Q1. A rotating object has a moment of inertia I and an angular velocity ω. What is its rotational kinetic energy? (2020)
Solution:
Rotational kinetic energy K.E. = (1/2)Iω².
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Q2. A solid sphere of radius R rolls without slipping down an incline of height h. What is its speed at the bottom of the incline? (2021)
Solution:
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. For a solid sphere, v = √(5gh/7). Thus, speed at the bottom is √(3gh).
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Q3. A disc of radius R and mass M is rotating about its axis with an angular velocity ω. What is its rotational kinetic energy? (2020)
Solution:
Rotational kinetic energy K.E. = (1/2)Iω². For a disc, I = (1/2)MR², thus K.E. = (1/4)MR²ω².
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Q4. A uniform rod of length L is pivoted at one end and released from rest. What is the angular speed just before it hits the ground? (2019)
Solution:
Using conservation of energy, potential energy converts to rotational kinetic energy. Angular speed ω = √(3g/L).
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Q5. A particle moves in a circular path of radius r with a constant speed v. What is the centripetal acceleration? (2022)
Solution:
Centripetal acceleration a_c = v²/r.
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Q6. A flywheel has a moment of inertia I and is rotating with an angular velocity ω. If a torque τ is applied, what is the angular acceleration? (2021)
Solution:
From Newton's second law for rotation, τ = Iα, thus α = τ/I.
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Q7. A solid sphere of radius R rolls without slipping down an inclined plane of height h. What is the speed of the center of mass at the bottom of the incline? (2021)
Solution:
Using conservation of energy, potential energy at height h = kinetic energy at the bottom. For a solid sphere, v = √(3gh).
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Q8. A mass m is attached to a string of length L and is swung in a vertical circle. What is the tension in the string at the top of the circle? (2023)
Solution:
At the top, T + mg = mv²/L. T = mv²/L - mg.
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Q9. A cylinder rolls down an incline without slipping. If its mass is M and radius is R, what is the acceleration of its center of mass? (2020)
Solution:
For rolling without slipping, a_cm = (2/3)g sin(θ).
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Q10. A uniform rod of length L is pivoted at one end and released from rest. What is the angular speed when it reaches the vertical position? (2019)
Solution:
Using conservation of energy, potential energy converts to rotational kinetic energy. Angular speed ω = √(3g/L).
