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Circular Motion question

2023 · 6 Apr · Shift 2 · Q61
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  5. /2023 · 6 Apr · Shift 2 · Q61

Circular Motion question

2023 · 6 Apr · Shift 2 · Q61

JEE MainPhysicsCircular MotionMCQ+4 / −1
A child of mass 5 kg5 \mathrm{~kg}5 kg is going round a merry-go-round that makes 1 rotation in 3.14 s3.14 \mathrm{~s}3.14 s. The radius of the merry-go-round is 2 m2 \mathrm{~m}2 m. The centrifugal force on the child will be
  1. A
    50 N
  2. B
    80 N
  3. C
    100 N
  4. D
    40 N
View written solutionFree

Correct answer: D

  1. Given data

    • Mass of child: m=5 kgm = 5\,\text{kg}m=5kg
    • Time period of rotation: T=3.14 sT = 3.14\,\text{s}T=3.14s
    • Radius: r=2 mr = 2\,\text{m}r=2m
  2. Formula for centrifugal force The magnitude of centrifugal force (same as centripetal force in magnitude, in the rotating frame) is F=mω2rF = m\omega^2 rF=mω2r where angular speed ω=2πT\omega = \frac{2\pi}{T}ω=T2π​

  3. Calculate angular speed ω=2π3.14\omega = \frac{2\pi}{3.14}ω=3.142π​ Since π≈3.14\pi \approx 3.14π≈3.14, ω=2×3.143.14=2 rad/s\omega = \frac{2\times 3.14}{3.14} = 2\,\text{rad/s}ω=3.142×3.14​=2rad/s

  4. Calculate force F=mω2r=5×(2)2×2F = m\omega^2 r = 5 \times (2)^2 \times 2F=mω2r=5×(2)2×2 F=5×4×2=40 NF = 5 \times 4 \times 2 = 40\,\text{N}F=5×4×2=40N

  5. Match with options

    • A: 50 N50\,\text{N}50N
    • B: 80 N80\,\text{N}80N
    • C: 100 N100\,\text{N}100N
    • D: 40 N40\,\text{N}40N

    Therefore, the correct option is D.

  6. Comparison with stored answer Stored correct answer: D

    My derived answer also gives D.

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