Question

A horizontal disc rotates with a constant angular velocity ω = 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* = 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.

Answer 1

calculation for centripetal force, acting on small body is $F_{1}=m\omega^{2}r=0.5\times6^{2}\times0.3=5.4N$

Force perpendicular to the centripetal force due to velocity of small body relative to the disc will be given by $F_{2}=2mv^*\omega=2\times0.5\times6=3N$

Normal force in vertical direction is $F_{3}=mg=0.5\times 10=5N$

The net force on small body, $F_{Net}=\sqrt{5.4^{2}+3^{2}+5^{2}}=\sqrt{63.16}=7.95N$

$=8N$