Solution: Sphere Rolling on Rails
Problem Statement
A sphere of radius \( R \) is rolling without slipping on two horizontal and parallel rails that are not at the same level. The distance between the rails is equal to the radius of the sphere (\( R \)). If the speed of the center of the sphere is \( v \), find the greatest speed of a point on the sphere.
Diagram: Instantaneous Axis of Rotation (IAR)
1. Identifying the Instantaneous Axis of Rotation (IAR)
Since the sphere rolls without slipping on the two rails, the velocity of the contact points on the sphere must be zero.
Therefore, the line joining the two points of contact acts as the **Instantaneous Axis of Rotation (IAR)**.
2. Geometry of the Contact Points
Let the radius of the sphere be \( R \). The distance between the rails is given as \( R \).
The two contact points and the center of the sphere form an isosceles triangle. Since the chord length (distance between rails) is \( R \) and the sides connecting to the center are also \( R \) (radius), the triangle formed by the two contact points and the center is Equilateral.
The angle subtended by the rails at the center is \( 60^\circ \). The perpendicular distance \( d \) from the center of the sphere to the IAR (chord) is: \[ d = \sqrt{R^2 – (R/2)^2} = \frac{\sqrt{3}}{2}R \]
The angle subtended by the rails at the center is \( 60^\circ \). The perpendicular distance \( d \) from the center of the sphere to the IAR (chord) is: \[ d = \sqrt{R^2 – (R/2)^2} = \frac{\sqrt{3}}{2}R \]
3. Angular Velocity (\( \omega \))
The velocity of the center of mass \( v \) is related to the angular velocity \( \omega \) by the distance to the IAR.
\[ v = \omega \times (\text{distance from C to IAR}) \]
\[ v = \omega \left( \frac{\sqrt{3}}{2}R \right) \]
\[ \omega = \frac{2v}{\sqrt{3}R} \]
4. Finding the Point of Maximum Speed
The speed of any point on a rotating body is \( v_p = \omega r \), where \( r \) is the perpendicular distance from the IAR. To maximize speed, we need to maximize \( r \).
The point farthest from the chord (IAR) is the point diametrically opposite to the midpoint of the chord on the sphere’s surface.
Distance from center to IAR is \( \frac{\sqrt{3}}{2}R \). Distance from center to surface is \( R \). Total max distance \( r_{max} \) from IAR: \[ r_{max} = R + \frac{\sqrt{3}}{2}R = R \left( 1 + \frac{\sqrt{3}}{2} \right) \]
Distance from center to IAR is \( \frac{\sqrt{3}}{2}R \). Distance from center to surface is \( R \). Total max distance \( r_{max} \) from IAR: \[ r_{max} = R + \frac{\sqrt{3}}{2}R = R \left( 1 + \frac{\sqrt{3}}{2} \right) \]
5. Calculating Maximum Velocity
\[ v_{max} = \omega \cdot r_{max} \]
Substitute \( \omega \) and \( r_{max} \):
\[ v_{max} = \left( \frac{2v}{\sqrt{3}R} \right) \times R \left( 1 + \frac{\sqrt{3}}{2} \right) \]
\[ v_{max} = \frac{2v}{\sqrt{3}} \left( 1 + \frac{\sqrt{3}}{2} \right) \]
\[ v_{max} = v \left( \frac{2}{\sqrt{3}} + 1 \right) \]
$$ v_{max} = v \left( 1 + \frac{2}{\sqrt{3}} \right) $$
