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A solid sphere rolls without slipping on the roof. The ratio of its rotational kinetic energy and its total kinetic energy is
- a)2/S
- b)4/5
- c)2/7
- d)3/7
Correct answer is option 'C'. Can you explain this answer?
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A solid sphere rolls without slipping on the roof. The ratio of its ro...
Solution:
When a solid sphere rolls without slipping on a surface, its total kinetic energy is given by the sum of its translational kinetic energy and rotational kinetic energy.
Let the mass of the sphere be m, radius be R and velocity be v.
Translational kinetic energy = (1/2)mv^2
Rotational kinetic energy = (1/2)Iω^2, where I is the moment of inertia of the sphere and ω is the angular velocity of the sphere.
For a solid sphere, moment of inertia I = (2/5)mr^2, where r is the radius of the sphere.
Since the sphere is rolling without slipping, its translational velocity v is related to its angular velocity ω as v = ωR.
Therefore, the total kinetic energy of the rolling sphere is given by:
K = (1/2)mv^2 + (1/2)Iω^2
= (1/2)mv^2 + (1/2)(2/5)mr^2(ω/R)^2
= (1/2)mv^2 + (1/5)mv^2
= (7/10)mv^2
The ratio of rotational kinetic energy to total kinetic energy is given by:
(rotational kinetic energy) / (total kinetic energy)
= [(1/2)Iω^2] / [(1/2)mv^2 + (1/2)Iω^2]
= [(1/2)(2/5)mr^2(ω/R)^2] / [(1/2)mv^2 + (1/2)(2/5)mr^2(ω/R)^2]
= (2/5)(ω/R)^2 / [v^2 + (2/5)(ω/R)^2]
= (2/5)(ω/Rv)^2 / [1 + (2/5)(ω/Rv)^2]
Substituting ω = v/R, we get:
(rotational kinetic energy) / (total kinetic energy)
= (2/5)(v/Rv)^2 / [1 + (2/5)(v/Rv)^2]
= (2/5)(1/1)^2 / [1 + (2/5)(1/1)^2]
= 2/7
Therefore, the correct option is (C) 2/7.
When a solid sphere rolls without slipping on a surface, its total kinetic energy is given by the sum of its translational kinetic energy and rotational kinetic energy.
Let the mass of the sphere be m, radius be R and velocity be v.
Translational kinetic energy = (1/2)mv^2
Rotational kinetic energy = (1/2)Iω^2, where I is the moment of inertia of the sphere and ω is the angular velocity of the sphere.
For a solid sphere, moment of inertia I = (2/5)mr^2, where r is the radius of the sphere.
Since the sphere is rolling without slipping, its translational velocity v is related to its angular velocity ω as v = ωR.
Therefore, the total kinetic energy of the rolling sphere is given by:
K = (1/2)mv^2 + (1/2)Iω^2
= (1/2)mv^2 + (1/2)(2/5)mr^2(ω/R)^2
= (1/2)mv^2 + (1/5)mv^2
= (7/10)mv^2
The ratio of rotational kinetic energy to total kinetic energy is given by:
(rotational kinetic energy) / (total kinetic energy)
= [(1/2)Iω^2] / [(1/2)mv^2 + (1/2)Iω^2]
= [(1/2)(2/5)mr^2(ω/R)^2] / [(1/2)mv^2 + (1/2)(2/5)mr^2(ω/R)^2]
= (2/5)(ω/R)^2 / [v^2 + (2/5)(ω/R)^2]
= (2/5)(ω/Rv)^2 / [1 + (2/5)(ω/Rv)^2]
Substituting ω = v/R, we get:
(rotational kinetic energy) / (total kinetic energy)
= (2/5)(v/Rv)^2 / [1 + (2/5)(v/Rv)^2]
= (2/5)(1/1)^2 / [1 + (2/5)(1/1)^2]
= 2/7
Therefore, the correct option is (C) 2/7.
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Community Answer
A solid sphere rolls without slipping on the roof. The ratio of its ro...
Condition of no slipping is v= rw
where w=angular velocity. and moment of inertia( I )of solid sphere is 2/5 ×m×r^2
so rotational KE of solid sphere= 1/2×I×w^2 = 1/2×2/5m×r^2×w^2 = m×r^2×w^2/5
and linear KE = 1/2 mv^2= 1/2 m×r^2×w^2
total KE=rotational+linear=7/5×m×r^2×w^2
so ratio of rotational KE/total KE = 2/7
where w=angular velocity. and moment of inertia( I )of solid sphere is 2/5 ×m×r^2
so rotational KE of solid sphere= 1/2×I×w^2 = 1/2×2/5m×r^2×w^2 = m×r^2×w^2/5
and linear KE = 1/2 mv^2= 1/2 m×r^2×w^2
total KE=rotational+linear=7/5×m×r^2×w^2
so ratio of rotational KE/total KE = 2/7
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A solid sphere rolls without slipping on the roof. The ratio of its rotational kinetic energy and its total kinetic energy isa)2/Sb)4/5c)2/7d)3/7Correct answer is option 'C'. Can you explain this answer?
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