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The ratio of rotational kinetic energy and translational kinetic energy of a rolling circular disc is:
A. 4:1
B. 2:1
C. 1:4
D. 1:2
Answer
463.8k+ views
Hint: In order to solve this problem we need to know the moment of inertia of the disc about the axis passing through the center and perpendicular to the disc and then we have to us the relation between angular velocity.=, translational velocity and radius of the circle. Then diving the Rotational KE with Translational KE to get the ratio. Doing this will give you the right answer.
Complete answer:
We need to find the ratio of rotational kinetic energy and translational kinetic energy of a rolling circular disc.
The rotational KE of the Disc is $\dfrac{1}{2}\left( {I{\omega ^2}} \right)$ I is the moment of inertia and $\omega $ is the angular momentum.
We also know that moment of inertia of disc about the center and perpendicular to the disc is $I = \dfrac{{M{R^2}}}{2}$
We also know that $\omega = \dfrac{V}{R}$
On putting the values in the equation of rotational KE we get the new equation as:
$ \Rightarrow \dfrac{1}{2}\left( {\dfrac{{M{R^2}}}{2}\left( {\dfrac{{{V^2}}}{{{R^2}}}} \right)} \right) = \dfrac{{M{V^2}}}{4}$
So, Rotational KE is $\dfrac{{M{V^2}}}{4}$………………(1)
We also know that Translational Kinetic energy = $\dfrac{1}{2}\left( {M{V^2}} \right)$………………….(2)
Dividing (1) from (2) we get the new equation as:
$\dfrac{{2M{V^2}}}{{4M{V^2}}} = \dfrac{1}{2}$.
So, the ratio is 1:2.
Therefore the correct option is D.
Note:
When to get to solve such problems you need to know that moment of inertia of disc about the center and perpendicular to the disc is $I = \dfrac{{M{R^2}}}{2}$ and The rotational KE of the Disc is $\dfrac{1}{2}\left( {I{\omega ^2}} \right)$ I is the moment of inertia and $\omega $ is the angular momentum and $\omega = \dfrac{V}{R}$. The moment of inertia, otherwise known as the mass moment of inertia, angular mass or rotational inertia, of a rigid body is a quantity that determines the torque needed for a desired angular acceleration about a rotational axis; similar to how mass determine the force needed for a desired acceleration. Knowing this will help further to solve more such problems.
Complete answer:
We need to find the ratio of rotational kinetic energy and translational kinetic energy of a rolling circular disc.
The rotational KE of the Disc is $\dfrac{1}{2}\left( {I{\omega ^2}} \right)$ I is the moment of inertia and $\omega $ is the angular momentum.
We also know that moment of inertia of disc about the center and perpendicular to the disc is $I = \dfrac{{M{R^2}}}{2}$
We also know that $\omega = \dfrac{V}{R}$
On putting the values in the equation of rotational KE we get the new equation as:
$ \Rightarrow \dfrac{1}{2}\left( {\dfrac{{M{R^2}}}{2}\left( {\dfrac{{{V^2}}}{{{R^2}}}} \right)} \right) = \dfrac{{M{V^2}}}{4}$
So, Rotational KE is $\dfrac{{M{V^2}}}{4}$………………(1)
We also know that Translational Kinetic energy = $\dfrac{1}{2}\left( {M{V^2}} \right)$………………….(2)
Dividing (1) from (2) we get the new equation as:
$\dfrac{{2M{V^2}}}{{4M{V^2}}} = \dfrac{1}{2}$.
So, the ratio is 1:2.
Therefore the correct option is D.
Note:
When to get to solve such problems you need to know that moment of inertia of disc about the center and perpendicular to the disc is $I = \dfrac{{M{R^2}}}{2}$ and The rotational KE of the Disc is $\dfrac{1}{2}\left( {I{\omega ^2}} \right)$ I is the moment of inertia and $\omega $ is the angular momentum and $\omega = \dfrac{V}{R}$. The moment of inertia, otherwise known as the mass moment of inertia, angular mass or rotational inertia, of a rigid body is a quantity that determines the torque needed for a desired angular acceleration about a rotational axis; similar to how mass determine the force needed for a desired acceleration. Knowing this will help further to solve more such problems.
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