Question

In a children's park a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rod always remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod. Let the mass of each kid be 15 kg, the distance between the points of the rod where the two kids hold it be 3.0 m and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the force of friction exerted by the rod on one of the kids.
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Answer 1

The Force of the Friction, exerted by the Rod on one of the kids is $\begin{equation}10 \pi^{2} \mathrm{N}$

Explanation:

Mass of the each  kid (m) = 15 kg

Diameter = 3 m

$Radius =\frac{diameter}{2}$

Radius of the Rod = $3/2$ = 1.5 m.  

The rod rotates at the rate of 20 revolutions per minute

Frequency = 20 Revolutions/minute.

1 minute = 60 second  

Frequency $=\frac{20}{60}{s}^{-1}$

$f = \frac{1}{3}{s}^{-1}$

Angular Velocity = 2πf

$\begin{equation}\begin{aligned}&=2 \times \pi \times \frac{1}{3}\\&=\frac{2 \pi}{3}\end{aligned}$

Now, we know that while rotation, the Centripetal force will balance the Friction. So that kids keeps on moving.

$\begin{equation}\begin{aligned}&m r w^{2}=\mu m g\\&m r w^{2}=f\\&f=15 \times 1.5 \times\left(\frac{2 \pi}{3}\right)^{2}\\&f=15 \times 1.5 \times \frac{2^{2} \times 3.14^{2}}{3^{2}}\\&f=15 \times 1.5 \times \frac{4 \times 9.85}{9}\\&f=15 \times 1.5 \times \frac{39.4}{9}\\&f=15 \times 1.5 \times 4.38\\&f=22.5 \times 4.38\\&f=95.55 N\end{aligned}\end{equation}$

$\begin{equation}\text { Or } \mathrm{f}=10 \pi^{2} \mathrm{N}$

The Force of the Friction, exerted by the Rod on one of the kids is  $\begin{equation}10 \pi^{2} \mathrm{N}$

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

$10 \pi^{2} \: N$