Verified Solution • 2025–26 Session moderate

QJEEAdv16Ph27eb: JEE Advanced 2016 Physics - Class 11 JEEAdv16Ph27eb Solution: A frame of the reference that is accelerated with respect to an …

A frame of the reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω\omega is an example of a non-inertial frame of reference. The relationship between the force F→\overrightarrow F rot experienced by a particle of mass m moving on the rotating disc and the force F→\overrightarrow F in experienced by the particle in an inertial frame of reference is,

F→\overrightarrow F rot = F→\overrightarrow F in + 2m ( v→\overrightarrow v rot ×\times ω→\overrightarrow \omega ) + m ( ω→\overrightarrow \omega ×\times r→\overrightarrow r ) ×\times ω→\overrightarrow \omega ,

where, vrot is the velocity of the particle in the rotating frame of reference and r is the position vector of the particle with respect to the centre of the disc.

JEE Advanced 2016 Paper 2 Offline Physics - Rotational Motion Question 39 English
Now, consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed ω\omega about its vertical axis through its centre. We assign a coordinate system with the origin at the centre of the disc, the X-axis along the slot, the Y-axis perpendicular to the slot and the Z-axis along the rotation axis ( ω\omega = ω\omega k^\widehat k ). A small block of mass m is gently placed in the slot at r = (R/2) i^\widehat i at t = 0 and is constrained to move only along the slot.

The distance r of the block at time t is

Multiple Choice Options

  • A R2cos⁡2ωt{R \over 2}\cos 2\omega t
  • B R2cos⁡ωt{R \over 2}\cos \omega t
  • C R2(eωt+e−ωt){R \over 2}({e^{\omega t}} + {e^{ - \omega t}})
  • D R2(e2ωt+e−2ωt){R \over 2}({e^{2\omega t}} + {e^{ - 2\omega t}})

Step-by-Step Verified Solution

The given particle of mass m experiences a centrifugal force:

F = ma = mr ω\omega 2

⇒\Rightarrow a = r ω\omega 2

⇒\Rightarrow dvdt=rω2{{dv} \over {dt}} = r{\omega ^2}

or, (dvdr)(drdt)=rω2\left( {{{dv} \over {dr}}} \right)\left( {{{dr} \over {dt}}} \right) = r{\omega ^2} ( ∵\because drdt=v{{{dr} \over {dt}}}=v )

⇒vdv=rω2dr\Rightarrow vdv = r{\omega ^2}dr

∫0vvdv=ω2∫R/2rrdr\int\limits_0^v {vdv = {\omega ^2}\int\limits_{R/2}^r {rdr} } ( ∵\because at t = 0, r=R2r = {R \over 2} )

⇒v22=ω2(r22)R/2r\Rightarrow {{{v^2}} \over 2} = {\omega ^2}\left( {{{{r^2}} \over 2}} \right)_{R/2}^r

⇒v=ωr2−R24\Rightarrow v = \omega \sqrt {{r^2} - {{{R^2}} \over 4}}

JEE Advanced 2016 Paper 2 Offline Physics - Rotational Motion Question 39 English Explanation

Now, v=drdtv = {{dr} \over {dt}}

Therefore, drdt=ωr2−R24{{dr} \over {dt}} = \omega \sqrt {{r^2} - {{{R^2}} \over 4}}

or, ∫R/2rdrr2−R24=ω∫0tdt\int\limits_{R/2}^r {{{dr} \over {\sqrt {{r^2} - {{{R^2}} \over 4}} }} = \omega \int\limits_0^t {dt} } ...... (1)

To solve this integral, we substitute r=R2sec⁡θr = {R \over 2}\sec \theta ; therefore,

dr=R2sec⁡θtan⁡θdθdr = {R \over 2}\sec \theta \tan \theta d\theta

⇒∫π2sec⁡−1(2r/R)(r/2)sec⁡θtan⁡θdθR24tan⁡2θ=ωt\Rightarrow \int\limits_{{\pi \over 2}}^{{{\sec }^{ - 1}}(2r/R)} {{{(r/2)\sec \theta \tan \theta d\theta } \over {\sqrt {{{{R^2}} \over 4}{{\tan }^2}\theta } }} = \omega t}

⇒ωt=∫π/2sec⁡−1(2r/R)sec⁡θdθ\Rightarrow \omega t = \int\limits_{\pi /2}^{{{\sec }^{ - 1}}(2r/R)} {\sec \theta d\theta }

=[ln⁡∣sec⁡θ+tan⁡θ∣]sec⁡−1(2r/R)= {\left[ {\ln \left| {\sec \theta + \tan \theta } \right|} \right]^{{{\sec }^{ - 1}}(2r/R)}}

=[ln⁡∣sec⁡θ+sec⁡2θ−1∣]π/2sec⁡−1(2r/R)= \left[ {\ln \left| {\sec \theta + \sqrt {{{\sec }^2}\theta - 1} } \right|} \right]_{\pi /2}^{{{\sec }^{ - 1}}(2r/R)}

⇒ωt=ln⁡[2rR+4r2−R2R]\Rightarrow \omega t = \ln \left[ {{{2r} \over R} + {{\sqrt {4{r^2} - {R^2}} } \over R}} \right]

Let 2rR=x{{2r} \over R} = x . Therefore,

eω=x+x2−1{e^\omega } = x + \sqrt {{x^2} - 1}

⇒(eωt−x)2=x2−1\Rightarrow ({e^{\omega t}} - x)^2 = {x^2} - 1

⇒e2ωt−2xeωt+x2=x2−1\Rightarrow {e^{2\omega t}} - 2x{e^{\omega t}} + x^2 = x^2 - 1

⇒e2ωt−2xeωt+1=0\Rightarrow {e^{2\omega t}} - 2x{e^{\omega t}} + 1 = 0

⇒x=e2ωt+12eωt=eωt+e−ωt2\Rightarrow x = {{{e^{2\omega t}} + 1} \over {2{e^{\omega t}}}} = {{{e^{\omega t}} + {e^{ - \omega t}}} \over 2}

On replacing x by 2rR{{2r} \over R} , we get

2rR=eωt+e−ωt2{{2r} \over R} = {{{e^{\omega t}} + {e^{ - \omega t}}} \over 2}

or, r=R4[eωt+e−ωt]r = {R \over 4}[{e^{\omega t}} + {e^{ - \omega t}}]

Hence, option (c) is correct.

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