Mineral Processing and Extractive Metallurgy · Mineral beneficiation · 2 marks · Numerical answer
In a ball mill, balls should leave the wall at C and follow CDA as drawn, with ∠BOC = 120°. Assume no slip; O is the mill axis, OB is parallel to gravity, inner diameter 3.26 m and g = 9.8 m/s². Find rotational speed in rpm, to 1 decimal place.
At detachment the normal reaction vanishes, so the inward gravity component alone supplies centripetal acceleration. C is 60° from the upward vertical, giving $\omega^2r=g\cos60^\circ=g/2$. With r = 3.26/2 = 1.63 m, $\omega=\sqrt{9.8/(2\times1.63)}=1.734$ rad/s. Speed is $60\omega/(2\pi)=16.6$ rpm.
