Answer
Verified
439.8k+ views
Hint: The total energy is the sum of translational and rotational kinetic energy. Ratio is calculated by dividing rotational and total kinetic energy.
Step by step answer: A circular disc is rolling down the inclined plane i.e. without slipping. When the body (circular disc) is rolling down the body, it has both rotational and translational kinetic energy i.e., rotational energy as the body is rotating while coming down the inclined plane and translational energy because it is coming straight from the inclined plane.
Let the mass of the circular disc be m, velocity be v and angular velocity (the velocity possessed by a body during rotation) be ω and radius of the disc be r.
The moment of inertia (I) is defined as the sum of the products of the masses of all the particles of the body and square of the distances from the axis of rotation i.e. $I = \dfrac{1}{2}m{r^2}$ [since the disc is rotating about an axis passing through the centre]
Rotational Kinetic energy (R)=$\dfrac{1}{2}I{\omega ^2}$ [I is moment of inertia]
$\Rightarrow$ $\dfrac{1}{2}\left( {\dfrac{1}{2} \times m{r^2}} \right){\omega ^2} = \dfrac{1}{4}\left( {m{r^2}{\omega ^2}} \right)$
The moment of inertia is defined as the sum of the products of the masses of all the particles of the body and square of the distances from the axis of rotation.
Translational kinetic energy (T)= $\dfrac{1}{2}\left( {m{v^2}} \right)$
$\Rightarrow$ $\dfrac{1}{2}m{\left( {r\omega } \right)^2}$ $\left[ {v = r\omega } \right]$
$\Rightarrow$ $\dfrac{1}{2}\left( {m{r^2}{\omega ^2}} \right)$
The total energy = $R + T$
$\Rightarrow$ $\dfrac{1}{4}\left( {m{r^2}{\omega ^2}} \right) + \dfrac{1}{2}\left( {m{r^2}{\omega ^2}} \right) = \left( {\dfrac{1}{4} + \dfrac{1}{2}} \right)\left( {m{r^2}{\omega ^2}} \right)$
$\therefore $ $\dfrac{3}{4}\left( {m{r^2}{\omega ^2}} \right)$
Ratio of rotational and total kinetic energy = $\dfrac{{\dfrac{1}{4}m{r^2}{\omega ^2}}}{{\dfrac{3}{4}m{r^2}{\omega ^2}}}$ = $\dfrac{1}{3}$
Therefore, option B is correct.
Note: When a body rolls down an inclined plane, both translational and kinetic energy contributes to the motion. So, the total energy is being calculated by adding both these energies.
Step by step answer: A circular disc is rolling down the inclined plane i.e. without slipping. When the body (circular disc) is rolling down the body, it has both rotational and translational kinetic energy i.e., rotational energy as the body is rotating while coming down the inclined plane and translational energy because it is coming straight from the inclined plane.
Let the mass of the circular disc be m, velocity be v and angular velocity (the velocity possessed by a body during rotation) be ω and radius of the disc be r.
The moment of inertia (I) is defined as the sum of the products of the masses of all the particles of the body and square of the distances from the axis of rotation i.e. $I = \dfrac{1}{2}m{r^2}$ [since the disc is rotating about an axis passing through the centre]
Rotational Kinetic energy (R)=$\dfrac{1}{2}I{\omega ^2}$ [I is moment of inertia]
$\Rightarrow$ $\dfrac{1}{2}\left( {\dfrac{1}{2} \times m{r^2}} \right){\omega ^2} = \dfrac{1}{4}\left( {m{r^2}{\omega ^2}} \right)$
The moment of inertia is defined as the sum of the products of the masses of all the particles of the body and square of the distances from the axis of rotation.
Translational kinetic energy (T)= $\dfrac{1}{2}\left( {m{v^2}} \right)$
$\Rightarrow$ $\dfrac{1}{2}m{\left( {r\omega } \right)^2}$ $\left[ {v = r\omega } \right]$
$\Rightarrow$ $\dfrac{1}{2}\left( {m{r^2}{\omega ^2}} \right)$
The total energy = $R + T$
$\Rightarrow$ $\dfrac{1}{4}\left( {m{r^2}{\omega ^2}} \right) + \dfrac{1}{2}\left( {m{r^2}{\omega ^2}} \right) = \left( {\dfrac{1}{4} + \dfrac{1}{2}} \right)\left( {m{r^2}{\omega ^2}} \right)$
$\therefore $ $\dfrac{3}{4}\left( {m{r^2}{\omega ^2}} \right)$
Ratio of rotational and total kinetic energy = $\dfrac{{\dfrac{1}{4}m{r^2}{\omega ^2}}}{{\dfrac{3}{4}m{r^2}{\omega ^2}}}$ = $\dfrac{1}{3}$
Therefore, option B is correct.
Note: When a body rolls down an inclined plane, both translational and kinetic energy contributes to the motion. So, the total energy is being calculated by adding both these energies.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
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
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE