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# A solid cylinder of uniform density of radius $2cm$ has mass of $50g$. It its length is $10cm$, calculate its moment of inertia about 1) Its own axis of rotation passing through the center,2) An axis passing through its centre and perpendicular to its length.

Last updated date: 09th Sep 2024
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Hint: Moment of inertia of a cylinder is related to its mass, length and radius. All the values are given in the question. So we can directly substitute it in the equations and calculate the moment of inertia of the cylinder.
Formula used:
$I=\dfrac{M{{R}^{2}}}{2}$
$I=M\left( \dfrac{{{R}^{2}}}{4}+\dfrac{{{L}^{2}}}{12} \right)$

Radius of the cylinder, $R=2cm$
Mass of the cylinder, $M=50g$
Length of the cylinder, $L=10cm$

1) The moment of inertia of a cylinder about its own axis passing through its centre is given as,
$I=\dfrac{M{{R}^{2}}}{2}$ ------- 1
Substitute the given values in equation 1 we get,
$I=\dfrac{\left( 50\times {{10}^{-3}} \right){{\left( 2\times {{10}^{-2}} \right)}^{2}}}{2}={{10}^{-5}}kg{{m}^{2}}$

2) The moment of inertia of a cylinder about an axis passing through its centre and perpendicular to the length is given as,
$I=M\left( \dfrac{{{R}^{2}}}{4}+\dfrac{{{L}^{2}}}{12} \right)$ --------- 2
Substituting the values given, in equation 2 we get,
$I=50\times {{10}^{-3}}\left( \dfrac{{{\left( 2\times {{10}^{-2}} \right)}^{2}}}{4}+\dfrac{{{\left( 10\times {{10}^{-2}} \right)}^{2}}}{12} \right)$
$I=50\times {{10}^{-3}}\times 0.0001+0.00083$
$I=46.5\times {{10}^{-6}}kg{{m}^{2}}$

If two disks have the same mass but one is solid and the other has all the mass around the rim, then the disks would have different moments of inertia. Hence, while calculating the moment of inertia of objects, we only have to examine the orbital motion of small point-like bodies, where all the mass of the object is concentrated at one particular point at a given radius $r$.