Answer
405k+ views
Hint:since the inclined planes have same altitude and length it means total energy (rotational kinetic energy plus translational kinetic energy) will be same so the object which has more linear velocity at the bottom will reach first or the object which has more translational kinetic energy at the bottom of the inclined plane(assuming both objects disk and sphere performs pure rolling)
Complete step by step answer:
For an object rolling in an inclined plane,torque acting on it will be only due to friction.
Therefore,$\tau = fR$ where $f$ is friction and R is radius of the object.
Let the moment of inertia of the object be $I$. Then angular acceleration will be,
$\alpha = \dfrac{\tau }{I}\\
\Rightarrow\alpha = \dfrac{{fR}}{I}$
And forces on the object will be gravitational force and friction
Acceleration along the incline will be $a = \dfrac{{mg\sin \theta - f}}{m}$
For pure rolling,
$a = \alpha R$
$ \Rightarrow \dfrac{{mg\sin \theta - f}}{m} = \dfrac{{fR \times R}}{I}$
$ \Rightarrow g\sin \theta = \dfrac{{f{R^2}}}{I} + \dfrac{f}{m}$
$ \Rightarrow f = \dfrac{{mgI\sin \theta }}{{I + m{R^2}}}\\
\Rightarrow f= \dfrac{{mg\sin \theta }}{{1 + \dfrac{{m{R^2}}}{I}}}$
Once we know the value of friction we can find the value of acceleration.
Therefore,
$a = \dfrac{{mg\sin \theta - f}}{m}\\
\Rightarrow a= \dfrac{{mg\sin \theta - \dfrac{{mgI\sin \theta }}{{I + m{R^2}}}}}{m}\\
\Rightarrow a= g\sin \theta (1 - \dfrac{I}{{I + m{R^2}}})\\
\Rightarrow a= g\sin \theta (\dfrac{{m{R^2}}}{{I + m{R^2}}})$
If we write moment of inertia in terms of radius of gyration
i.e; $I = m{K^2}$
then value of acceleration $a = \dfrac{{g\sin \theta }}{{1 + \dfrac{{{K^2}}}{{{R^2}}}}}$
As we can see from the above equation,larger the value of K smaller the acceleration and vice versa.
For disc K is R and for sphere K is $\sqrt {\dfrac{2}{5}} $R. Since the value of K is smaller in sphere it means acceleration of sphere will be more therefore the sphere will reach first.
Hence Option B is correct.
Note:as here you can see acceleration does not depend upon the mass of the object it only depends upon value of K and value of K depends upon the geometry of the object. What it tells us is that no matter what the mass and size of an object is, as long as the geometry of the objects is the same it will take the same time to reach the ground in an inclined plane.
Complete step by step answer:
For an object rolling in an inclined plane,torque acting on it will be only due to friction.
Therefore,$\tau = fR$ where $f$ is friction and R is radius of the object.
Let the moment of inertia of the object be $I$. Then angular acceleration will be,
$\alpha = \dfrac{\tau }{I}\\
\Rightarrow\alpha = \dfrac{{fR}}{I}$
And forces on the object will be gravitational force and friction
Acceleration along the incline will be $a = \dfrac{{mg\sin \theta - f}}{m}$
For pure rolling,
$a = \alpha R$
$ \Rightarrow \dfrac{{mg\sin \theta - f}}{m} = \dfrac{{fR \times R}}{I}$
$ \Rightarrow g\sin \theta = \dfrac{{f{R^2}}}{I} + \dfrac{f}{m}$
$ \Rightarrow f = \dfrac{{mgI\sin \theta }}{{I + m{R^2}}}\\
\Rightarrow f= \dfrac{{mg\sin \theta }}{{1 + \dfrac{{m{R^2}}}{I}}}$
Once we know the value of friction we can find the value of acceleration.
Therefore,
$a = \dfrac{{mg\sin \theta - f}}{m}\\
\Rightarrow a= \dfrac{{mg\sin \theta - \dfrac{{mgI\sin \theta }}{{I + m{R^2}}}}}{m}\\
\Rightarrow a= g\sin \theta (1 - \dfrac{I}{{I + m{R^2}}})\\
\Rightarrow a= g\sin \theta (\dfrac{{m{R^2}}}{{I + m{R^2}}})$
If we write moment of inertia in terms of radius of gyration
i.e; $I = m{K^2}$
then value of acceleration $a = \dfrac{{g\sin \theta }}{{1 + \dfrac{{{K^2}}}{{{R^2}}}}}$
As we can see from the above equation,larger the value of K smaller the acceleration and vice versa.
For disc K is R and for sphere K is $\sqrt {\dfrac{2}{5}} $R. Since the value of K is smaller in sphere it means acceleration of sphere will be more therefore the sphere will reach first.
Hence Option B is correct.
Note:as here you can see acceleration does not depend upon the mass of the object it only depends upon value of K and value of K depends upon the geometry of the object. What it tells us is that no matter what the mass and size of an object is, as long as the geometry of the objects is the same it will take the same time to reach the ground in an inclined plane.
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)