# 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) $\dfrac{{173}}{{320}}m{R^2}$

Answer

Verified

290.4k+ views

**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 $\dfrac{2}{5}m{R^2}$

Last updated date: 06th Jun 2023

•

Total views: 290.4k

•

Views today: 2.46k

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Name the Largest and the Smallest Cell in the Human Body ?

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

How do you define least count for Vernier Calipers class 12 physics CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main