Answer
424.2k+ views
Hint: For a small radius dr, we have to consider a small distribution of mass dm. We need to formulate for small change in moment of inertia and integrate it along smaller radius (a) and bigger radius (b). After obtaining a moment of inertia of the disc I, we can find the radius of gyration.
Formula used:
Radius of gyration, $k=\sqrt{\dfrac{I}{m}}$
Complete answer:
Radius of gyration of a body about an axis of rotation is defined as the radial distance to a point which would have a moment of inertia which is equal to the moment of inertia of the body’s actual distribution of mass. This is conditional to the total mass of the body being concentrated.
Let us consider the moment of inertia of the disc to be I
dI is a partial derivative of moment of inertia.
$\begin{align}
& dI=(dm){{r}^{2}} \\
& =(\sigma dA){{r}^{2}} \\
& =(\dfrac{{{\sigma }_{0}}}{r}2\pi rdr){{r}^{2}} \\
& =({{\sigma }_{0}}2\pi ){{r}^{2}}dr \\
& I=\int{dI=\int_{a}^{b}{{{\sigma }_{0}}}}2\pi {{r}^{2}}dr \\
& ={{\sigma }_{0}}2\pi (\dfrac{{{b}^{3}}-{{a}^{3}}}{3}) \\
& m=\int{dm}=\int{\sigma dA} \\
& ={{\sigma }_{0}}2\pi \int_{a}^{b}{dr} \\
& m={{\sigma }_{0}}2\pi (b-a) \\
\end{align}$
Radius of gyration of the disc is
$\begin{align}
& k=\sqrt{\dfrac{I}{m}}=\sqrt{\dfrac{({{b}^{3}}-{{a}^{3}})}{3(b-a)}} \\
& =\sqrt{(\dfrac{{{a}^{2}}+{{b}^{2}}+ab}{3})} \\
\end{align}$
Hence the radius of gyration of the disc is $\sqrt{(\dfrac{{{a}^{2}}+{{b}^{2}}+ab}{3})}$
Additional Information:
The dynamics and kinematics of rotation around a fixed axis of a rigid body are algebraically much easier than those for free rotation of a rigid body. They are completely analogous to those of linear motion along a single fixed direction, which is not true for free rotation of a rigid body.
Note:
A rigid body is an object of limited reach in which all the distances between the component particles doesn’t change. Perfect rigid bodies do not exist. External forces can deform any solid. So we define a rigid body as an object that can be deformed under large forces.
Formula used:
Radius of gyration, $k=\sqrt{\dfrac{I}{m}}$
Complete answer:
Radius of gyration of a body about an axis of rotation is defined as the radial distance to a point which would have a moment of inertia which is equal to the moment of inertia of the body’s actual distribution of mass. This is conditional to the total mass of the body being concentrated.
Let us consider the moment of inertia of the disc to be I
dI is a partial derivative of moment of inertia.
![seo images](https://www.vedantu.com/question-sets/39b51a74-926a-4618-846b-fa014d7dd18c2670276650085532220.png)
$\begin{align}
& dI=(dm){{r}^{2}} \\
& =(\sigma dA){{r}^{2}} \\
& =(\dfrac{{{\sigma }_{0}}}{r}2\pi rdr){{r}^{2}} \\
& =({{\sigma }_{0}}2\pi ){{r}^{2}}dr \\
& I=\int{dI=\int_{a}^{b}{{{\sigma }_{0}}}}2\pi {{r}^{2}}dr \\
& ={{\sigma }_{0}}2\pi (\dfrac{{{b}^{3}}-{{a}^{3}}}{3}) \\
& m=\int{dm}=\int{\sigma dA} \\
& ={{\sigma }_{0}}2\pi \int_{a}^{b}{dr} \\
& m={{\sigma }_{0}}2\pi (b-a) \\
\end{align}$
Radius of gyration of the disc is
$\begin{align}
& k=\sqrt{\dfrac{I}{m}}=\sqrt{\dfrac{({{b}^{3}}-{{a}^{3}})}{3(b-a)}} \\
& =\sqrt{(\dfrac{{{a}^{2}}+{{b}^{2}}+ab}{3})} \\
\end{align}$
Hence the radius of gyration of the disc is $\sqrt{(\dfrac{{{a}^{2}}+{{b}^{2}}+ab}{3})}$
Additional Information:
The dynamics and kinematics of rotation around a fixed axis of a rigid body are algebraically much easier than those for free rotation of a rigid body. They are completely analogous to those of linear motion along a single fixed direction, which is not true for free rotation of a rigid body.
Note:
A rigid body is an object of limited reach in which all the distances between the component particles doesn’t change. Perfect rigid bodies do not exist. External forces can deform any solid. So we define a rigid body as an object that can be deformed under large forces.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)