Question

A ball of mass 3.0 kg is tied to the end of a 75-cm length of string. The ball and string are swung in a circle in a vertical plane at constant speed v, which is the minimum speed necessary to keep the string taut at all times. The center of the circle is above the floor.

Suddenly, the string is released at the instant the ball is at point P, located half way from the top to bottom on the descending arc. After what span of time does the ball hit the floor?

Answer 1

Tension in the string will be minimum when ball is at highest position

So here we can say

$T + mg = F_{net}$

here we have

$T + mg = \frac{mv^2}{R}$

now for minimum possible speed

T = 0

$mg = \frac{mv^2}{R}$

by solving above

$v = \sqrt{Rg}$

here we have

$R = 75 cm$

$v = \sqrt{0.75\times 9.8}$

$v = 2.7 m/s$

now when string is at mid position from top to bottom it is released

so now time taken to hit the floor is given as

$y = v_i t + \frac{1}{2}at^2$

$0.75 = 2.7 t + 4.9 t^2$

by solving above equation

t = 0.20 s