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Rotational Motion question

2025 · Shift 1 · Q33
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  5. /2025 · Shift 1 · Q33

Rotational Motion question

2025 · Shift 1 · Q33

JEE AdvancedPhysicsRotational MotionMCQ+3 / −1
The center of a disk of radius rrr and mass mmm is attached to a spring of spring constant kkk, inside a ring of radius R>rR>rR>r as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following the Hooke's law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as T=2πωT=\frac{2 \pi}{\omega}T=ω2π​. The correct expression for ω\omegaω is (ggg is the acceleration due to gravity): JEE Advanced 2025 Paper 1 Online Physics - Rotational Motion Question 1 English
  1. A
    23(gR−r+km)\sqrt{\frac{2}{3} \left( \frac{g}{R - r} + \frac{k}{m} \right)}32​(R−rg​+mk​)​
  2. B
    2g3(R−r)+km\sqrt{\frac{2g}{3(R - r)} + \frac{k}{m}}3(R−r)2g​+mk​​
  3. C
    16(gR−r+km)\sqrt{\frac{1}{6} \left( \frac{g}{R - r} + \frac{k}{m} \right)}61​(R−rg​+mk​)​
  4. D
    14(gR−r+km)\sqrt{\frac{1}{4} \left( \frac{g}{R - r} + \frac{k}{m} \right)}41​(R−rg​+mk​)​
View written solutionFree

Correct answer: A

Method 1: Using Conservation of Energy

  1. Define Coordinates and Variables: Let θ\thetaθ be the small angular displacement of the center of the disk from its equilibrium position (the bottom of the ring). The center of the disk moves along a circular path of radius Reff=R−rR_{eff} = R - rReff​=R−r. The arc length displacement of the center of mass (C.M.) is s=(R−r)θs = (R - r)\thetas=(R−r)θ. The velocity of the C.M. is vcm=dsdt=(R−r)dθdt=(R−r)θ˙v_{cm} = \frac{ds}{dt} = (R - r)\frac{d\theta}{dt} = (R - r)\dot{\theta}vcm​=dtds​=(R−r)dtdθ​=(R−r)θ˙.

  2. Kinetic Energy (KE): The total kinetic energy is the sum of the translational kinetic energy of the C.M. and the rotational kinetic energy about the C.M.

    • Translational KE: KEtrans=12mvcm2=12m(R−r)2θ˙2KE_{trans} = \frac{1}{2}mv_{cm}^2 = \frac{1}{2}m(R - r)^2\dot{\theta}^2KEtrans​=21​mvcm2​=21​m(R−r)2θ˙2.
    • For the disk rolling without slipping, the condition is vcm=ωdiskrv_{cm} = \omega_{disk} rvcm​=ωdisk​r, where ωdisk\omega_{disk}ωdisk​ is the angular velocity of the disk about its center. So, ωdisk=vcmr=(R−r)θ˙r\omega_{disk} = \frac{v_{cm}}{r} = \frac{(R - r)\dot{\theta}}{r}ωdisk​=rvcm​​=r(R−r)θ˙​.
    • The moment of inertia of the disk about its C.M. is Icm=12mr2I_{cm} = \frac{1}{2}mr^2Icm​=21​mr2.
    • Rotational KE: KErot=12Icmωdisk2=12(12mr2)((R−r)θ˙r)2=14mr2(R−r)2θ˙2r2=14m(R−r)2θ˙2KE_{rot} = \frac{1}{2}I_{cm}\omega_{disk}^2 = \frac{1}{2} \left( \frac{1}{2}mr^2 \right) \left( \frac{(R - r)\dot{\theta}}{r} \right)^2 = \frac{1}{4}mr^2 \frac{(R-r)^2\dot{\theta}^2}{r^2} = \frac{1}{4}m(R - r)^2\dot{\theta}^2KErot​=21​Icm​ωdisk2​=21​(21​mr2)(r(R−r)θ˙​)2=41​mr2r2(R−r)2θ˙2​=41​m(R−r)2θ˙2.
    • Total KE: KE=KEtrans+KErot=12m(R−r)2θ˙2+14m(R−r)2θ˙2=34m(R−r)2θ˙2KE = KE_{trans} + KE_{rot} = \frac{1}{2}m(R - r)^2\dot{\theta}^2 + \frac{1}{4}m(R - r)^2\dot{\theta}^2 = \frac{3}{4}m(R - r)^2\dot{\theta}^2KE=KEtrans​+KErot​=21​m(R−r)2θ˙2+41​m(R−r)2θ˙2=43​m(R−r)2θ˙2.

  3. Potential Energy (PE): Let the potential energy be zero at the equilibrium position (θ=0\theta=0θ=0).

    • Gravitational PE: The C.M. rises by a height h=(R−r)−(R−r)cos⁡θ=(R−r)(1−cos⁡θ)h = (R - r) - (R - r)\cos\theta = (R - r)(1 - \cos\theta)h=(R−r)−(R−r)cosθ=(R−r)(1−cosθ). For small θ\thetaθ, cos⁡θ≈1−θ22\cos\theta \approx 1 - \frac{\theta^2}{2}cosθ≈1−2θ2​, so h≈(R−r)θ22h \approx (R - r)\frac{\theta^2}{2}h≈(R−r)2θ2​. PEg=mgh≈mg(R−r)θ22=12mg(R−r)θ2PE_g = mgh \approx mg(R - r)\frac{\theta^2}{2} = \frac{1}{2}mg(R - r)\theta^2PEg​=mgh≈mg(R−r)2θ2​=21​mg(R−r)θ2.
    • Spring PE: The spring is stretched along the arc, so the extension is s=(R−r)θs = (R-r)\thetas=(R−r)θ. PEsp=12ks2=12k(R−r)2θ2PE_{sp} = \frac{1}{2}ks^2 = \frac{1}{2}k(R - r)^2\theta^2PEsp​=21​ks2=21​k(R−r)2θ2.
    • Total PE: PE=PEg+PEsp=12mg(R−r)θ2+12k(R−r)2θ2=12[mg(R−r)+k(R−r)2]θ2PE = PE_g + PE_{sp} = \frac{1}{2}mg(R - r)\theta^2 + \frac{1}{2}k(R - r)^2\theta^2 = \frac{1}{2}[mg(R - r) + k(R-r)^2]\theta^2PE=PEg​+PEsp​=21​mg(R−r)θ2+21​k(R−r)2θ2=21​[mg(R−r)+k(R−r)2]θ2.

  4. Equation of Motion: The total mechanical energy E=KE+PEE = KE + PEE=KE+PE is conserved. E=34m(R−r)2θ˙2+12[mg(R−r)+k(R−r)2]θ2E = \frac{3}{4}m(R - r)^2\dot{\theta}^2 + \frac{1}{2}[mg(R - r) + k(R-r)^2]\theta^2E=43​m(R−r)2θ˙2+21​[mg(R−r)+k(R−r)2]θ2. For a conservative system, dEdt=0\frac{dE}{dt} = 0dtdE​=0. ddt(34m(R−r)2θ˙2+12[mg(R−r)+k(R−r)2]θ2)=0\frac{d}{dt} \left( \frac{3}{4}m(R - r)^2\dot{\theta}^2 + \frac{1}{2}[mg(R - r) + k(R-r)^2]\theta^2 \right) = 0dtd​(43​m(R−r)2θ˙2+21​[mg(R−r)+k(R−r)2]θ2)=0 34m(R−r)2(2θ˙θ¨)+12[mg(R−r)+k(R−r)2](2θθ˙)=0\frac{3}{4}m(R - r)^2(2\dot{\theta}\ddot{\theta}) + \frac{1}{2}[mg(R - r) + k(R-r)^2](2\theta\dot{\theta}) = 043​m(R−r)2(2θ˙θ¨)+21​[mg(R−r)+k(R−r)2](2θθ˙)=0 Dividing by θ˙\dot{\theta}θ˙ (for non-trivial motion): 32m(R−r)2θ¨+[mg(R−r)+k(R−r)2]θ=0\frac{3}{2}m(R - r)^2\ddot{\theta} + [mg(R - r) + k(R-r)^2]\theta = 023​m(R−r)2θ¨+[mg(R−r)+k(R−r)2]θ=0 θ¨+mg(R−r)+k(R−r)232m(R−r)2θ=0\ddot{\theta} + \frac{mg(R - r) + k(R-r)^2}{\frac{3}{2}m(R - r)^2} \theta = 0θ¨+23​m(R−r)2mg(R−r)+k(R−r)2​θ=0 This is the equation for Simple Harmonic Motion (SHM), θ¨+ω2θ=0\ddot{\theta} + \omega^2\theta = 0θ¨+ω2θ=0.

  5. Calculate Angular Frequency (ω\omegaω): ω2=mg(R−r)+k(R−r)232m(R−r)2\omega^2 = \frac{mg(R - r) + k(R-r)^2}{\frac{3}{2}m(R - r)^2}ω2=23​m(R−r)2mg(R−r)+k(R−r)2​ ω2=mg(R−r)32m(R−r)2+k(R−r)232m(R−r)2\omega^2 = \frac{mg(R - r)}{\frac{3}{2}m(R - r)^2} + \frac{k(R-r)^2}{\frac{3}{2}m(R - r)^2}ω2=23​m(R−r)2mg(R−r)​+23​m(R−r)2k(R−r)2​ ω2=2g3(R−r)+2k3m\omega^2 = \frac{2g}{3(R - r)} + \frac{2k}{3m}ω2=3(R−r)2g​+3m2k​ ω2=23(gR−r+km)\omega^2 = \frac{2}{3} \left( \frac{g}{R - r} + \frac{k}{m} \right)ω2=32​(R−rg​+mk​) ω=23(gR−r+km)\omega = \sqrt{\frac{2}{3} \left( \frac{g}{R - r} + \frac{k}{m} \right)}ω=32​(R−rg​+mk​)​

Method 2: Using Torque

  1. Forces and Torques: Consider the forces acting on the disk. The net tangential force provides the acceleration of the C.M. and the friction provides the torque for rotation. Tangential equation of motion for the C.M.: mat=∑Ftma_t = \sum F_tmat​=∑Ft​ m(R−r)θ¨=−mgsin⁡θ−Fsp−ffrictionm(R-r)\ddot{\theta} = -mg\sin\theta - F_{sp} - f_{friction}m(R−r)θ¨=−mgsinθ−Fsp​−ffriction​ For small θ\thetaθ, sin⁡θ≈θ\sin\theta \approx \thetasinθ≈θ. The spring force is Fsp=ks=k(R−r)θF_{sp} = ks = k(R-r)\thetaFsp​=ks=k(R−r)θ. m(R−r)θ¨=−mgθ−k(R−r)θ−ffrictionm(R-r)\ddot{\theta} = -mg\theta - k(R-r)\theta - f_{friction}m(R−r)θ¨=−mgθ−k(R−r)θ−ffriction​ (1)

  2. Rotational Motion: Torque about the C.M. is provided by the friction force. τcm=Icmα\tau_{cm} = I_{cm} \alphaτcm​=Icm​α ffriction⋅r=(12mr2)ϕ¨f_{friction} \cdot r = (\frac{1}{2}mr^2) \ddot{\phi}ffriction​⋅r=(21​mr2)ϕ¨​ From the no-slip condition, (R−r)θ˙=rϕ˙(R-r)\dot{\theta} = r\dot{\phi}(R−r)θ˙=rϕ˙​, so (R−r)θ¨=rϕ¨(R-r)\ddot{\theta} = r\ddot{\phi}(R−r)θ¨=rϕ¨​, which gives ϕ¨=(R−r)rθ¨\ddot{\phi} = \frac{(R-r)}{r}\ddot{\theta}ϕ¨​=r(R−r)​θ¨. ffriction⋅r=(12mr2)(R−r)rθ¨  ⟹  ffriction=12m(R−r)θ¨f_{friction} \cdot r = (\frac{1}{2}mr^2) \frac{(R-r)}{r}\ddot{\theta} \implies f_{friction} = \frac{1}{2}m(R-r)\ddot{\theta}ffriction​⋅r=(21​mr2)r(R−r)​θ¨⟹ffriction​=21​m(R−r)θ¨ (2)

  3. Combine Equations: Substitute the expression for ffrictionf_{friction}ffriction​ from (2) into (1): m(R−r)θ¨=−mgθ−k(R−r)θ−12m(R−r)θ¨m(R-r)\ddot{\theta} = -mg\theta - k(R-r)\theta - \frac{1}{2}m(R-r)\ddot{\theta}m(R−r)θ¨=−mgθ−k(R−r)θ−21​m(R−r)θ¨ m(R−r)θ¨+12m(R−r)θ¨=−(mg+k(R−r))θm(R-r)\ddot{\theta} + \frac{1}{2}m(R-r)\ddot{\theta} = -(mg + k(R-r))\thetam(R−r)θ¨+21​m(R−r)θ¨=−(mg+k(R−r))θ 32m(R−r)θ¨=−(mg+k(R−r))θ\frac{3}{2}m(R-r)\ddot{\theta} = -(mg + k(R-r))\theta23​m(R−r)θ¨=−(mg+k(R−r))θ θ¨+mg+k(R−r)32m(R−r)θ=0\ddot{\theta} + \frac{mg + k(R-r)}{\frac{3}{2}m(R-r)} \theta = 0θ¨+23​m(R−r)mg+k(R−r)​θ=0

  4. Calculate Angular Frequency (ω\omegaω): ω2=mg+k(R−r)32m(R−r)=mg32m(R−r)+k(R−r)32m(R−r)\omega^2 = \frac{mg + k(R-r)}{\frac{3}{2}m(R-r)} = \frac{mg}{\frac{3}{2}m(R-r)} + \frac{k(R-r)}{\frac{3}{2}m(R-r)}ω2=23​m(R−r)mg+k(R−r)​=23​m(R−r)mg​+23​m(R−r)k(R−r)​ ω2=2g3(R−r)+2k3m=23(gR−r+km)\omega^2 = \frac{2g}{3(R-r)} + \frac{2k}{3m} = \frac{2}{3} \left( \frac{g}{R-r} + \frac{k}{m} \right)ω2=3(R−r)2g​+3m2k​=32​(R−rg​+mk​) ω=23(gR−r+km)\omega = \sqrt{\frac{2}{3} \left( \frac{g}{R - r} + \frac{k}{m} \right)}ω=32​(R−rg​+mk​)​

Both methods yield the same result. Comparing this with the given options, it matches option A.

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