Answer to Question #274955 in Mechanics | Relativity for bailouyin

Question

If the moment of inertia of a tire is 0. 63 kg.m² and its weight is 24.5 N, what is the diameter of the tire? If the weight is doubled, what will be the moment of inertia?

Answer 1

$I=mr^2$

$r=\sqrt{I/m}=\sqrt{0.63/(24.5/g)}=0.50$ m

diameter of the tire:

$d=2r=2\cdot0.50=1$ m

If the weight is doubled:

$I=2mr^2=2\cdot24.5\cdot0.5^2/g=1.25$ kg$\cdot$ m²