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Rotational Motion question

2007 · Shift 1 · Q56
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Rotational Motion question

2007 · Shift 1 · Q56

JEE AdvancedPhysicsRotational MotionMCQ+3 / −1
Two discs AAA and BBB are mounted coaxially on a vertical axle. The discs have moment of inertia III and 2I2 I2I respectively about the common axis. Disc AAA is imparted an initial angular velocity 2ω2 \omega2ω using the entire potential energy of a spring compressed by a distance x1x_1x1​. Disc BBB is imparted an angular velocity ω\omegaω by a spring having the same spring constant and compressed by a distance x2x_2x2​. Both the discs rotate in the clockwise direction.The ratio x1x2{{{x_1}} \over {{x_2}}}x2​x1​​ is
  1. A
    2
  2. B
    12\frac{1}{2}21​
  3. C
    2\sqrt22​
  4. D
    12\frac{1}{\sqrt2}2​1​
View written solutionFree

Correct answer: C

Step-by-Step Solution:

  1. Principle of Energy Conservation The problem states that the entire potential energy of a compressed spring is used to impart angular velocity to each disc. This is a direct application of the principle of conservation of energy, where the initial potential energy of the spring is converted into the final rotational kinetic energy of the disc.

  2. Formulas for Energy

    • The potential energy (UUU) stored in a spring with spring constant kkk compressed by a distance xxx is given by: U=12kx2U = {1 \over 2} k x^2U=21​kx2
    • The rotational kinetic energy (KrotK_{rot}Krot​) of a body with moment of inertia III rotating with angular velocity ω\omegaω is given by: Krot=12Iω2K_{rot} = {1 \over 2} I \omega^2Krot​=21​Iω2
  3. Analysis for Disc A

    • Moment of inertia of disc A, IA=II_A = IIA​=I.
    • Angular velocity of disc A, ωA=2ω\omega_A = 2\omegaωA​=2ω.
    • The spring is compressed by a distance x1x_1x1​.
    • According to the conservation of energy: Potential Energy of spring 1 = Rotational Kinetic Energy of Disc A U1=KAU_1 = K_AU1​=KA​ 12kx12=12IAωA2{1 \over 2} k x_1^2 = {1 \over 2} I_A \omega_A^221​kx12​=21​IA​ωA2​
    • Substituting the given values: 12kx12=12I(2ω)2{1 \over 2} k x_1^2 = {1 \over 2} I (2\omega)^221​kx12​=21​I(2ω)2 12kx12=12I(4ω2){1 \over 2} k x_1^2 = {1 \over 2} I (4\omega^2)21​kx12​=21​I(4ω2) 12kx12=2Iω2⋯(1){1 \over 2} k x_1^2 = 2 I \omega^2 \quad \cdots(1)21​kx12​=2Iω2⋯(1)
  4. Analysis for Disc B

    • Moment of inertia of disc B, IB=2II_B = 2IIB​=2I.
    • Angular velocity of disc B, ωB=ω\omega_B = \omegaωB​=ω.
    • The spring used has the same spring constant kkk and is compressed by a distance x2x_2x2​.
    • According to the conservation of energy: Potential Energy of spring 2 = Rotational Kinetic Energy of Disc B U2=KBU_2 = K_BU2​=KB​ 12kx22=12IBωB2{1 \over 2} k x_2^2 = {1 \over 2} I_B \omega_B^221​kx22​=21​IB​ωB2​
    • Substituting the given values: 12kx22=12(2I)(ω)2{1 \over 2} k x_2^2 = {1 \over 2} (2I) (\omega)^221​kx22​=21​(2I)(ω)2 12kx22=Iω2⋯(2){1 \over 2} k x_2^2 = I \omega^2 \quad \cdots(2)21​kx22​=Iω2⋯(2)
  5. Calculate the Ratio x1/x2x_1 / x_2x1​/x2​

    • To find the ratio, we can divide Equation (1) by Equation (2): 12kx1212kx22=2Iω2Iω2{{{1 \over 2} k x_1^2} \over {{1 \over 2} k x_2^2}} = {{2 I \omega^2} \over {I \omega^2}}21​kx22​21​kx12​​=Iω22Iω2​
    • The terms 12k{1 \over 2}k21​k on the left side and Iω2I\omega^2Iω2 on the right side cancel out: x12x22=2{{x_1^2} \over {x_2^2}} = 2x22​x12​​=2
    • Taking the square root of both sides gives the ratio of the compression distances: x1x2=2{{x_1} \over {x_2}} = \sqrt{2}x2​x1​​=2​
  6. Conclusion The ratio x1/x2x_1 / x_2x1​/x2​ is 2\sqrt{2}2​. This corresponds to option C.

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