Question

A $40kg$ child makes $10$ revolutions in $29.3$ seconds, moving on a circular path. The radius of the path taken by the child is $2.9$ meters. Then, centripetal force acting on him is:

• A. $5.32N$
• B. $53.2N$
• C. $533.6N$
• D. $267N$

Answer 1

The correct option is C $533.6N$

Given, $m=40kg$ and
Radius of circular path($r$) = $2.9m$
Time taken to complete $1$ revolution
i.e ${}^{\prime }{T}^{\prime }=\frac{29.3}{10}=2.93s$
$T=\frac{2\pi }{\omega }\Rightarrow \omega =\frac{2\pi }{T}$

Now speed of boy
$v=r\omega =\frac{2\pi r}{T}=\frac{2\times 3.14\times 2.9}{2.93}=6.22m/s$

To find centripetal acceleration, applying the formula:
$a_c=\frac{v^2}{r}=\frac{(6.22{)}^2}{2.9}=13.34m/s^2$
$\therefore F_{centripetal}=m\times a_c$
$\Rightarrow F_c=40\times 13.34=533.6N$