Question

A body of mass 0.4 kg is revolved in a horizontal circle of radius 5 m. if it performs 120 rpm, the centripetal force acting on it is nearly equal to​

Answer 1

Answer:

(120 ×2×3.14)/60 =754/60 =13.

Centripetal force=mv^2/r ={(0.4) (13)^2}/5 =68/5=14.

Answer 2

Given : Mass of body, m = 0.4 kg

            radius of Horizontal circle = 5 m

             f = 120 rpm.

To Find : Centripetal force acting

Solution :

m = 0.4 kg

r = 5 m

f = 120 rpm

Thus, angular speed is

$\[\omega = 2\pi f = 2\pi \times \frac{{120}}{{60}} = 4\pi \]$ rad/s

Now, centripetal force is given as

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

$\[F = 0.4 \times {(4\pi )^2} \times 5 = 32{\pi ^2}\]$ N

Hence Centripetal force acting on is $\[32{\pi ^2}N\]$