Answer to Question #280345 in Mechanics | Relativity for Vayops

Question

A particle of mass 2.5kg is suspended from a fixed point by a light inextensible string of length 1.8m.the particle moves in a horizontal circle on a smooth floor. calculate the angular speed of the particle if the tension in the string is 28 N

Answer 1

Given:

$m=2.5\:\rm kg$

$l=1.8\:\rm m$

$F=28\:\rm N$

The Newton's second law says

$ma=F$

Using definition of the centripetal acceleration, we obtain

$m\frac{v^2}{l}=F$

Finally

$v=\sqrt{Fl/m}=\sqrt{28*1.8/2.5}=4.5\:\rm m/s$