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Indian Navy SSR (Sailor) System of Particles and Rotational Motion — practice questions

Explore 4 free System of Particles and Rotational Motion questions in Indian Navy SSR (Sailor) Physics. Open each question or exercise to read its content, reference answers and available explanation.

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System of Particles and Rotational Motion questions and exercises with answers

Showing questions 1–4 of 4.

Parallel axis theorem states I_parallel = I_cm + Md². Here d is:Read the answer and worked solution →Two point masses 3 kg each at the ends of a massless rod of length 4 m. I about axis through center perpendicular to rod:Read the answer and worked solution →For a thin rod (mass M, length L) about axis through ONE END perpendicular to rod, I equals:Read the answer and worked solution →A wheel rotating at 50 rad/s decelerates uniformly to rest in 10 s. Total angular displacement:Read the answer and worked solution →

Worked practice questions

Parallel axis theorem states I_parallel = I_cm + Md². Here d is:

  1. Time period
  2. The radius
  3. Perpendicular distance between two parallel axes
  4. Total mass
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Correct answer: C. Perpendicular distance between two parallel axes

Parallel axis theorem: I about any axis = I about parallel axis through center of mass + M × d², where d is the perpendicular distance between the axes.

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Two point masses 3 kg each at the ends of a massless rod of length 4 m. I about axis through center perpendicular to rod:

  1. 48
  2. 36
  3. 24
  4. 12
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Correct answer: C. 24

Each mass at distance 2 m from center. We have I = sum m × r². That is I = 3 × 2² + 3 × 2². That is I = 12 + 12. So I = 24 kg m².

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For a thin rod (mass M, length L) about axis through ONE END perpendicular to rod, I equals:

  1. ML²/12
  2. ML²
  3. ML²/2
  4. ML²/3
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Correct answer: D. ML²/3

By parallel-axis theorem: I_end = I_cm + M(L/2)² = ML²/12 + ML²/4 = ML²/12 + 3ML²/12 = 4ML²/12 = ML²/3.

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