Question

A horizontal disc rotates with a constant angular velocity $\omega =6.0rad/s$ about a vertical axis passing through its centre. A small body of mass $m=0.50kg$ moves along a diameter of the disc with a velocity $v^{\prime }=50cm/s$ 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 $r=30cm$ from the rotation axis.

Answer 1

Centripetal force on small body is $F_1=m{\omega }^{2}r=0.5\times 6^2\times 0.3=5.4N$

Force normal to the centripetal force due to velocity of small body relative to the disc is $F_2=2m{v}^{{}^{\prime }}\omega =2\times 0.5\times 0.5\times 6=3N$

Normal force in vertical direction is $N=mg=0.5\times 10=5N$

So net force on small body is, $F_n=\sqrt{F_1^2+F_2^2+N^2}=\sqrt{63.16}=7.95N\approx 8N$