Answer
Verified
407.7k+ views
Hint: The moment of inertia of a point object about any axis is equal to the product of its mass and the square of the perpendicular distance between object and axis. To calculate the moment of inertia of any continuous mass system we break down the object into point mass and add moment of inertia of all point masses to get the moment of inertia of the entire system. We can also break the continuous mass system into a system of already known moment of inertia (for e.g. breaking discs into rings, etc.).
Complete step by step answer:
To calculate the moment of inertia of the remaining disc we will divide the disc into thin semicircular rings. In order to do so we will go to distance r (from centre O) and further by $dr$(where $dr\to 0$ ) take entire semicircular ring between r and $dr$
Now, our aim will be to first calculate the moment of inertia of this semicircular ring. We can clearly observe that every point of this semicircular ring is situated at equal distance from the axis, which is equal to the radius of the semicircular ring which is equal to r. So to calculate the moment of inertia of this semicircular ring we can simply multiply its mass to square of its radius. Since the disc has uniform mass distribution, we can get mass of a semicircular ring by multiplying its mass by surface mass density (i.e. mass per unit area).
Surface mass density of the disc will be its total mass divided by its total area i.e. $\dfrac{2M}{\pi {{R}_{2}}^{2}}$. To calculate the area of a semicircular ring we can assume the semicircular ring is taken out from the disc and spread out. So now the semicircular ring will convert into a rectangle with length equal to its perimeter i.e. $\pi r$ and breadth will be the width of the semicircular ring i.e. $dr$. So the area of the semicircular ring becomes a product of length and breadth which is $\pi rdr$ .
Now moment of inertia of semicircular ring will be product its mass and square of its radius i.e.
$\dfrac{2Mrdr}{{{R}_{2}}^{2}}\times {{r}^{2}}=\dfrac{2M{{r}^{3}}dr}{{{R}_{2}}^{2}}$
This is the moment of inertia of a semicircular ring at distance r, now to calculate moment of inertia of complete disc we will add all such small semicircular rings covering the entire remaining disc. This can be achieved by integration and to cover the entire disc limit of r should be from ${{R}_{1}}$ to${{R}_{2}}$.
So moment of inertia of remaining disc is
\[\int\limits_{{{R}_{1}}}^{{{R}_{2}}}{\dfrac{2M{{r}^{3}}dr}{{{R}_{2}}^{2}}}=\dfrac{2M}{{{R}_{2}}^{2}}\int\limits_{{{R}_{1}}}^{{{R}_{2}}}{{{r}^{3}}dr}\]
On integration we will get
$\dfrac{M}{2{{R}_{2}}^{2}}\left( {{R}_{2}}^{4}-{{R}_{1}}^{4} \right)$
Hence the correct option is A.
Note: $dr$ here is taken to be tending to zero so that we can assume the taken portion to be a ring. Often students make a mistake while calculating the value of surface mass density, while measuring it we should check that given mass is of original disc or mass of disc after being cut and accordingly divide it with area.
Complete step by step answer:
To calculate the moment of inertia of the remaining disc we will divide the disc into thin semicircular rings. In order to do so we will go to distance r (from centre O) and further by $dr$(where $dr\to 0$ ) take entire semicircular ring between r and $dr$
Now, our aim will be to first calculate the moment of inertia of this semicircular ring. We can clearly observe that every point of this semicircular ring is situated at equal distance from the axis, which is equal to the radius of the semicircular ring which is equal to r. So to calculate the moment of inertia of this semicircular ring we can simply multiply its mass to square of its radius. Since the disc has uniform mass distribution, we can get mass of a semicircular ring by multiplying its mass by surface mass density (i.e. mass per unit area).
Surface mass density of the disc will be its total mass divided by its total area i.e. $\dfrac{2M}{\pi {{R}_{2}}^{2}}$. To calculate the area of a semicircular ring we can assume the semicircular ring is taken out from the disc and spread out. So now the semicircular ring will convert into a rectangle with length equal to its perimeter i.e. $\pi r$ and breadth will be the width of the semicircular ring i.e. $dr$. So the area of the semicircular ring becomes a product of length and breadth which is $\pi rdr$ .
Now moment of inertia of semicircular ring will be product its mass and square of its radius i.e.
$\dfrac{2Mrdr}{{{R}_{2}}^{2}}\times {{r}^{2}}=\dfrac{2M{{r}^{3}}dr}{{{R}_{2}}^{2}}$
This is the moment of inertia of a semicircular ring at distance r, now to calculate moment of inertia of complete disc we will add all such small semicircular rings covering the entire remaining disc. This can be achieved by integration and to cover the entire disc limit of r should be from ${{R}_{1}}$ to${{R}_{2}}$.
So moment of inertia of remaining disc is
\[\int\limits_{{{R}_{1}}}^{{{R}_{2}}}{\dfrac{2M{{r}^{3}}dr}{{{R}_{2}}^{2}}}=\dfrac{2M}{{{R}_{2}}^{2}}\int\limits_{{{R}_{1}}}^{{{R}_{2}}}{{{r}^{3}}dr}\]
On integration we will get
$\dfrac{M}{2{{R}_{2}}^{2}}\left( {{R}_{2}}^{4}-{{R}_{1}}^{4} \right)$
Hence the correct option is A.
Note: $dr$ here is taken to be tending to zero so that we can assume the taken portion to be a ring. Often students make a mistake while calculating the value of surface mass density, while measuring it we should check that given mass is of original disc or mass of disc after being cut and accordingly divide it with area.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE