
A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation (\[{e_{sphere}}/{e_{cylinder}}\]) will be
A. 3:1
B. 2:3
C. 1:5
D. 1:4
Answer
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Hint: This question belongs to the chapter called Rotational Dynamics. As in this question, we need two cases: one for the solid sphere and case two for the solid cylinder. We know the basic value of mass to be m and radius r. The angular velocity of the sphere is \[\omega \] and the cylinder is 2\[\omega \]. We need to keep the inertia formula to find our required data.
Complete step-by-step answer:
Case one- Solid sphere
We know the mass to be m and radius to be r
Angular velocity of the sphere is \[\omega \]
So, inertia of solid sphere = \[\dfrac{2}{3}m{r^2}\]
Now kinetic energy of the sphere = \[{E_s} = \dfrac{1}{2}{I_s}{\omega ^2}\]
Placing the value of I in the formula we get,
\[ \Rightarrow {E_s} = \dfrac{1}{5}m{r^2}{\omega ^2}\]
Case two. Solid cylinder
We know the mass to be m and radius to be r
Angular velocity of the cylinder is 2\[\omega \].
So, inertia of solid sphere = \[\dfrac{{m{R^2}}}{2}\]
Now kinetic energy of the sphere = \[\dfrac{1}{2}I2{\omega ^2}\]
Placing the value of I in the formula we get,
\[ \Rightarrow {E_y} = \dfrac{1}{2} \times \dfrac{{m{r^2}}}{2} \times 4{\omega ^2}\]= \[m{r^2}{\omega ^2}\]
Now, placing the required condition (\[{e_{sphere}}/{e_{cylinder}}\]), we get,
\[ \Rightarrow {e_{sphere}}/{e_{cylinder}}\] = \[\dfrac{1}{5}\]
The ratio of their kinetic energies of rotation (\[{e_{sphere}}/{e_{cylinder}}\]) will be 1:5.
Note- An object's kinetic energy (KE) is the force of its motions. The work needed to speed a corpus of a particular mass from rest into its specified speed is defined. After this energy has been gained during its acceleration, the body maintains this cinematic energy without changes in speed. The body performs the same job as the present pace is decelerated to a state of rest.
Complete step-by-step answer:
Case one- Solid sphere
We know the mass to be m and radius to be r
Angular velocity of the sphere is \[\omega \]
So, inertia of solid sphere = \[\dfrac{2}{3}m{r^2}\]
Now kinetic energy of the sphere = \[{E_s} = \dfrac{1}{2}{I_s}{\omega ^2}\]
Placing the value of I in the formula we get,
\[ \Rightarrow {E_s} = \dfrac{1}{5}m{r^2}{\omega ^2}\]
Case two. Solid cylinder
We know the mass to be m and radius to be r
Angular velocity of the cylinder is 2\[\omega \].
So, inertia of solid sphere = \[\dfrac{{m{R^2}}}{2}\]
Now kinetic energy of the sphere = \[\dfrac{1}{2}I2{\omega ^2}\]
Placing the value of I in the formula we get,
\[ \Rightarrow {E_y} = \dfrac{1}{2} \times \dfrac{{m{r^2}}}{2} \times 4{\omega ^2}\]= \[m{r^2}{\omega ^2}\]
Now, placing the required condition (\[{e_{sphere}}/{e_{cylinder}}\]), we get,
\[ \Rightarrow {e_{sphere}}/{e_{cylinder}}\] = \[\dfrac{1}{5}\]
The ratio of their kinetic energies of rotation (\[{e_{sphere}}/{e_{cylinder}}\]) will be 1:5.
Note- An object's kinetic energy (KE) is the force of its motions. The work needed to speed a corpus of a particular mass from rest into its specified speed is defined. After this energy has been gained during its acceleration, the body maintains this cinematic energy without changes in speed. The body performs the same job as the present pace is decelerated to a state of rest.
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