A car moves uniformly along a horizontal since curvey `= a sin (x//alpha)`, when a and are certain constnat The coefficient of friction between the wheels and the road is equal to `mu`. At what velocity will the car ride without sliding ?

Answer

To solve the problem, we need to determine the maximum velocity at which a car can move along a curved path without sliding off due to the effects of friction. The path is defined by the equation \( y = a \sin\left(\frac{x}{\alpha}\right) \), where \( a \) and \( \alpha \) are constants, and the coefficient of friction between the wheels and the road is \( \mu \).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Car:**
- The car experiences gravitational force \( mg \) acting downwards.
- The frictional force \( F_f \) acts towards the center of the curve, preventing the car from sliding outwards.

2. **Centripetal Force Requirement:**
- For the car to move in a circular path, the required centripetal force \( F_c \) must be provided by the frictional force. Thus, we have:
\[ F_c = \frac{mv^2}{r} \]
- Where \( v \) is the velocity of the car and \( r \) is the radius of curvature of the path.

3. **Frictional Force:**
- The maximum frictional force is given by:
\[ F_f = \mu mg \]
- For the car to not slide, the centripetal force must be less than or equal to the frictional force:
\[ \frac{mv^2}{r} \leq \mu mg \]

4. **Simplifying the Inequality:**
- Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{v^2}{r} \leq \mu g \]
- Rearranging gives us:
\[ v^2 \leq \mu g r \]
- Therefore, the maximum velocity \( v_{\text{max}} \) is:
\[ v_{\text{max}} = \sqrt{\mu g r} \]

5. **Finding the Radius of Curvature:**
- The radius of curvature \( r \) for the given curve \( y = a \sin\left(\frac{x}{\alpha}\right) \) can be calculated using the formula:
\[ r = \frac{(1 + (dy/dx)^2)^{3/2}}{|d^2y/dx^2|} \]
- First, we need to compute \( dy/dx \) and \( d^2y/dx^2 \):
- \( dy/dx = \frac{a}{\alpha} \cos\left(\frac{x}{\alpha}\right) \)
- \( d^2y/dx^2 = -\frac{a}{\alpha^2} \sin\left(\frac{x}{\alpha}\right) \)

6. **Substituting into the Radius of Curvature Formula:**
- Substitute \( dy/dx \) and \( d^2y/dx^2 \) into the radius of curvature formula:
\[ r = \frac{\left(1 + \left(\frac{a}{\alpha} \cos\left(\frac{x}{\alpha}\right)\right)^2\right)^{3/2}}{\left| -\frac{a}{\alpha^2} \sin\left(\frac{x}{\alpha}\right) \right|} \]

7. **Finding Minimum Radius of Curvature:**
- The minimum radius occurs at \( x = \frac{\pi}{2} \alpha \), which simplifies to:
\[ r_{\text{min}} = \frac{\alpha^2}{a} \]

8. **Final Velocity Expression:**
- Substitute \( r_{\text{min}} \) back into the velocity equation:
\[ v_{\text{max}} = \sqrt{\mu g \cdot \frac{\alpha^2}{a}} = \alpha \sqrt{\frac{\mu g}{a}} \]

### Conclusion:
The maximum velocity at which the car can ride without sliding is given by:
\[ v_{\text{max}} = \alpha \sqrt{\frac{\mu g}{a}} \]