# Question #85a19

Oct 24, 2016

For circular motion of a body of mass $m$ moving with velocity $v$ in a circle of radius $r$, centripetal force is given by the expression
${F}_{c} = \frac{m {v}^{2}}{r}$

Using the centrifugal force conditions, the tension in the string at the top of path can be written as
${T}_{\text{top}} = \frac{m {v}^{2}}{r} - m g$
where $g = 9.81 m {s}^{-} 2$ is acceleration due to gravity.

Inserting given values we get
$4.0 = \frac{0.30 {v}^{2}}{0.7} - 0.30 \times 9.81$
Rearranging we get
$\frac{0.30 {v}^{2}}{0.7} = 4.0 + 0.30 \times 9.81$
$\implies v = 4.0 m {s}^{-} 1$, rounded to one decimal place.