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MCQEasyJEE 2023Uniform Circular Motion

JEE Physics 2023 Question with Solution

A child of mass 5 kg5 \, \text{kg} is going round a merry-go-round that makes 11 rotation in 3.14 s3.14 \, \text{s}. The radius of the merry-go-round is 2 m2 \, \text{m}. The centrifugal force on the child will be :

  • A

    40 N40 \, \text{N}

  • B

    100 N100 \, \text{N}

  • C

    80 N80 \, \text{N}

  • D

    50 N50 \, \text{N}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: m=5 kgm = 5 \, \text{kg}, R=2 mR = 2 \, \text{m}, time for 11 revolution t=3.14 s=π st = 3.14 \, \text{s} = \pi \, \text{s}.

Find: The centrifugal force on the child.

For 11 revolution, the angular displacement is

θ=2π rad\theta = 2\pi \, \text{rad}

So, the angular speed is

ω=θt=2ππ=2 rad/s\omega = \frac{\theta}{t} = \frac{2\pi}{\pi} = 2 \, \text{rad/s}

Now use the centrifugal force formula:

F=mRω2F = mR\omega^2

Substituting the given values,

F=5×2×22=40 NF = 5 \times 2 \times 2^2 = 40 \, \text{N}

Therefore, the centrifugal force on the child is 40 N40 \, \text{N}. The correct option is A.

Common mistakes

  • Using linear speed directly without first finding angular speed. This is wrong because the given data is in time period form. First compute ω=2πT\omega = \frac{2\pi}{T}, then use F=mRω2F = mR\omega^2.

  • Taking 3.14 s3.14 \, \text{s} as a random decimal and not recognizing it as approximately π s\pi \, \text{s}. This can make the simplification harder. Rewrite the period as π s\pi \, \text{s} before calculating ω\omega.

  • Using the wrong radius dependence, such as dividing by RR instead of multiplying by RR. This is wrong because with angular speed, centripetal or centrifugal force is F=mRω2F = mR\omega^2, not mω2R\frac{m\omega^2}{R}.

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