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Find the moment of inertia about axis 1 (diameter of hemisphere) of solid hemisphere $(m,R)$.
(A) $\dfrac{{m{R^2}}}{5}$
(B) $\dfrac{2}{5}m{R^2}$
(C) $\dfrac{2}{3}m{R^2}$
(D) $\dfrac{{173}}{{320}}m{R^2}$

Last updated date: 20th Jun 2024
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Hint First of all get the mass $dm$ of the disc by using the expression:
$dm = \left( {\dfrac{m}{{{V_H}}}} \right) \times {V_D}$
where, $m$ is the mass of hemisphere
${V_H}$ is the volume of hemisphere
${V_D}$ is the volume of disc
Now, use the formula
${I_{yy'}} = {I_{cm}} + m{r^2}$
Then, integrate it with the limit from $0$ to $R$
After solving we will get the value of ${I_{yy'}}$ then, use the value of $\rho = \dfrac{{3m}}{{2\pi {R^3}}}$
Then, if $I$ displace them parallel to the axis and again join them to complete a sphere
$2I = \dfrac{2}{5}m{R^2}$

Complete Step by Step Solution
Suppose that hemisphere is constructed by small discs
Mass $dm$of disc= $dm = \left( {\dfrac{m}{{{V_H}}}} \right) \times {V_D}$
  \therefore dm = \left( {\dfrac{m}{{\dfrac{2}{3}\pi {R^3}}}} \right)\left( {\pi {y^2}dx} \right) \\
  dm = \rho \left( {\pi {y^2}dx} \right) \\
Because, $\rho = \dfrac{m}{{\dfrac{2}{3}\pi {R^3}}}$

According to parallel and perpendicular axis theorem we get
${I_{yy'}} = {I_{cm}} + m{r^2}$
$\therefore d{I_{yy'}} = (dm)\dfrac{{{y^2}}}{4} + (dm){x^2}$
Now, integrating both sides
  {I_{yy'}} = \int\limits_0^R {(dm)\dfrac{{{y^2}}}{4} + } \int\limits_0^R {dm} {x^2} \\
   \Rightarrow \int\limits_0^R {(\rho \pi {y^2}dx)\dfrac{{{y^2}}}{4} + } \int\limits_0^R {(\rho \pi {y^2}dx){x^2}} \\
Now, put the limits
  {I_{yy'}} = \dfrac{{\rho \pi }}{4}\left[ {\dfrac{{{R^5}}}{5} + {R^5} - 2{R^2}\dfrac{{{R^3}}}{3}} \right] + \left[ {\rho \pi {R^5}\left( {\dfrac{{{R^3}}}{3}} \right) - \rho \pi \left( {\dfrac{{{R^5}}}{5}} \right)} \right] \\
   = \rho \pi {R^5}\left[ {\dfrac{1}{4} + \dfrac{1}{{20}} - \dfrac{1}{4} \times \dfrac{2}{3}} \right] + \rho \pi {R^5}\left[ {\dfrac{1}{3} - \dfrac{1}{5}} \right] \\
   = \rho \pi {R^5}\left( {\dfrac{4}{{15}}} \right) \\
Now, putting the value of $\rho $ in the above equation
  {I_{yy'}} = \dfrac{{3m}}{{2\pi {R^3}}}\pi {R^5}\dfrac{4}{{15}} = \dfrac{2}{5}m{R^2} \\
  \therefore I = \dfrac{2}{5}m{R^2} \\
But if $I$displaces the parallel to axis and join them to complete the sphere then,
$2I = \dfrac{2}{5}m{R^2}$
After cancelling 2 on both sides we get
$I = \dfrac{{m{R^2}}}{5}$

Hence, option (A) is the correct answer

Note Moment of Inertia of continuous bodies is $\dfrac{2}{5}m{R^2}$
Moment of Inertia of disc about centre of mass is $\dfrac{{m{R^2}}}{2}$
Moment of inertia of sphere about centre of mass is $\dfrac{2}{5}m{R^2}$