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Question
Answer in Brief:
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 a 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 center. Calculate x if the wheel now rotates at 80 rpm.
Solution
Given:
f1 = 100 rpm,
f2 = 80 rpm,
M = 10 kg,
R = 0.4 m,
m = 1.6 kg
To find:
Value of x
Solution:
I1 = Iwheel = `1/2"MR"^2 = 1/2(10)(0.4)^2 = 0.8` kg.m2
The MI of the wheel and the lump of clay is
I2 = Iwheel + mx2
By the principle of conservation of angular momentum,
I1ω1 = I2ω2
∴ I1(2πf1) = I2(2πf2)
∴ I2 = Iwheel + mx2 = `"f"_1/"f"_2 "I"_1 = "f"_1/"f"_2 "I"_"wheel"`
∴ `"mx"^2 = ("f"_1/"f"_2 - 1)"I"_"wheel" = (100/80 - 1)(0.8)`
`= (5/4 - 1)(0.8) = 0.2` kg.m2
v = x2 = `0.2/1.6 = 1/8`
∴ x = `1/sqrt8`m = 0.3536 m
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