Respuesta :
Answer:
option E
Explanation:
given,
Rotational Kinetic Energy, KE_r = [tex]\dfrac{1}{2}I\omega^2[/tex]
Moment of inertia of the solid,
[tex]I = \dfrac{2}{5}MR^2[/tex]
[tex]\omega = \dfrac{V}{R}[/tex]
now,
[tex]KE_r = \dfrac{1}{2}I\omega^2[/tex]
[tex]KE_r = \dfrac{1}{2}\times \dfrac{2}{5}MR^2 (\dfrac{V}{R})^2[/tex]
[tex]KE_r =\dfrac{1}{5}MV^2[/tex]......(1)
transnational kinetic energy
[tex]KE_t =\dfrac{1}{2}MV^2[/tex]
Total kinetic energy
[tex]KE = \dfrac{1}{2}MV^2 + \dfrac{1}{5} MV^2[/tex]
[tex]KE = \dfrac{7}{10}MV^2[/tex]
ratio of rotational kinetic energy to the total kinetic energy
[tex]\dfrac{KE_r}{KE_t}=\dfrac{\dfrac{1}{5}MV^2}{\dfrac{7}{10}MV^2}[/tex]
[tex]\dfrac{KE_r}{KE_t}=\dfrac{2}{7}[/tex]
hence, the correct answer is option E
The ratio between the rotational kinetic energy about the center of the sphere and the sphere’s total kinetic energy is (E) 2/7
Rotational kinetic energy of sphere
The rotational kinetic energy of the sphere K = 1/2Iω² where
- I = rotational inertia of sphere = 2/5MR² where
- M = mass of sphere and
- R = radius of sphere,
- ω = angular acceleration of sphere = V/R where
- V = speed of sphere.
So, K = 1/2Iω²
K = 1/2 × 2/5MR² × V²/R²
K = MV²/5
Translational kinetic energy of sphere
The translational kinetic energy of the sphere K' = 1/2MV² where
- M = mass of sphere and
- V = speed of sphere
Total kinetic energy of sphere
The total kinetic energy of the sphere K" = K + K'
= MV²/5 + MV²/2
= 7MV²/10
Ratio of rotational kinetic energy to total kinetic energy
The ratio of rotational to total kinetic energy of the sphere is K/K" = MV²/5 ÷ 7MV²/10
= 1/5 × 10/7
= 2/7
So, the ratio between the rotational kinetic energy about the center of the sphere and the sphere’s total kinetic energy is (E) 2/7
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