A wheel of radius R is rolling without sliding uniformly on a horizontal surface. Find the radius of curvature of the path of a point on its circumference when it is at highest point in its path.

Answer

To find the radius of curvature of the path of a point on the circumference of a wheel when it is at the highest point in its path, we can follow these steps:

### Step 1: Understand the Motion of the Wheel
The wheel is rolling without sliding on a horizontal surface. This means that at the point of contact with the ground, the velocity is zero, and the wheel rotates about its center of mass.

**Hint:** Remember that rolling without slipping means the point of contact does not slide over the surface.

### Step 2: Identify the Velocity at the Highest Point
At the highest point of the wheel, the point on the circumference has a velocity that is the sum of the translational velocity of the center of mass (V) and the tangential velocity due to rotation (rω). Since the wheel rolls uniformly, we have the relation \(V = rω\).

**Hint:** The total velocity at the highest point is the sum of the translational and rotational components.

### Step 3: Calculate the Total Velocity
At the highest point, the total velocity \(V_{total}\) can be expressed as:
\[ V_{total} = V + rω = V + V = 2V \]

**Hint:** Substitute \(V\) with \(rω\) to find the total velocity at the highest point.

### Step 4: Define the Radius of Curvature
The radius of curvature (ρ) is defined in terms of the centripetal acceleration. The centripetal acceleration \(a_c\) for an object moving in a circular path is given by:
\[ a_c = \frac{V_{total}^2}{ρ} \]

**Hint:** Remember that centripetal acceleration is necessary for circular motion.

### Step 5: Set Up the Equation for Centripetal Force
At the highest point, the centripetal force is provided by the normal force acting on the point. Therefore, we can equate the centripetal acceleration to the expression involving the radius of curvature:
\[ \frac{(2V)^2}{ρ} = \frac{4V^2}{ρ} \]

**Hint:** This equation relates the total velocity at the highest point to the radius of curvature.

### Step 6: Relate Centripetal Acceleration to Normal Force
At the highest point, the centripetal force is equal to the normal force acting on the point. Thus, we can express the normal force as:
\[ N = \frac{4V^2}{ρ} \]

**Hint:** The normal force acts as the centripetal force at the highest point.

### Step 7: Solve for the Radius of Curvature
From the previous steps, we can derive the radius of curvature:
\[ ρ = \frac{V^2}{a_c} \]
Substituting \(V_{total} = 2V\):
\[ ρ = \frac{(2V)^2}{g} = \frac{4V^2}{g} \]

Since the centripetal acceleration is also given by \(g\) at the highest point, we can conclude:
\[ ρ = 4R \]

**Hint:** The radius of curvature at the highest point is four times the radius of the wheel.

### Final Result
The radius of curvature of the path of a point on the circumference of the wheel when it is at the highest point is:
\[ ρ = 4R \]