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Rotational Motion question

2020 · 5 Sep · Shift 2 · Q52
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  5. /2020 · 5 Sep · Shift 2 · Q52

Rotational Motion question

2020 · 5 Sep · Shift 2 · Q52

JEE MainPhysicsRotational MotionMCQ+4 / −1
A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with a time period T1 and, (ii) back and forth in a direction perpendicular to its plane, with a period T2. The ratio T1T2{{{T_1}} \over {{T_2}}}T2​T1​​ will be :
  1. A
    23{{\sqrt 2 } \over 3}32​​
  2. B
    23{2 \over {\sqrt 3 }}3​2​
  3. C
    23{2 \over 3}32​
  4. D
    32{3 \over {\sqrt 2 }}2​3​
View written solutionFree

Correct answer: B

  1. Model the ring as a physical pendulum

A ring of radius RRR is suspended from a point on its rim by a nail. For small oscillations, the time period of a physical pendulum is

T=2πIPMgℓT = 2\pi \sqrt{\frac{I_P}{Mg\ell}}T=2πMgℓIP​​​

where:

  • IPI_PIP​ = moment of inertia about the suspension point,
  • ℓ\ellℓ = distance from suspension point to center of mass.

For the ring, the center of mass is at its center, so

ℓ=R.\ell = R.ℓ=R.

Thus,

T=2πIPMgR.T = 2\pi \sqrt{\frac{I_P}{MgR}}.T=2πMgRIP​​​.

So the ratio T1T2\dfrac{T_1}{T_2}T2​T1​​ depends only on the corresponding moments of inertia about the nail.

  1. Case (i): Oscillation in its own plane

Here the ring swings in its plane, so the axis of rotation is perpendicular to the plane of the ring and passes through the nail.

For a ring, moment of inertia about its center and perpendicular to its plane is

IC(⊥)=MR2.I_C^{(\perp)} = MR^2.IC(⊥)​=MR2.

Using parallel axis theorem for an axis through the nail:

I1=IC(⊥)+MR2=MR2+MR2=2MR2.I_1 = I_C^{(\perp)} + MR^2 = MR^2 + MR^2 = 2MR^2.I1​=IC(⊥)​+MR2=MR2+MR2=2MR2.

Therefore,

T1=2π2MR2MgR=2π2Rg.T_1 = 2\pi \sqrt{\frac{2MR^2}{MgR}} = 2\pi \sqrt{\frac{2R}{g}}.T1​=2πMgR2MR2​​=2πg2R​​.
  1. Case (ii): Oscillation perpendicular to its plane

Now the ring moves back and forth in a direction perpendicular to its plane. So the ring rotates about a diameter through the nail lying in the plane of the ring.

For a ring, moment of inertia about any diameter through its center is

IC(diameter)=12MR2.I_C^{(\text{diameter})} = \frac{1}{2}MR^2.IC(diameter)​=21​MR2.

Again applying parallel axis theorem to the parallel axis through the nail:

I2=12MR2+MR2=32MR2.I_2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2.I2​=21​MR2+MR2=23​MR2.

Hence,

T2=2π32MR2MgR=2π3R2g.T_2 = 2\pi \sqrt{\frac{\frac{3}{2}MR^2}{MgR}} = 2\pi \sqrt{\frac{3R}{2g}}.T2​=2πMgR23​MR2​​=2π2g3R​​.
  1. Find the ratio
T1T2=I1I2=2MR232MR2=43=23.\frac{T_1}{T_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{2MR^2}{\frac{3}{2}MR^2}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}.T2​T1​​=I2​I1​​​=23​MR22MR2​​=34​​=3​2​.
  1. Compare with options
T1T2=23\frac{T_1}{T_2} = \frac{2}{\sqrt{3}}T2​T1​​=3​2​

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

This matches our derived answer.

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