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A solid sphere is rolling without slipping on a horizontal surface. The ratio of its rotational kinetic energy to its translational kinetic energy is
- a)2/9
- b)2/7
- c)2/5
- d)7/2
Correct answer is option 'C'. Can you explain this answer?
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Solution:
Concepts: Rolling motion, Rotational kinetic energy, Translational kinetic energy
Rolling motion is a combination of both rotational and translational motions. When a solid sphere is rolling without slipping on a horizontal surface, both types of kinetic energy are involved.
Rotational kinetic energy:
The rotational kinetic energy of a rolling sphere is given by the expression:
K_rot = (1/2)Iω^2,
where I is the moment of inertia of the sphere and ω is its angular velocity.
Translational kinetic energy:
The translational kinetic energy of the sphere is given by the expression:
K_trans = (1/2)mv^2,
where m is the mass of the sphere and v is its linear velocity.
Ratio of rotational kinetic energy to translational kinetic energy:
The ratio of rotational kinetic energy to translational kinetic energy is given by:
K_rot/K_trans = (1/2)Iω^2 / (1/2)mv^2
= (I/m)(ω/v)^2
= (2/5)(v/R)^2
where R is the radius of the sphere and v/R is the linear speed of the sphere.
The moment of inertia of a solid sphere is given by the expression:
I = (2/5)mr^2
where r is the radius of the sphere.
Substituting the value of I in the above expression, we get:
K_rot/K_trans = (2/5)(v/R)^2
= 2/5 * 1
= 2/5
Therefore, the correct option is C) 2/5.
Conclusion:
The ratio of rotational kinetic energy to translational kinetic energy for a rolling sphere is 2/5.
Concepts: Rolling motion, Rotational kinetic energy, Translational kinetic energy
Rolling motion is a combination of both rotational and translational motions. When a solid sphere is rolling without slipping on a horizontal surface, both types of kinetic energy are involved.
Rotational kinetic energy:
The rotational kinetic energy of a rolling sphere is given by the expression:
K_rot = (1/2)Iω^2,
where I is the moment of inertia of the sphere and ω is its angular velocity.
Translational kinetic energy:
The translational kinetic energy of the sphere is given by the expression:
K_trans = (1/2)mv^2,
where m is the mass of the sphere and v is its linear velocity.
Ratio of rotational kinetic energy to translational kinetic energy:
The ratio of rotational kinetic energy to translational kinetic energy is given by:
K_rot/K_trans = (1/2)Iω^2 / (1/2)mv^2
= (I/m)(ω/v)^2
= (2/5)(v/R)^2
where R is the radius of the sphere and v/R is the linear speed of the sphere.
The moment of inertia of a solid sphere is given by the expression:
I = (2/5)mr^2
where r is the radius of the sphere.
Substituting the value of I in the above expression, we get:
K_rot/K_trans = (2/5)(v/R)^2
= 2/5 * 1
= 2/5
Therefore, the correct option is C) 2/5.
Conclusion:
The ratio of rotational kinetic energy to translational kinetic energy for a rolling sphere is 2/5.
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A solid sphere is rolling without slipping on a horizontal surface. The ratio of its rotational kinetic energy to its translational kinetic energy is a)2/9b)2/7 c)2/5d)7/2Correct answer is option 'C'. Can you explain this answer?
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