Speed**:
- Taking the square root of both sides, we find the maximum angular speed \( \omega_{\text{max}} \):
\[ \omega_{\text{max}} = \sqrt{\frac{\mu g}{L}} \]

### Part (b): Angular Speed When the Block Slips

1. **Understanding the Scenario**:
- The ruler is rotated with a uniform angular acceleration \( \alpha \). The angular speed \( \omega \) increases from 0 to some value before the block slips.

2. **Tangential and Centripetal Acceleration**:
- The tangential acceleration \( a_t \) is given by:
\[ a_t = \alpha L \]
- The centripetal acceleration \( a_c \) is:
\[ a_c = \omega^2 L \]

3. **Net Acceleration**:
- The net acceleration \( a_{\text{net}} \) acting on the block can be found using the Pythagorean theorem since \( a_t \) and \( a_c \) are perpendicular:
\[ a_{\text{net}} = \sqrt{a_t^2 + a_c^2} = \sqrt{(\alpha L)^2 + (\omega^2 L)^2} \]

4. **Condition for Slipping**:
- For the block to slip, the net force must equal the maximum frictional force:
\[ m a_{\text{net}} = \mu mg \]
- Thus, we have:
\[ a_{\text{net}} = \mu g \]

5. **Setting Up the Equation**:
- Substituting the expression for \( a_{\text{net}} \):
\[ \sqrt{(\alpha L)^2 + (\omega^2 L)^2} = \mu g \]

6. **Squaring Both Sides**:
- Squaring gives:
\[ (\alpha L)^2 + (\omega^2 L)^2 = (\mu g)^2 \]

7. **Solving for Angular Speed \( \omega \)**:
- Rearranging the equation:
\[ (\omega^2 L)^2 = (\mu g)^2 - (\alpha L)^2 \]
- Thus:
\[ \omega^2 = \frac{(\mu g)^2 - (\alpha L)^2}{L^2} \]
- Finally, taking the square root:
\[ \omega = \sqrt{\frac{(\mu g)^2 - (\alpha L)^2}{L^2}} = \frac{\sqrt{(\mu g)^2 - (\alpha L)^2}}{L} \]

### Summary of Solutions:
- **Maximum Angular Speed**:
\[ \omega_{\text{max}} = \sqrt{\frac{\mu g}{L}} \]
- **Angular Speed at Slipping**:
\[ \omega = \frac{\sqrt{(\mu g)^2 - (\alpha L)^2}}{L} \]