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MCQEasyJEE 2023Circular Motion Dynamics

JEE Physics 2023 Question with Solution

A coin placed on a rotating table just slips when it is placed at a distance of 1 cm1 \, \text{cm} from the center. If the angular velocity of the table is halved, it will just slip when placed at a distance of _____ from the centre:

  • A

    6 cm6 \, \text{cm}

  • B

    4 cm4 \, \text{cm}

  • C

    2 cm2 \, \text{cm}

  • D

    1 cm1 \, \text{cm}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The coin just slips at r1=1 cmr_1 = 1 \, \text{cm} for angular velocity ω1\omega_1. The angular velocity is then halved, so ω2=ω12\omega_2 = \frac{\omega_1}{2}.

Find: The new slipping distance r2r_2 from the centre.

At the point of just slipping, friction provides the required centripetal force:

fr=mω2rf_r = m\omega^2 r

For the same coin and same surface, the limiting friction remains the same at the slipping condition. Therefore,

mω2r=constantm\omega^2 r = \text{constant}

So,

ω2r=constant\omega^2 r = \text{constant}

Hence,

ω12r1=ω22r2\omega_1^2 r_1 = \omega_2^2 r_2

Substituting ω2=ω12\omega_2 = \frac{\omega_1}{2} and r1=1 cmr_1 = 1 \, \text{cm},

ω12(1)=(ω12)2r2\omega_1^2 (1) = \left(\frac{\omega_1}{2}\right)^2 r_2 ω12=ω124r2\omega_1^2 = \frac{\omega_1^2}{4} r_2 r2=4 cmr_2 = 4 \, \text{cm}

Therefore, the coin will just slip at 4 cm4 \, \text{cm} from the centre. The correct option is B.

Common mistakes

  • Using mω2r2=constantm\omega^2 r^2 = \text{constant} is incorrect because centripetal force depends linearly on rr, not on r2r^2. Use mω2r=constantm\omega^2 r = \text{constant} at the limiting condition.

  • Assuming the slipping distance becomes half when angular velocity is halved is wrong. Since r∝1ω2r \propto \frac{1}{\omega^2}, halving ω\omega makes the distance four times, not two times.

  • Forgetting that limiting friction is the same for the same coin and table leads to the wrong relation. At just slipping, equate the required centripetal force to the maximum static friction.

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