Subjective Type

A horizontal disc rotates with a constant angular velocity $$\omega=6.0\:rad/s$$ about a vertical axis passing through its centre. A small body of mass $$m=0.50\:kg$$ moves along a diameter of the disc with a velocity $$v^\prime=50\:cm/s$$ which is constant relative to the disc. Find the force in newtons that the disc exerts on the body at the moment when it is located at the distance $$r=30\:cm$$ from the rotation axis.

Solution

Centripetal force on small body is $$F_1 = m\omega^2r=0.5\times6^2\times0.3=5.4N$$
Force normal to the centripetal force due to velocity of small body relative to the disc is $$F_2 = 2mv^{'} \omega=2\times0.5\times0.5\times6=3N$$
Normal force in vertical direction is $$N=mg=0.5\times10=5N$$
So net force on small body is, $$F_n = \sqrt{F_1^2+F_2^2+N^2}=\sqrt{63.16}=7.95N \approx 8 N$$

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