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JEE Advanced 2016Paper 2
Exam
JEE Advanced
Year
2016
Paper
Paper 2
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2016 · Paper 2 · Physics · Laws of Motion

Paragraph: A frame of reference that is accelerated with respect to...

Paragraph: A frame of 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 F⃗rot \vec{F}_{\text{rot }} experienced by a particle of mass mm moving on the rotating disc and the force F⃗in \vec{F}_{\text{in }} experienced by the particle in an inertial frame of reference is F⃗rot=F⃗in+2m(v⃗rot×ω⃗)+m(ω⃗×r⃗)×ω⃗\vec{F}_{\mathrm{rot}}=\vec{F}_{\mathrm{in}}+2 m\left(\vec{v}_{\mathrm{rot}} \times \vec{\omega}\right)+m(\vec{\omega} \times \vec{r}) \times \vec{\omega}, where v⃗rot \vec{v}_{\text{rot }} is the velocity of the particle in the rotating frame of reference and r⃗\vec{r} is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius RR rotating counter-clockwise with a constant angular speed ω\omega about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the xx-axis along the slot, the yy-axis perpendicular to the slot and the zz-axis along the rotation axis (ω⃗=ωk^).(\vec{\omega}=\omega \hat{k}) . A small block of mass mm is gently placed in the slot at r⃗=(R/2)i^\vec{r}=(R / 2) \hat{i} at t=0t=0 and is constrained to move only along the slot. Paragraph: A frame of reference that is accelerated with respect to an inertial frame of reference is called a

Question: The distance rr of the block at time tt is

Options

  1. AR4(e2\omegat+e−2\omegat)\dfrac{R}{4}\left({e}^{2\omegat} + {e}^{- 2\omegat}\right)
  2. BR2cos⁡2\omegat\dfrac{R}{2}\cos 2\omegat
  3. CR2cos\omegat\dfrac{R}{2}cos \omegat
  4. DR4(e\omegat+e−\omegat)\dfrac{R}{4}\left({e}^{\omegat} + {e}^{- \omegat}\right)

Correct answer

D. R4(e\omegat+e−\omegat)\dfrac{R}{4}\left({e}^{\omegat} + {e}^{- \omegat}\right)

Step-by-step solution

Force on block along slot =mω2r=ma=m(vdvdr)= m{\omega}^{2}r = ma = m\left(\dfrac{vdv}{dr}\right)
∫0vvdv=∫R2rω2rdr\mathop{∫}_{0}^{v}vdv = \mathop{∫}_{\dfrac{R}{2}}^{r}{\omega}^{2}rdr
v22=ω22(r2−R24)⇒v=ωr2−R24=drdt\dfrac{{v}^{2}}{2} = \dfrac{{\omega}^{2}}{2}\left({r}^{2} - \dfrac{{R}^{2}}{4}\right) ⇒ v = \omega\sqrt{{r}^{2} - \dfrac{{R}^{2}}{4}} = \dfrac{dr}{dt}
⇒∫R4rdrr2−R24=∫0t\omegadt⇒ \mathop{∫}_{\dfrac{R}{4}}^{r}\dfrac{dr}{\sqrt{{r}^{2} - \dfrac{{R}^{2}}{4}}} = \mathop{∫}_{0}^{t}\omegadt
ln⁡(r+r2−R24R2)−ln⁡(R2+R24−R24R2)=\omegat\ln \left(\dfrac{r + \sqrt{{r}^{2} - \dfrac{{R}^{2}}{4}}}{\dfrac{R}{2}}\right) - \ln \left(\dfrac{\dfrac{R}{2} + \sqrt{\dfrac{{R}^{2}}{4} - \dfrac{{R}^{2}}{4}}}{\dfrac{R}{2}}\right) = \omegat
⇒r+r2−R24=R2e\omegat⇒ r + \sqrt{{r}^{2} - \dfrac{{R}^{2}}{4}} = \dfrac{R}{2}e ^ \omegat
⇒r2−R24=R24e2\omegat+r2−2rR2e\omegat⇒ {r}^{2} - \dfrac{{R}^{2}}{4} = \dfrac{{R}^{2}}{4}{e}^{2\omegat} + {r}^{2} - 2r\dfrac{R}{2}{e}^{\omegat}
⇒r=R24e2\omegat+R24Re\omegat=R4⇒ r = \dfrac{\dfrac{{R}^{2}}{4}{e}^{2\omegat} + \dfrac{{R}^{2}}{4}}{R{e}^{\omegat}} = \dfrac{R}{4}\left({e}^{\omegat} + {e}^{- \omegat}\right)
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