Question

a stone of mass 0.1 kg tied to one end of a string1.0 m long is revolved in a horizontal circle at the rate of 10/pi revolutions per sec. calculate the tension of the string.

Answer 1

mass of stone tied to one end of a string, m = 0.1 kg
angular speed , $\omega=\frac{10}{\pi}rpm$
or, $\omega=\frac{10}{\pi}\times2\pi\:rad/s$
$\omega=20rad/s$

We know, $v=\omega r$
so, v = 20 × 1 = 20m/s

tension of the string = centripetal force
T = mv²/r
= {(0.1) × (20)²}/1
= 0.1 × 400
= 40N

hence, tension of the string is 40N.

Answer 2

Explanation:

here is the answer;)))