A solid sphere rolls without slipping down an inclined plane. What fraction of its total energy is rotational kinetic energy?
This question evaluates the energy distribution in rolling motion.
Design multiple-choice questions (MCQs) for CBSE Class 11 Physics, Chapter 6: “System of Particles and Rotational Motion,” based on the NCERT syllabus for the 2024-2025 academic year. The questions should cover important topics like center of mass, torque, angular momentum and rotational dynamics. Include a blend of theoretical and numerical problems that assess students’ understanding and problem-solving skills. Vary the difficulty level to challenge students at all stages of learning. Ensure the questions are precise, easy to understand, and aligned with NCERT guidelines, aiding students in effective exam preparation and strengthening their grasp of rotational motion concepts.
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When a solid sphere rolls without slipping down an inclined plane, its total energy is distributed between translational kinetic energy and rotational kinetic energy. The fraction of its total energy that is rotational kinetic energy can be derived as follows:
1. Total Energy (E):
E = KEₜᵣₐₙₛₗₐₜᵢₒₙₐₗ + KEᵣₒₜₐₜᵢₒₙₐₗ
2. Translational Kinetic Energy:
KEₜᵣₐₙₛₗₐₜᵢₒₙₐₗ = 1/2Mv², where M is the mass and v is the velocity of the center of mass.
3. Rotational Kinetic Energy:
KEᵣₒₜₐₜᵢₒₙₐₗ = 1/2Iω², where I is the moment of inertia and ω is the angular velocity. For a solid sphere, I = 2/5MR² and ω = v/R. Substituting ω, KEᵣₒₜₐₜᵢₒₙₐₗ = 1/2.2/5MR². v²/R² = 1/5Mv².
4. Total Energy Expression:
E = 1/2Mv² + 1/5Mv² = 7/10Mv².
5. Fraction of Rotating Energy:
Fraction of Rotation Energy = KEᵣₒₜₐₜᵢₒₙₐₗ/E = (1/5Mv²) / (7/10 Mv²) = 2/7.
The fraction of the total energy that is rotational kinetic energy is 2/7.
This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6 System of Particles and Rotational Motion. Give answer according to your understanding.
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