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 textsphere / E textcylinder) Will be

Question

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 textsphere / E textcylinder) Will be (A)  2: 3 (B) 1: 5 (C) 1: 4 (D)  3: 1

Tardigrade Admin · Accepted Answer

K ⋅ E =(1/2) I ω2 ( K ⋅ E text sphere / K ⋅ E text cylinder )=( I text sphere / I text cylinder ) ((ω text sphere /ω text cylinder ))2 =((2/5) MR 2/( MR 2)2)((ω/2 ω))2 =(4/5) × (1/4)=1: 5

$2 : 3$

$1 : 5$

$1 : 4$

$3 : 1$

$K \cdot E = \frac{1}{2} I ω^{2}$
$\frac{K \cdot E _{sphere}}{K \cdot E _{cylinder}} = \frac{I _{sphere}}{I _{cylinder}} (\frac{ω _{sphere}}{ω _{cylinder}})^{2}$
$= \frac{\frac{2}{5} M R ^{2}}{\frac{M R ^{2}}{2}} (\frac{ω}{2 ω})^{2}$
$= \frac{4}{5} \times \frac{1}{4} = 1 : 5$

Q. 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 (Esphere​/Ecylinder​) Will be

2:3

1:5

1:4

3:1

K⋅E=21​Iω2 K⋅Ecylinder ​K⋅Esphere ​​=Icylinder ​Isphere ​​(ωcylinder ​ωsphere ​​)2 =2MR2​52​MR2​(2ωω​)2 =54​×41​=1:5

$2 : 3$

$1 : 5$

$1 : 4$

$3 : 1$

$K \cdot E = \frac{1}{2} I ω^{2}$
$\frac{K \cdot E _{sphere}}{K \cdot E _{cylinder}} = \frac{I _{sphere}}{I _{cylinder}} (\frac{ω _{sphere}}{ω _{cylinder}})^{2}$
$= \frac{\frac{2}{5} M R ^{2}}{\frac{M R ^{2}}{2}} (\frac{ω}{2 ω})^{2}$
$= \frac{4}{5} \times \frac{1}{4} = 1 : 5$