A stone of mass 1 kg is tied at one end of string of length 1 m. It is whirled in a vertical circle at constant speed of $4\ ms^{- 1}$. The tension in the string is 6 N when the stone is at $(g = 10ms^{- 2})$

Question

Dr. Abubacker · Accepted Answer

(d) When the string makes an angle $	heta$ with the vertical, then
$T - mg\ cos\ 	heta = \frac{mv^{2}}{r}$
Substituting the values, we obtain
$6 - (1)(10)\cos{	heta = \frac{1 	imes (4)^{2}}{1}}$
or $6 - 10$ $cos\ 	heta$=16
or $cos\ 	heta = - 1 = \cos{180{^\circ}}$
$	herefore\  = 180{^\circ}$

Q. A stone of mass 1 kg is tied at one end of string of length 1 m. It is whirled in a vertical circle at constant speed of $4\ ms^{- 1}$. The tension in the string is 6 N when the stone is at $(g = 10ms^{- 2})$

Explanation

Reviewed by Dr. Abubacker

Want more practice?