Question

A solid cylinder of uniform density of radius 2cm has mass of 50g. If its length is 12cm, calculate its moment of inertia about an axis passing through its centre and perpendicular to its length.

Answer 1

Hope you'll understand my handwriting...

Answer 2

Moment of inertia about an axis is $\bold{6.50\times 10^{-5} \ kg.m^{2}}$

Given:

Radius = R = 2 cm

Mass = M = 50 g

Length = L = 12 cm

To find:

Moment of inertia = ?

Formula used:

$\bold{Moment \ of \ inertia = { I }_{ Cm } =M(\frac{{ R }^{ 2 }}{4}+\frac{{ L }^{ 2 }}{12})}$

Solution:

Moment of inertia is:

$\bold{{ I }_{ Cm } =M(\frac{{ R }^{ 2 }}{4}+\frac{{ L }^{ 2 }}{12})}$

$\bold{I_{Cm}=50\times { 10 }^{ -3 }\left[ \cfrac { { \left( 2\times { 10 }^{ -2 } \right) }^{ 2 } }{ 4 } +\cfrac { { \left( 12\times { 10 }^{ -2 } \right) }^{ 2 } }{ 12 } \right]}$

$\bold{I_{Cm}=50\times { 10 }^{ -3 }\times \left[ 0.0001+0.0012 \right]}$

$\bold{I_{Cm}=5\times { 10 }^{ -2 }\times 13\times { 10 }^{ -4 }}$

$\bold{I_{Cm}=65\times { 10 }^{ -6 }=6.50\times { 10 }^{ -5 } \ kg.{ m }^{ 2 }}$