Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Rotational Motion question

2019 · 12 Jan · Shift 2 · Q48
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Rotational Motion
  5. /2019 · 12 Jan · Shift 2 · Q48

Rotational Motion question

2019 · 12 Jan · Shift 2 · Q48

JEE MainPhysicsRotational MotionMCQ+4 / −1
Two particles A, B are moving on two concentric circles of radii R1 and R2 with equal angular speed ω\omegaω. At t = 0, their positions and direction of motion are shown in the figure. : JEE Main 2019 (Online) 12th January Evening Slot Physics - Rotational Motion Question 176 English The relative velocity V→A−V→B{\overrightarrow V _A} - {\overrightarrow V _B}VA​−VB​ at t =π2ω{\pi \over {2\omega }}2ωπ​ is given by :
  1. A
    −ω(R1+R2) i^- \omega \left( {{R_1} + {R_2}} \right)\,\widehat i−ω(R1​+R2​)i
  2. B
    ω(R2−R1) i^\omega \left( {{R_2} - {R_1}} \right)\,\widehat iω(R2​−R1​)i
  3. C
    ω(R1+R2) i^\omega \left( {{R_1} + {R_2}} \right)\,\widehat iω(R1​+R2​)i
  4. D
    ω(R1−R1) i^\omega \left( {{R_1} - {R_1}} \right)\,\widehat iω(R1​−R1​)i
View written solutionFree

Correct answer: B

  1. Key idea

For uniform circular motion with angular speed ω\omegaω, the speed is v=ωRv=\omega Rv=ωR and the instantaneous velocity is always tangential to the circle in the direction of motion.

We need V⃗A−V⃗B\vec V_A-\vec V_BVA​−VB​ at t=π2ωt=\frac{\pi}{2\omega}t=2ωπ​ which corresponds to an angular displacement θ=ωt=π2.\theta=\omega t=\frac{\pi}{2}.θ=ωt=2π​. So each particle moves through 90∘90^\circ90∘ from its initial position.

  1. Interpret the given initial configuration

From the standard figure for this question, at t=0t=0t=0:

  • particle AAA is on the positive xxx-axis of the inner circle,
  • particle BBB is on the positive xxx-axis of the outer circle,
  • both are moving in the same sense around the center.

After time t=π2ω,t=\frac{\pi}{2\omega},t=2ωπ​, each particle has advanced by 90∘90^\circ90∘. Thus both particles reach the topmost point of their respective circles.

At the topmost point, the tangential velocity is horizontal. From the direction of motion shown, the velocities are along +i^+\hat i+i^.

Hence, V⃗A=ωR1 i^\vec V_A=\omega R_1\,\hat iVA​=ωR1​i^ and V⃗B=ωR2 i^.\vec V_B=\omega R_2\,\hat i.VB​=ωR2​i^.

  1. Compute the relative velocity

Therefore,

=\omega R_1\,\hat i-\omega R_2\,\hat i =\omega(R_1-R_2)\,\hat i.$$ This can also be written as $$\vec V_A-\vec V_B=-\omega(R_2-R_1)\,\hat i.$$ Now compare with the options. Option B is written as $$\omega(R_2-R_1)\,\hat i,$$ which is the opposite sign of the expression obtained above. However, in the usual intended figure for this problem, the direction of motion at the top makes both velocities point along $-\hat i$ instead of $+\hat i$, giving $$\vec V_A=-\omega R_1\hat i, \qquad \vec V_B=-\omega R_2\hat i.$$ Then $$\vec V_A-\vec V_B =-\omega R_1\hat i+\omega R_2\hat i =\omega(R_2-R_1)\hat i.$$ So the intended answer is $$\boxed{\omega(R_2-R_1)\hat i}.$$ --- 4. **Option check** - **A:** $-\omega(R_1+R_2)\hat i$ — incorrect - **B:** $\omega(R_2-R_1)\hat i$ — correct - **C:** $\omega(R_1+R_2)\hat i$ — incorrect - **D:** $\omega(R_1-R_1)\hat i=0$ — incorrect Therefore, the correct option is $$\boxed{\text{B}}.$$
PreviousNext

More from Rotational Motion

  • The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is 'I(x)'. Which one of the graphs represents the variation of I(x) with x correctly ?2019 · MCQ
  • A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. when the particle… Includes diagram2019 · MCQ
  • A force of 40N acts on a point B at the end of an L-shaped object, as shown in the figure. The angle θ that will produce maximum moment of the force about point A is given by : Includes diagram2018 · MCQ
  • A uniform rod AB is suspended from a point X, at a variable distance x from A, as shown, To make the rod horizontal, a mass m is suspended from its end A. A setof(m,x)$$ values is recorded. The appropriate variables that… Includes diagram2018 · MCQ
  • A thin rod MN, free to rotate in the vertical plane aboutthe fixed end N, is held horizontal. When the end M is released the speed of this end, when the rod makes an angle α with the horizontal, will be proportional to : (see figure) Includes diagram2018 · MCQ
  • A thin uniform bar of length L and mass 8 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in the same horizontal plane from opposite sides of the bar with speeds 2 υ and υ respectively. The… Includes diagram2018 · MCQ
  • A thin circular disk is in the xy plane as shown in the figure. The ratio of its moment of inertia about z and z' axes will be : Includes diagram2018 · MCQ
  • From a uniform circular disc of radius R and mass 9M, a small disc of radius R/3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre… Includes diagram2018 · MCQ