Calculate the moment of inertia of each of the following uniform objects about t
ID: 1467732 • Letter: C
Question
Calculate the moment of inertia of each of the following uniform objects about the axes indicated. Consult Table Moments of Inertia of Various Bodies in the Textbook as needed.
A. A thin 2.40-kg rod of length 80.0 cm , about an axis perpendicular to it and passing through one end.
B.A thin 2.40-kg rod of length 80.0 cm , about an axis perpendicular to it and passing through its center.
C.A thin 2.40-kg rod of length 80.0 cm , about an axis parallel to the rod and passing through it.
D.A 2.00-kg sphere 23.0 cm in diameter, about an axis through its center, if the sphere is solid.
E.A 2.00-kg sphere 23.0 cm in diameter, about an axis through its center, if the sphere is a thin-walled hollow shell.
F.An 5.00-kg cylinder, of length 17.0 cm and diameter 26.0 cm , about the central axis of the cylinder, if the cylinder is thin-walled and hollow.
G.An 5.00-kg cylinder, of length 17.0 cm and diameter 26.0 cm , about the central axis of the cylinder, if the cylinder is solid.
Explanation / Answer
a) M.I of rod about axis at end of rod = M.I about axis at centre + M* (L/2)^2
= M*L^2/12 + ML^2/4 = M*L^2 / 3
=2.4*(0.8)^2/3 =0.512kg.m^2
b) M.I of rod about axis at centre of rod = M*L^2/12 =2.4*(0.8)^2/12 =0.128 kg.m^2
c) As the rod is thin and radius is not given, it can't be assumed as cylinder. Hence, M.I for a thin rod about an axis parallel to the rod and passing through it is Zero.
d) M.I of solid sphere about axis at centre = 2*m*r^2/5 = 2*2*(0.23/2)^2/5 =0.01058kg.m^2
e)M.I of hollow sphere about axis at centre = 2*m*r^2/3 = 2*2*(0.23/2)^2 /3 =0.0176325kg.m^2
f)M.I of hollow cylinder about axis at centre = m*r^2 = 5*0.13^2 = 0.0845kg/m^2
g)M.I of solid cylinder about axis at centre = m*r^2 /2 = 5*0.13^2 / 2 = 0.04225 kg/m^2
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.