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Rotational Motion question

2023 · Shift 1 · Q47
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Rotational Motion question

2023 · Shift 1 · Q47

JEE AdvancedPhysicsRotational MotionNumerical+4 / −1
Two point-like objects of masses 20 gm20 ~\mathrm{gm}20 gm and 30 gm30 ~\mathrm{gm}30 gm are fixed at the two ends of a rigid massless rod of length 10 cm10 \mathrm{~cm}10 cm. This system is suspended vertically from a rigid ceiling using a thin wire attached to its center of mass, as shown in the figure. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is 1.2×10−8 N m rad−11.2 \times 10^{-8} \mathrm{~N} \mathrm{~m} ~\mathrm{rad}^{-1}1.2×10−8 N m rad−1. The angular frequency of the oscillations in n×10−3 rad s−1n \times 10^{-3} ~\mathrm{rad} ~\mathrm{s}^{-1}n×10−3 rad s−1. The value of nnn is ‾\underline{\hspace{2cm}}​ . JEE Advanced 2023 Paper 1 Online Physics - Rotational Motion Question 7 English
Numerical answer
View written solutionFree

Correct answer: 10

  1. For a torsional pendulum

The angular frequency is

ω=CI\omega = \sqrt{\frac{C}{I}}ω=IC​​

where:

  • C=1.2×10−8 N m rad−1C = 1.2 \times 10^{-8}\, \text{N m rad}^{-1}C=1.2×10−8N m rad−1 is the torsional constant,
  • III is the moment of inertia of the suspended body about the vertical axis through the point of suspension (here, through the center of mass).
  1. Convert given data to SI units

m1=20 g=0.02 kg,m2=30 g=0.03 kgm_1 = 20\,\text{g} = 0.02\,\text{kg}, \qquad m_2 = 30\,\text{g} = 0.03\,\text{kg}m1​=20g=0.02kg,m2​=30g=0.03kg

Rod length:

L=10 cm=0.10 mL = 10\,\text{cm} = 0.10\,\text{m}L=10cm=0.10m

Total mass:

M=m1+m2=0.05 kgM = m_1 + m_2 = 0.05\,\text{kg}M=m1​+m2​=0.05kg

  1. Find distances of the masses from the center of mass

Let the distances from the center of mass be r1r_1r1​ for m1m_1m1​ and r2r_2r2​ for m2m_2m2​. Then,

r1+r2=0.10r_1 + r_2 = 0.10r1​+r2​=0.10

and from center of mass condition,

m1r1=m2r2m_1 r_1 = m_2 r_2m1​r1​=m2​r2​

So,

0.02r1=0.03r20.02 r_1 = 0.03 r_20.02r1​=0.03r2​

r1=32r2r_1 = \frac{3}{2}r_2r1​=23​r2​

Using in r1+r2=0.10r_1+r_2=0.10r1​+r2​=0.10,

32r2+r2=0.10\frac{3}{2}r_2 + r_2 = 0.1023​r2​+r2​=0.10

52r2=0.10\frac{5}{2}r_2 = 0.1025​r2​=0.10

r2=0.04 m,r1=0.06 mr_2 = 0.04\,\text{m}, \qquad r_1 = 0.06\,\text{m}r2​=0.04m,r1​=0.06m

  1. Moment of inertia about the center of mass
I=m1r12+m2r22I = m_1 r_1^2 + m_2 r_2^2I=m1​r12​+m2​r22​ I=0.02(0.06)2+0.03(0.04)2I = 0.02(0.06)^2 + 0.03(0.04)^2I=0.02(0.06)2+0.03(0.04)2 I=0.02(0.0036)+0.03(0.0016)I = 0.02(0.0036) + 0.03(0.0016)I=0.02(0.0036)+0.03(0.0016) I=7.2×10−5+4.8×10−5I = 7.2 \times 10^{-5} + 4.8 \times 10^{-5}I=7.2×10−5+4.8×10−5 I=1.2×10−4 kg m2I = 1.2 \times 10^{-4}\,\text{kg m}^2I=1.2×10−4kg m2
  1. Compute angular frequency
ω=1.2×10−81.2×10−4=10−4\omega = \sqrt{\frac{1.2\times 10^{-8}}{1.2\times 10^{-4}}} = \sqrt{10^{-4}}ω=1.2×10−41.2×10−8​​=10−4​ ω=10−2 rad s−1\omega = 10^{-2}\,\text{rad s}^{-1}ω=10−2rad s−1

Given that

ω=n×10−3 rad s−1\omega = n \times 10^{-3}\,\text{rad s}^{-1}ω=n×10−3rad s−1

So,

10−2=10×10−310^{-2} = 10 \times 10^{-3}10−2=10×10−3

Hence,

n=10n = 10n=10

  1. Comparison with stored answer

Derived answer: 101010

Stored correct answer: 101010

They match.

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