Verified Solution • 2025–26 Session moderate

QJEEAdv16Ph6dc5: JEE Advanced 2016 Physics - Class 11 JEEAdv16Ph6dc5 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 40 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 net reaction of the disc on the block is

Multiple Choice Options

  • A mω2Rsin⁡ωtj^−mgk^m{\omega ^2}R\sin \omega t\widehat j - mg\widehat k
  • B 12mω2R(eωt−e−ωt)j^+mgk^{1 \over 2}m{\omega ^2}R({e^{\omega t}} - {e^{ - \omega t}})\widehat j + mg\widehat k
  • C 12mω2R(e2ωt−e−2ωt)j^+mgk^{1 \over 2}m{\omega ^2}R({e^{2\omega t}} - {e^{ - 2\omega t}})\widehat j + mg\widehat k
  • D −mω2Rcos⁡ωrj^−mgk^- m{\omega ^2}R\cos \omega r\widehat j - mg\widehat k

Step-by-Step Verified Solution

The rotational quantity of velocity given by

vrot=drdt=R4[ωeωt−ωe−ωt]{v_{rot}} = {{dr} \over {dt}} = {R \over 4}[\omega {e^{\omega t}} - \omega {e^{ - \omega t}}]

vrot=ωR4[eωt−e−ωt]{v_{rot}} = {{\omega R} \over 4}[{e^{\omega t}} - {e^{ - \omega t}}] ...... (1)

The force experienced by the block in the inertial frame is

F→in=−mω2ri^−mgk^{\overrightarrow F _{in}} = - m{\omega ^2}r\widehat i - mg\widehat k ..... (2)

Therefore, the force experienced by the block in the rotating frame is

F→rot=F→in+2m(v→rot×ωk^)+m(ωk^×ri^)×ωk^{\overrightarrow F _{rot}} = {\overrightarrow F _{in}} + 2m({\overrightarrow v _{rot}} \times \omega \widehat k) + m(\omega \widehat k \times r\widehat i) \times \omega \widehat k

F→rot=(−mω2ri^−mgk^)+2mωRr[eωt−e−ωt]i^×ωk^+mω2ri^{\overrightarrow F _{rot}} = ( - m{\omega ^2}r\widehat i - mg\widehat k) + 2m{{\omega R} \over r}[{e^{\omega t}} - {e^{ - \omega t}}]\widehat i \times \omega \widehat k + m{\omega ^2}r\widehat i

=−mω2R2[eωt−e−ωt]j^−mgk^= - {{m{\omega ^2}R} \over 2}[{e^{\omega t}} - {e^{ - \omega t}}]\widehat j - mg\widehat k

Therefore, the net reaction of the disc on the block is

F→reaction−F→rot=mω2R2[eωt−eωt]j^+mgk^{\overrightarrow F _{reaction}} - {\overrightarrow F _{rot}} = {{m{\omega ^2}R} \over 2}[{e^{\omega t}} - {e^{\omega t}}]\widehat j + mg\widehat k

Hence, option (B) is correct.

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