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

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

MI of the wheel and lump of the clay is

I= Iwheel + mx2

By the principle of conservation of angular momentum

I1w= I2w2

I1(2πf1) = I2(2πf2)

I= Iwheel + mx=$$\frac{f_1}{f_2}I_1=\frac{f_1}{f_2}I_{wheel}$$

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

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

∴ x$$\frac{0.2}{1.6}=\frac{ 1}{8}$$

∴ x= $$1\over\sqrt 8$$ = 0.3536 m

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