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A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity ωo. A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is K initially, its final kinetic energy will be
  • a)
    2 K
  • b)
    K/ 2 
  • c)
    K
  • d)
    K/4
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A horizontal circular plate is rotating about a vertical axis passing ...
KEYCONCPET :  (K.E.)rotation =L2/2I
Here , L = constant
∴ (K.E.)rotational × I = constant.
When I is doubled, K.E.rotational becomes half.
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A horizontal circular plate is rotating about a vertical axis passing ...
Understanding the Problem
A man is sitting at the center of a rotating circular plate with an angular velocity of ω₀. When he stretches out his hands, the moment of inertia (I) of the system doubles. We need to analyze how this change affects the kinetic energy (K) of the system.
Initial Kinetic Energy
- The initial kinetic energy (K) is given by the formula:
K = (1/2) * I₀ * ω₀²
where I₀ is the initial moment of inertia.
Final Moment of Inertia
- When the man stretches his hands, the moment of inertia doubles:
I_final = 2 * I₀
Final Kinetic Energy Calculation
- The angular momentum (L) of the system is conserved since no external torque acts on it:
L_initial = L_final
- This implies:
I₀ * ω₀ = I_final * ω_final
- Substituting I_final:
I₀ * ω₀ = 2 * I₀ * ω_final
- Solving for the final angular velocity ω_final:
ω_final = (1/2) * ω₀
Final Kinetic Energy
- The final kinetic energy (K_final) is:
K_final = (1/2) * I_final * ω_final²
- Substituting for I_final and ω_final:
K_final = (1/2) * (2 * I₀) * ((1/2) * ω₀)²
- Simplifying this gives:
K_final = (1/2) * (2 * I₀) * (1/4) * ω₀²
K_final = (1/4) * I₀ * ω₀²
Since K = (1/2) * I₀ * ω₀², we find that:
K_final = K / 2
Conclusion
Thus, the final kinetic energy of the system after the man stretches out his hands is K / 2. The correct answer is option 'B'.
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A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity ωo. A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is K initially, its final kinetic energy will bea)2 Kb)K/ 2c)Kd)K/4Correct answer is option 'B'. Can you explain this answer?
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A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity ωo. A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is K initially, its final kinetic energy will bea)2 Kb)K/ 2c)Kd)K/4Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity ωo. A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is K initially, its final kinetic energy will bea)2 Kb)K/ 2c)Kd)K/4Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity ωo. A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is K initially, its final kinetic energy will bea)2 Kb)K/ 2c)Kd)K/4Correct answer is option 'B'. Can you explain this answer?.
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