A solid sphere rolls down an inclined plane without slipping. The fraction of its total energy associated with its translation motion is :Correct answer is '0.714'. Can you explain this answer?

Question

Answer

Given data is insufficient

Answer

Explanation:

The total energy of the solid sphere is given by the sum of its translational kinetic energy and rotational kinetic energy. When the sphere is rolling without slipping, the fraction of its total energy associated with its translation motion can be calculated as follows:

- Translational kinetic energy: The translational kinetic energy of the sphere is given by $ \frac{1}{2}mv^2 $, where m is the mass of the sphere and v is its linear velocity.

- Rotational kinetic energy: The rotational kinetic energy of the sphere is given by $ \frac{1}{2}I\omega^2 $, where I is the moment of inertia of the sphere and ω is its angular velocity.

- Total kinetic energy: The total kinetic energy of the sphere is the sum of its translational and rotational kinetic energies, i.e., $ KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $.

- Fraction of total energy associated with translation: The fraction of the total energy associated with translation can be calculated as $ \frac{KE_{translational}}{KE_{total}} = \frac{\frac{1}{2}mv^2}{\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2} $.

Given that the sphere is rolling without slipping, the linear velocity v is related to the angular velocity ω by $ v = r\omega $, where r is the radius of the sphere. Using this relationship and the expression for the moment of inertia of a solid sphere, the fraction of total energy associated with translation can be calculated as 0.714.

Therefore, the fraction of the total energy associated with the translation motion of the solid sphere is 0.714.

Answer

The correct answer is: 0.714

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