Answer
Verified
441k+ views
Hint: First step will be considering a disc shaped component parallel to the base of the solid hemisphere. Then we will find the moment of inertia of that disc about its center of mass. Then by using parallel axis theorem, we will find the moment of inertia about the axis 1. Then upon integrating the equations we got over the range 0 to R, we will get the required result.
Complete step by step solution:
Let us consider a circular disc-shaped component of the solid hemisphere at a distance x from the axis 1 with radius y and width dx.
As the mass and radius of this hemispherical solid is m and R.
So, mass of the disc, $dm=\dfrac{\text{mass of hemisphere}}{\text{volume of hemisphere}}\times \text{volume of the disc}$
$\implies dm=\left(\dfrac{m}{\dfrac{2}{3}\pi R^3}\right)(\pi y^2 dx)$
Let us take $\rho = \dfrac{m}{\dfrac{2}{3}\pi R^3}$.
Therefore, $dm=\rho(\pi y^2 d)$ ………. (i)
Now, moment of inertia of the solid hemisphere about the axis 1 can be given by the parallel axis theorem as,
$I_1 = I_{cm}+mx^2$, where $I_{cm}$ is the moment of inertia of the disk about its centre of mass.
Therefore, $dI_1 =(dm)\dfrac{y^2}{4}+(dm)x^2$
Now we will integrate both sides of the equation for the range 0 to R.
$\implies I_1 = \int_{0}^{R}{(dm)\dfrac{y^2}{4}}+\int_{0}^{R}{dmx^2}$
Putting the value of dm from equation (i)
$\implies I_1 =\int_{0}^{R}{(\rho \pi y^2 dx)\dfrac{y^2}{4}}+\int_{0}^{R}{(\rho \pi y^2 dx)x^2}$
$= \dfrac{\rho \pi}{4}\int_{0}^{R}(R^2 -x^2)dx+ \rho y\int_{0}^{R}(R^2 X^2 -x^4)dx$
$= \dfrac{\rho \pi}{4}\left[R^5 +\dfrac{R^5}{5}-2R^2\dfrac{R^3}{3}\right]+\left[\rho \pi R^2 \left(\dfrac{R^3}{3}\right)-\rho \pi \left(\dfrac{R^5}{5}\right)\right]$
$=\dfrac{ \rho \pi R^5}{4} \left[1+\dfrac{1}{5}-\dfrac{2}{3}\right]+\rho \pi R^5 \left[\dfrac{1}{3}-\dfrac{1}{5}\right]$
$= \rho \pi R^5 \left(\dfrac{4}{15}\right)$
Since, $\rho = \dfrac{m}{\dfrac{2}{3}\pi R^3}$.
$\therefore I_1 = \dfrac{3m}{2\pi R^3}.\pi R^5{4}{15}=\dfrac{2}{5}mR^2.$
Hence, option b is the correct answer.
Note: Moment of inertia of a disc is a basic thing applied here and should be remembered. One may forget to use the parallel axis theorem while solving such types of questions. The axis of solid hemisphere about which the moment of inertia has been asked should be parallel to the axis of the disc.
Complete step by step solution:
Let us consider a circular disc-shaped component of the solid hemisphere at a distance x from the axis 1 with radius y and width dx.
As the mass and radius of this hemispherical solid is m and R.
So, mass of the disc, $dm=\dfrac{\text{mass of hemisphere}}{\text{volume of hemisphere}}\times \text{volume of the disc}$
$\implies dm=\left(\dfrac{m}{\dfrac{2}{3}\pi R^3}\right)(\pi y^2 dx)$
Let us take $\rho = \dfrac{m}{\dfrac{2}{3}\pi R^3}$.
Therefore, $dm=\rho(\pi y^2 d)$ ………. (i)
Now, moment of inertia of the solid hemisphere about the axis 1 can be given by the parallel axis theorem as,
$I_1 = I_{cm}+mx^2$, where $I_{cm}$ is the moment of inertia of the disk about its centre of mass.
Therefore, $dI_1 =(dm)\dfrac{y^2}{4}+(dm)x^2$
Now we will integrate both sides of the equation for the range 0 to R.
$\implies I_1 = \int_{0}^{R}{(dm)\dfrac{y^2}{4}}+\int_{0}^{R}{dmx^2}$
Putting the value of dm from equation (i)
$\implies I_1 =\int_{0}^{R}{(\rho \pi y^2 dx)\dfrac{y^2}{4}}+\int_{0}^{R}{(\rho \pi y^2 dx)x^2}$
$= \dfrac{\rho \pi}{4}\int_{0}^{R}(R^2 -x^2)dx+ \rho y\int_{0}^{R}(R^2 X^2 -x^4)dx$
$= \dfrac{\rho \pi}{4}\left[R^5 +\dfrac{R^5}{5}-2R^2\dfrac{R^3}{3}\right]+\left[\rho \pi R^2 \left(\dfrac{R^3}{3}\right)-\rho \pi \left(\dfrac{R^5}{5}\right)\right]$
$=\dfrac{ \rho \pi R^5}{4} \left[1+\dfrac{1}{5}-\dfrac{2}{3}\right]+\rho \pi R^5 \left[\dfrac{1}{3}-\dfrac{1}{5}\right]$
$= \rho \pi R^5 \left(\dfrac{4}{15}\right)$
Since, $\rho = \dfrac{m}{\dfrac{2}{3}\pi R^3}$.
$\therefore I_1 = \dfrac{3m}{2\pi R^3}.\pi R^5{4}{15}=\dfrac{2}{5}mR^2.$
Hence, option b is the correct answer.
Note: Moment of inertia of a disc is a basic thing applied here and should be remembered. One may forget to use the parallel axis theorem while solving such types of questions. The axis of solid hemisphere about which the moment of inertia has been asked should be parallel to the axis of the disc.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Advantages and disadvantages of science
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
10 examples of evaporation in daily life with explanations
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
How do you graph the function fx 4x class 9 maths CBSE
Difference Between Plant Cell and Animal Cell