Question

Ina 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 holds the rod near the ends and thus rotate with the rod (in fig.). Let the mass of each kid is $15kg$, the between the points of the rod where the two kids holds it be $3.0m$ and suppose the 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.

Answer 1

Given,

Mass of kids, $m=15kg$

Length of rod, $L=3m$

Distance of kid from center, $d=1.5m$

$\omega =\frac{2\pi N}{60}=\frac{2\pi \times 20}{60}=\frac{2\pi }{3}rad/\sec$

Frictional force $F=md{\omega }^2=15\times \left(1.5\right)\times {\left(\frac{2\pi }{3}\right)}^2=10{\pi }^2$

Hence, Frictional force on one of the kids is 10${\pi }^2$