As shown in the figure, a small coin of mass $40 ext{ g}$ is placed on the horizontal surface of a

Question

As shown in the figure, a small coin of mass $40 	ext{ g}$ is placed on the horizontal surface of a rotating disc. The disc starts from rest and is given a constant angular acceleration $\alpha = 2 	ext{ rad/s}^2$. The coefficient of static friction between the coin and the disc is $\mu_s = 3/4$ and coefficient of kinetic friction is $\mu_k = 0.5$. The coin is placed at a distance $r = 1 	ext{ m}$ from the centre of the disc. The magnitude of the resultant force on the coin exerted by the disc just before it starts slipping on the disc is: (Take $g = 10 	ext{ m/s}^2$)

Professor PC Thomas Classes · Answer

Just before slipping, the static friction force $f_s$ reaches its limiting value $f_{s,	ext{max}} = \mu_s mg = (3/4) 	imes 0.04 	ext{ kg} 	imes 10 	ext{ m/s}^2 = 0.3 	ext{ N}$. The disc exerts two forces on the coin: the normal force $N = mg = 0.04 	imes 10 = 0.4 	ext{ N}$ (vertically upwards) and the static friction force $f_s$ (in the horizontal plane). The total resultant force exerted by the disc on the coin is $F_{res} = \sqrt{N^2 + f_s^2} = \sqrt{0.4^2 + 0.3^2} = 0.5 	ext{ N}$.