A flywheel used to prepare earthenware pots is set into rotation at 50 rpm. It is in the form of a disc of mass 5 kg and radius 0.2 m. A lump of clay (to be taken equivalent to a particle) of mass 0.8

Question

A flywheel used to prepare earthenware pots is set into rotation at 50 rpm. It is in the form of a disc of mass 5 kg and radius 0.2 m. A lump of clay (to be taken equivalent to a particle) of mass 0.8 kg falls on it and adheres to it at a certain distance 'x' from the centre. The value of 'x' if the wheel now rotates at 40 rpm, is

TestHub · Accepted Answer

The initial moment of inertia of the flywheel (disc) is $I_{1} = \frac{1}{2} M R^{2}$ .

Substituting the given values, $I_{1} = \frac{1}{2} \times 5 \times (0.2)^{2} = 0.1 kg m^{2}$ .

The initial angular velocity is $ω_{1} = 50 rpm$ .

When the lump of clay of mass $m = 0.8 kg$ falls and adheres at a distance $x$ , the new moment of inertia becomes $I_{2} = I_{1} + m x^{2} = 0.1 + 0.8 x^{2}$ .

The new angular velocity is $ω_{2} = 40 rpm$ .

By the principle of conservation of angular momentum, $I_{1} ω_{1} = I_{2} ω_{2}$ .

$0.1 \times 50 = (0.1 + 0.8 x^{2}) \times 40$

$5 = 4 + 32 x^{2}$

$32 x^{2} = 1$

$x^{2} = \frac{1}{32}$

$x = \frac{1}{32} m$

Answer: $\frac{1}{32}$

Physics - Rotation Question with Solution | TestHub

Options:

Doubts & Discussion