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JEE Advanced 2023 · Laws of Motion

A thin circular coin of mass 5 gm5\,\text{gm} and radius 4/3 cm4/3\,\text{cm} is initially in a horizontal xyxy-plane. The coin is tossed vertically upward (in +z+z direction) by applying an impulse of π2×10−2 N-s\sqrt{\frac{\pi}{2}}\times 10^{-2}\,\text{N-s} at a distance 2/3 cm2/3\,\text{cm} from its center. The coin spins about its diameter and moves along the +z+z direction. By the time the coin reaches back to its initial position, it completes nn rotations. The value of nn is ____. [Given: The acceleration due to gravity g=10 m s−2g = 10\,\text{m s}^{-2}]

Answer: 3030

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Written solution

Step 1: Relate linear impulse to initial vertical velocity.

The linear impulse JJ determines the initial vertical velocity vv.

J=mvJ = mv v=Jmv = \frac{J}{m}

Step 2: Relate angular impulse to angular velocity.

The angular impulse about the diameter is equal to the change in angular momentum.

Jd=IωJd = I\omega ω=JdI\omega = \frac{Jd}{I}

Step 3: Determine the moment of inertia.

The moment of inertia of a thin circular coin of mass mm and radius RR about its diameter is used to find the angular velocity.

I=14mR2I = \frac{1}{4}mR^2 ω=4JdmR2\omega = \frac{4Jd}{mR^2}

Step 4: Calculate the time of flight.

The time tt for the coin to return to its initial position is given by the vertical motion under gravity.

t=2vgt = \frac{2v}{g} t=2Jmgt = \frac{2J}{mg}

Step 5: Express the number of rotations.

The number of rotations nn is the total angle rotated divided by 2π2\pi.

n=ωt2πn = \frac{\omega t}{2\pi} n=4J2dπm2R2gn = \frac{4J^2 d}{\pi m^2 R^2 g}

Step 6: Substitute the given values to calculate nn.

Substitute m=5×10−3 kgm = 5 \times 10^{-3}\text{ kg}, R=43×10−2 mR = \frac{4}{3} \times 10^{-2}\text{ m}, J=π2×10−2 N⋅sJ = \sqrt{\frac{\pi}{2}} \times 10^{-2}\text{ N}\cdot\text{s}, d=23×10−2 md = \frac{2}{3} \times 10^{-2}\text{ m}, and g=10 m/s2g = 10\text{ m/s}^2.

n=4(π2×10−2)2(23×10−2)π(5×10−3)2(43×10−2)2(10)n = \frac{4 \left(\sqrt{\frac{\pi}{2}} \times 10^{-2}\right)^2 \left(\frac{2}{3} \times 10^{-2}\right)}{\pi \left(5 \times 10^{-3}\right)^2 \left(\frac{4}{3} \times 10^{-2}\right)^2 (10)} n=4(π2×10−4)(23×10−2)π(25×10−6)(169×10−4)(10)n = \frac{4 \left(\frac{\pi}{2} \times 10^{-4}\right) \left(\frac{2}{3} \times 10^{-2}\right)}{\pi \left(25 \times 10^{-6}\right) \left(\frac{16}{9} \times 10^{-4}\right) (10)} n=30n = 30

Answer: 30

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