Question

A body of mass 500gm is tied to one end of a string of length 1m and is whirled round in a horizontal circle making 2 revolutions per second. calculate the tension in the string.

Answer 1

The answer for this question is 76.88N

Answer 2

The given string has a tension of 76.88 N.

Given:

mass, m =500gm = 0.5 kg

Radius (the length of the horizontal circle) = 1m

Revolutions, N =2 rps  

[rps stands for revolutions per second]

To find:

Tension in the spring

Solution:

Centrifugal force is the force that rises from inertia of the body, which acts in a circular path on a moving body. So, the tension in the given string is the centrifugal force created whiles the revolution of the string.

Formula for this force is given by

$F=m \omega^{2} r$

Therefore to find ω,

$\omega_{\left(r a d . s^{-1}\right)}=2 \pi N_{(r p s)}$

$\omega=2 \pi \times 2$

ω= 12.4 radians

Then the tension,

$F=0.5 \times(12.4)^{2} \times 1$

$=0.5 \times 153.76 \times 1$

=76.88 N