A body is tied up by a string of length l and rotated in verticle circle at minimum speed it reaches at highest point and breakes in parabolic path find value of of horizontal range

The scenario you're describing involves a body tied to a string and rotated in a vertical circle, reaching the highest point before breaking free and following a parabolic path. To find the horizontal range of the body after it breaks free, we can break down the problem into several parts.

Understanding the Motion

When the body is tied to a string and rotated in a vertical circle, it experiences centripetal acceleration, and at the highest point, the tension in the string is at its minimum. For the body to just reach the highest point without falling back, it needs to have a specific minimum speed.

Speed at the Highest Point

At the highest point of the vertical circle, the forces acting on the body are gravity and the tension in the string. The minimum speed \( v \) at this point can be derived from the condition that the centripetal force required to keep the body moving in a circle is provided solely by its weight when the tension is zero. This leads to the equation:

• Weight = Centripetal Force:
• mg = \(\frac{mv^2}{l}\)

Here, \( m \) is the mass of the body, \( g \) is the acceleration due to gravity, and \( l \) is the length of the string. Simplifying this gives:

v = \sqrt{gl}

Breaking Free and Parabolic Motion

Once the body reaches the highest point and the string breaks, it will follow a projectile motion path. The key to finding the horizontal range lies in analyzing this parabolic motion. At the moment of breaking, the body has a horizontal velocity of \( v \) and is at a height of \( 2l \) (the diameter of the circle).

Time of Flight

The time \( t \) it takes for the body to hit the ground can be calculated using the formula for vertical motion:

y = v_{iy} t + \frac{1}{2} g t^2

Since the initial vertical velocity \( v_{iy} \) at the highest point is 0 (it’s only moving horizontally), we can simplify to:

2l = \frac{1}{2} g t^2

Solving for \( t \) gives:

t = \sqrt{\frac{4l}{g}}

Calculating Horizontal Range

The horizontal range \( R \) can be found using the horizontal component of velocity and the time of flight. The horizontal velocity \( v_{ix} \) at the moment of breaking is \( \sqrt{gl} \). Thus, the horizontal range is given by:

R = v_{ix} \cdot t

Substituting the values we have:

R = \sqrt{gl} \cdot \sqrt{\frac{4l}{g}} = 2l

Final Result

The horizontal range of the body after it breaks free from the string and follows a parabolic path is therefore:

R = 2l

This means that the body will land a distance of twice the length of the string away from the point directly below where it broke free. Understanding these principles of motion helps in visualizing and solving similar problems in physics effectively.