Question

A flywheel used to prepare earthenware pots is set into rotation at 100 rpm. It is in the form of a disc of mass 10 kg and radius 0.4 m. A lump of clay (to be taken equivalent to a particle) of mass 1.6 kg falls on it and adheres to it at a certain distance x from the centre. Calculate x if the wheel now rotates at 80 rpm.

Answer 1

Given : f₁ = 100 rpm, f₂= 80 rpm, M =10 kg, R = 0.4m, m = 1.6kg

I₁=I_wheel = \(\frac{1}{2} MR^2\)= \(\frac{1}{2}(10)(0.4)^2\) = 0.8 kg m²

MI of the wheel and lump of the clay is

I₂ = I_wheel + mx²

By the principle of conservation of angular momentum

I₁w₁ = I₂w₂

I₁(2πf₁) = I₂(2πf₂)

I₂ = I_wheel + mx² =\(\frac{f_1}{f_2}I_1=\frac{f_1}{f_2}I_{wheel}\) 

\(∴ mx^2 = (\frac{f1}{f2}−1)I_{wheel}\) 

\(∴ mx^2= (\frac{100}{80}−1)(0.8)= 0.2 kg\ m^2\)  

∴ x² = \(\frac{0.2}{1.6}=\frac{ 1}{8}\) 

∴ x= \(1\over\sqrt 8\) = 0.3536 m