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A horizontal disc rotates with a constant angular velocity $ω = 6.0 r a d / s$ 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 = 50 cm/s which is constant relative to the disc. Find the force that the disc exerts on the body at the moment when it is located at the distance r = 30 cm from the rotation axis is _______N.
(Round off to nearest integer)

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answer is 8.

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Detailed Solution

The body will experience three mutually perpendicular forces in the rotating table

Centrifugal force $m ω^{2} r .$

Normal reaction $N = m g$

Coriolis force $2 m ω v .$

The resultant force $= \sqrt{{(m ω^{2} r)}^{2} + {(m g)}^{2} + {(2 m ω v)}^{2}} = 8 N$

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