The Dynamics of a Bob in Vertical Circular Motion

Understanding the motion of a bob in vertical circular motion (VCM) is essential in various fields of physics, engineering, and even recreational activities like amusement park rides. One common question revolves around the point at which the tension in the string (if any) slackens as the bob moves through different points in the circle. Let's explore this intriguing problem and delve into the principles behind it.

Understanding the Basics of Vertical Circular Motion

Consider a bob bobbing in a vertical circular motion. At any point in the vertical circle, the bob experiences a combination of gravitational force and centripetal force due to the tension in the string. The gravitational force acts vertically downward, while the centripetal force acts horizontally towards the center of the circle, maintaining the circular path. If the tension in the string is zero, the bob will detach from the string and continue as a projectile.

Condition for Slackness

The critical point to consider here is the topmost point of the vertical circle. If the speed of the bob at the topmost point (point A) is less than a certain threshold, the tension in the string becomes zero, leading to slackness. The threshold speed, beyond which the string remains taut, is given by the equation v √(gR), where v is the speed, g is the gravitational acceleration, and R is the radius of the vertical circle.

Investigating the Given Problem: v √2gR

The problem states that the speed of the bob at point A (the topmost point) is √2gR. This is higher than the critical speed √gR. Therefore, the tension in the string is guaranteed to be non-zero at all points in the vertical circle. Let's explore this in more detail:

Deriving the Threshold Speed

At the topmost point (point A) of the vertical circle, the total energy of the bob can be expressed as:

Kinetic Energy (KE) Potential Energy (PE) Total Energy

KE (1/2)mv2

PE mgh, where h is the height, which is equal to R in this case.

For the bob to be in circular motion, the centripetal force must be balanced by the pull of gravity and the tension. Therefore, the tension in the string at the topmost point should be non-negative. If the speed is √2gR, the tension in the string is as follows:

T mg mv2/R

Substituting v √2gR into the equation:

T mg m(2gR)/R 2mg

T mg

This indicates that the tension in the string is non-zero and equal to mg. Therefore, there will be no slackness in the string as the bob moves through the vertical circle.

General Conditions for Slackness in the String

Let's generalize the condition for the bob to be in a state of slackness in the string. The bob will experience slack if the tension becomes zero at any point in the vertical circle. The tension in the string at any point in the vertical circle can be expressed as:

T mgcosθ mv2/R

where θ is the angle from the vertical. For the tension to be zero:

mgcosθ mv2/R

v2 gRcosθ

The minimum speed required to ensure no slackness is when the tension is just above zero. This occurs when the speed is the minimum for the bob to maintain circular motion. This speed is given by:

v √(gR)

For the given problem, the speed is √2gR, which is higher than √gR. Therefore, the tension in the string is non-zero throughout the vertical circle.

Conclusion

To summarize, the condition for slackness in a string during a vertical circular motion is when the speed at the topmost point is less than √gR. Given that the speed at the topmost point (point A) is √2gR, the string will not slacken at any point in the vertical circle. This ensures that the bob remains in continuous circular motion without losing contact with the string.

Frequently Asked Questions

What happens if the tension in the string becomes zero at the bottommost point?
The bob will no longer be constrained by the string, essentially detaching from it like a thrown projectile. It will continue with a velocity determined by the energy at that bottommost point. Can the speed of the bob be adjusted to maintain tension in the string?
Yes, by adjusting the initial speed or the gravitational acceleration, you can ensure that the string remains taut throughout the circular motion. The key is to ensure the speed at each point is equal to or greater than the threshold speed √gR. What happens to the bob if the string breaks?
Once the string breaks, the bob, being no longer constrained, will follow the path of motion determined by its velocity at the instant of breakage, essentially becoming a projectile.