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Rotational Motion question

2023 · Shift 2 · Q42
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  5. /2023 · Shift 2 · Q42

Rotational Motion question

2023 · Shift 2 · Q42

JEE AdvancedPhysicsRotational MotionNumerical+4 / −1
A thin circular coin of mass 5gm5 \mathrm{gm}5gm and radius 4/3 cm4 / 3 \mathrm{~cm}4/3 cm is initially in a horizontal xyx yxy-plane. The coin is tossed vertically up ( +z+z+z direction) by applying an impulse of π2×10−2 N\sqrt{\frac{\pi}{2}} \times 10^{-2} \mathrm{~N}2π​​×10−2 N-s at a distance 2/3 cm2 / 3 \mathrm{~cm}2/3 cm from its center. The coin spins about its diameter and moves along the +z+z+z direction. By the time the coin reaches back to its initial position, it completes nnn rotations. The value of nnn is ‾\underline{\hspace{2cm}}​. [Given: The acceleration due to gravity g=10 m s−2g=10 \mathrm{~m} \mathrm{~s}^{-2}g=10 m s−2 ] JEE Advanced 2023 Paper 2 Online Physics - Rotational Motion Question 8 English
Numerical answer
View written solutionFree

Correct answer: 30

  1. Given data
  • Mass of coin: m=5 gm=5×10−3 kgm = 5\text{ gm} = 5\times 10^{-3}\,\text{kg}m=5 gm=5×10−3kg
  • Radius: R=43 cm=43×10−2 mR = \dfrac{4}{3}\text{ cm} = \dfrac{4}{3}\times 10^{-2}\,\text{m}R=34​ cm=34​×10−2m
  • Impulse applied: J=π2×10−2 N sJ = \sqrt{\frac{\pi}{2}}\times 10^{-2}\,\text{N s}J=2π​​×10−2N s
  • Distance of line of action from center: d=23 cm=23×10−2 md = \frac{2}{3}\text{ cm} = \frac{2}{3}\times 10^{-2}\,\text{m}d=32​ cm=32​×10−2m
  • g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2

The impulse gives:

  • translational momentum to the center of mass,
  • angular momentum about the center due to offset ddd.
  1. Translational motion of the coin

Impulse equals change in linear momentum: J=mvJ = m vJ=mv So, v=Jm=π/2×10−25×10−3v = \frac{J}{m} = \frac{\sqrt{\pi/2}\times 10^{-2}}{5\times 10^{-3}}v=mJ​=5×10−3π/2​×10−2​ v=2π2 m/sv = 2\sqrt{\frac{\pi}{2}}\,\text{m/s}v=22π​​m/s

Time of flight (up and back to same level): T=2vgT = \frac{2v}{g}T=g2v​ T=2⋅2π/210T = \frac{2\cdot 2\sqrt{\pi/2}}{10}T=102⋅2π/2​​ T=25π2 sT = \frac{2}{5}\sqrt{\frac{\pi}{2}}\,\text{s}T=52​2π​​s

  1. Angular speed produced by the impulse

Angular impulse about the center: JdJdJd This equals change in angular momentum: Iω=JdI\omega = JdIω=Jd

Since the coin spins about a diameter, moment of inertia of a thin disc about a diameter is I=14mR2I = \frac{1}{4}mR^2I=41​mR2

Hence, ω=JdI=Jd(1/4)mR2=4JdmR2\omega = \frac{Jd}{I} = \frac{Jd}{(1/4)mR^2} = \frac{4Jd}{mR^2}ω=IJd​=(1/4)mR2Jd​=mR24Jd​

Now substitute: ω=4(π/2×10−2)(23×10−2)(5×10−3)(43×10−2)2\omega = \frac{4\left(\sqrt{\pi/2}\times 10^{-2}\right)\left(\frac{2}{3}\times 10^{-2}\right)}{(5\times 10^{-3})\left(\frac{4}{3}\times 10^{-2}\right)^2}ω=(5×10−3)(34​×10−2)24(π/2​×10−2)(32​×10−2)​

First simplify the geometric/mass factor: R2=(43×10−2)2=169×10−4R^2 = \left(\frac{4}{3}\times 10^{-2}\right)^2 = \frac{16}{9}\times 10^{-4}R2=(34​×10−2)2=916​×10−4

So, (5×10−3)R2=5×10−3⋅169×10−4=809×10−7(5\times 10^{-3})R^2 = 5\times 10^{-3}\cdot \frac{16}{9}\times 10^{-4} = \frac{80}{9}\times 10^{-7}(5×10−3)R2=5×10−3⋅916​×10−4=980​×10−7

Numerator: 4⋅π2×10−2⋅23×10−2=83π2×10−44\cdot \sqrt{\frac{\pi}{2}}\times 10^{-2}\cdot \frac{2}{3}\times 10^{-2} = \frac{8}{3}\sqrt{\frac{\pi}{2}}\times 10^{-4}4⋅2π​​×10−2⋅32​×10−2=38​2π​​×10−4

Thus, ω=83π/2×10−4809×10−7\omega = \frac{\frac{8}{3}\sqrt{\pi/2}\times 10^{-4}}{\frac{80}{9}\times 10^{-7}}ω=980​×10−738​π/2​×10−4​ ω=300π2 rad/s\omega = 300\sqrt{\frac{\pi}{2}}\,\text{rad/s}ω=3002π​​rad/s

  1. Number of rotations during the flight

Angle rotated during time TTT: θ=ωT\theta = \omega Tθ=ωT

So, θ=300π2⋅25π2\theta = 300\sqrt{\frac{\pi}{2}}\cdot \frac{2}{5}\sqrt{\frac{\pi}{2}}θ=3002π​​⋅52​2π​​ θ=300⋅25⋅π2\theta = 300\cdot \frac{2}{5}\cdot \frac{\pi}{2}θ=300⋅52​⋅2π​ θ=60π rad\theta = 60\pi\,\text{rad}θ=60πrad

Number of complete rotations: n=θ2π=60π2π=30n = \frac{\theta}{2\pi} = \frac{60\pi}{2\pi} = 30n=2πθ​=2π60π​=30

  1. Final answer

30\boxed{30}30​

The derived answer matches the stored correct answer.

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