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 as shown in figure. Let the mass of each kid be $15\text{kg}$, the distance between the points of the rod where the two kids hold it be $3.0\text{m}$ 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.

• A. $80\text{N}$
• B. $90\text{N}$
• C. $100\text{N}$
• D. $110\text{N}$

Answer 1

The correct option is C $100\text{N}$

Given, $d=3\text{m}$ and mass $m=15\text{kg}$

Angular speed, $\omega =20\text{rot/min}$

$\omega =\frac{20\times 2\pi }{60}=\frac{2\pi }{3}$

Distance from one boy to the axis of rotation $r=1.5\text{m}$.

Frictional force acting on the boy's hand is responsible for centipetal force.

$\therefore f=mr{\omega }^2=15\times \left(1.5\right)\times {\left(\frac{2\pi }{3}\right)}^2$

$\therefore f=10{\pi }^2\text{N}\approx 100\text{N}$

Hence, option (c) is the correct answer.