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Chapter 4: Problem 3

A small steel ball of radius \(r=1 \mathrm{mm}\) is placed on top of a horizontal pipe of outside radius \(R=50 \mathrm{mm}\) and begins to roll under the influence of gravity, Rolling resistance and air resistance are negligible. As the speed of the ball increases, it eventually leaves the surface of the pipe and becomes a projectile. Determine the speed and location at which the ball loses contact with the pipe.

Short Answer

Expert verified
The speed at which the ball loses contact with the pipe can be given by the relation \(v = \sqrt{gR \cos \theta}\), where \(\cos \theta = r / (r+R )\). The location at which the ball loses contact is at the angle \(\theta\).

Step by step solution

01

Understand the forces acting on the ball

As the ball rolls, its movements are influenced by two main forces: gravity (\(mg\), where \(m\) is the mass of ball and \(g\) is acceleration due to gravity) pulling it downwards, and the centripetal force (\(mv^2/R\), where \(v\) is the velocity of the ball and \(R\) is the radius of the pipe), which is needed to keep the ball moving in a circle.
02

Find the condition at which ball loses contact

The ball will lose contact with the pipe when the gravitational force (directed downwards and perpendicular to the pipe) equals the required centripetal force for the ball to remain in contact. At this point, the forces are balanced. The angle \(\theta\) that the ball makes with the horizontal when it loses contact satisfies the equation: \(\cos \theta = r / (r+R )\).
03

Calculate the speed

We need to find the speed of the ball when it loses contact with the pipe. This can be found using the equation: \( mg \cos \theta = mv^2/R\), where \(\theta\) is the angle when ball loses contact with the pipe. We can rewrite this equation to find the speed \(v\) as: \(v = \sqrt{gR \cos \theta}\).

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