Question

A stone tied to the end of a string of $1$ m long is whirled in a horizontal circle with a constant speed. If the stone makes $22$ revolutions in $44$s, what is the magnitude and direction of acceleration of the stone?

Answer 1

Given length of the string l= 1 m also angular velocity $\omega =\frac{22}{44}$ revolution/second.

We know linear velocity $v=\omega \times r$

Thus velocity $v=\frac{22}{44}\times 1$=0.5 m/s

And we know centripetal or radial acceleration $a_c=\frac{v^2}{r}$

Thus $a_c=0.25m/s^2$