Discuss the motion in a vertical circle. Find an expression for the minimum velocities at the lowest point and top point. Also find tension at these points?

When we talk about motion in a vertical circle, we're looking at how an object moves along a circular path while being influenced by gravity. This scenario is common in various real-world applications, such as roller coasters, pendulums, and even satellites in orbit. Understanding the forces at play, particularly at the highest and lowest points of the circular path, is crucial for analyzing the motion accurately.

Understanding Forces in Vertical Circular Motion

In vertical circular motion, two primary forces act on the object: gravitational force and tension (if a string or rod is involved). The gravitational force always acts downward, while tension can vary depending on the position of the object in the circle.

Key Points in the Circular Path

Let's consider an object of mass \( m \) moving in a vertical circle of radius \( r \). The two critical positions to analyze are:

• The **lowest point** of the circle
• The **highest point** of the circle

Velocity at the Lowest Point

At the lowest point of the circle, the object experiences the maximum tension due to both the gravitational force and the centripetal force required to keep it moving in a circle. The forces acting on the object can be expressed as follows:

The net force acting on the object at the lowest point is given by:

T - mg = \frac{mv^2}{r}

Where:

• \( T \) is the tension in the string
• \( m \) is the mass of the object
• \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
• \( v \) is the velocity at the lowest point
• \( r \) is the radius of the circular path

To find the minimum velocity at the lowest point, we can set the tension \( T \) to zero (which is the minimum condition for the object to stay in circular motion). Thus, we have:

0 - mg = \frac{mv^2}{r}

Rearranging this gives:

v^2 = rg

Taking the square root, we find:

v = \sqrt{rg}

Velocity at the Highest Point

At the highest point of the circle, the situation is slightly different. Here, both the gravitational force and the tension contribute to the centripetal force needed to keep the object moving in a circle. The equation at this point is:

T + mg = \frac{mv^2}{r}

Again, to find the minimum velocity at the highest point, we can set the tension \( T \) to zero:

0 + mg = \frac{mv^2}{r}

Rearranging gives:

v^2 = rg

Thus, the minimum velocity at the highest point is also:

v = \sqrt{rg}

Tension at Different Points

Now, let’s calculate the tension at both the lowest and highest points when the object is moving at the minimum velocity.

Tension at the Lowest Point

Using the minimum velocity expression \( v = \sqrt{rg} \) in the tension equation:

T - mg = \frac{m(\sqrt{rg})^2}{r}

This simplifies to:

T - mg = mg

Thus, we find:

T = 2mg

Tension at the Highest Point

For the highest point, using the same minimum velocity:

T + mg = \frac{m(\sqrt{rg})^2}{r}

This simplifies to:

T + mg = mg

Thus, we find:

T = 0

Summary of Findings

In summary, for an object moving in a vertical circle of radius \( r \):

• The minimum velocity at both the lowest and highest points is given by \( v = \sqrt{rg} \).
• The tension at the lowest point is \( T = 2mg \).
• The tension at the highest point is \( T = 0 \).

This analysis highlights the importance of understanding the forces at play in vertical circular motion, which can be applied to various physical systems. Whether you're designing a roller coaster or analyzing a pendulum, these principles remain fundamental to ensuring safe and effective motion.