Question

A body is just being revolved in a vertical circle of radius R such that lowest point of trajectory just touches the ground.The string breaks when the body is at the highest point.The horizontal distance covered by the body just after the string breaks is?

Answer 1

The “horizontal distance” covered by the body is 2r.

Given:

Radius of the circle = R

Solution:

At the highest point the force at that point shall be  

$\frac{m v^{2}}{R}=T+m g$   {at this point T=0 when string is broken}

The above equation is as per Newton’s second law of motion

Minimum velocity at highest point will be $v=\sqrt{r g}$

And to reach the ground time taken by the body will be  

$\mathrm{T}=\sqrt{\frac{2 h}{g}}$

So the horizontal distance that can be covered by the body will be given as,  

$v \times t=\sqrt{r g} \times \sqrt{\frac{2 h}{g}} \ldots \ldots \ldots \ldots(i)$

As we know that, (h) height is the sum of double time of the radius (r).

h=r+r=2 r

Then equation 1 becomes,

$=\sqrt{r g} \times \sqrt{\frac{2 \times 2 \times r}{g}}=2r$

The “horizontal distance” covered by the body is 2r.