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

2012 · Shift 2 · Q53
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 Advanced
  3. /Physics
  4. /Rotational Motion
  5. /2012 · Shift 2 · Q53

Rotational Motion question

2012 · Shift 2 · Q53

JEE AdvancedPhysicsRotational MotionMCQ+3 / −1
The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed ω\omegaω, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as it seen from the changed orientation of points P and Q). Both these motions have the same angular speed ω\omegaω in this case. IIT-JEE 2012 Paper 2 Offline Physics - Rotational Motion Question 31 English Comprehension 1 Now consider two similar systems as shown in the figure. Case (a) : The disc with its face vertical and parallel to x-z axis; Case (b) : The disc with its face making an angle of 45 ∘^\circ∘ with xy-plane and its horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed ω\omegaω about the z-axis. IIT-JEE 2012 Paper 2 Offline Physics - Rotational Motion Question 31 English Comprehension 2Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct?
  1. A
    It is vertical for both Cases (a) and (b).
  2. B
    It is vertical for Case (a); and is at 45 ∘^\circ∘ to the xz-plane and lies in the plane of the disc for Case (b).
  3. C
    It is horizontal for Case (a); and is 45 ∘^\circ∘ to the xz-plane and is normal to the plane of the disc for Case (b).
  4. D
    It is vertical for Case (a); and is 45 ∘^\circ∘ to the xz-plane and is normal to the plane of the disc for Case (b).
View written solutionFree

Correct answer: A

  1. Key idea: decomposition of rigid body motion

At any instant, the motion of the disc can be written as:

  • translation of its centre of mass (CM), and
  • rotation about an instantaneous axis through the CM.

If the body is being carried around the zzz-axis with angular velocity ω⃗=ω k^\vec\omega = \omega \,\hat{k}ω=ωk^ and its orientation relative to the lab frame remains fixed while rotating with the stick, then every line fixed in the body also rotates with angular velocity ω⃗=ω k^\vec\omega = \omega \,\hat{k}ω=ωk^.

So the instantaneous angular velocity of the disc is simply

ω⃗=ω k^.\vec\omega = \omega \,\hat{k}.ω=ωk^.

Hence the instantaneous axis through the CM must be parallel to k^\hat{k}k^, i.e. vertical.

  1. Why this works in general

For any point of the disc,

v⃗=v⃗CM+ω⃗inst×r⃗′,\vec v = \vec v_{CM} + \vec\omega_{\text{inst}} \times \vec r',v=vCM​+ωinst​×r′,

where r⃗′\vec r'r′ is the position of the point relative to the CM, and ω⃗inst\vec\omega_{\text{inst}}ωinst​ is the angular velocity of rotation about the CM.

Since the whole rigid system rotates about the fixed zzz-axis with angular speed ω\omegaω,

v⃗=ω k^×r⃗.\vec v = \omega \, \hat{k} \times \vec r.v=ωk^×r.

Also,

v⃗CM=ω k^×r⃗CM.\vec v_{CM} = \omega \, \hat{k} \times \vec r_{CM}.vCM​=ωk^×rCM​.

Subtracting,

v⃗−v⃗CM=ω k^×(r⃗−r⃗CM)=ω k^×r⃗′.\vec v - \vec v_{CM} = \omega \, \hat{k} \times (\vec r - \vec r_{CM}) = \omega \, \hat{k} \times \vec r'.v−vCM​=ωk^×(r−rCM​)=ωk^×r′.

Comparing with

v⃗−v⃗CM=ω⃗inst×r⃗′,\vec v - \vec v_{CM} = \vec\omega_{\text{inst}} \times \vec r',v−vCM​=ωinst​×r′,

we get

ω⃗inst=ω k^.\vec\omega_{\text{inst}} = \omega \, \hat{k}.ωinst​=ωk^.

Thus, regardless of how the disc is tilted, the instantaneous axis through the CM is vertical.

  1. Case (a)

The disc face is vertical and parallel to the xzxzxz-plane.

Even though the plane of the disc is vertical, the body as a whole is rotating about the zzz-axis. Therefore the angular velocity vector is vertical:

ω⃗inst=ω k^.\vec\omega_{\text{inst}} = \omega \, \hat{k}.ωinst​=ωk^.

So the instantaneous axis through the CM is vertical.

  1. Case (b)

The disc face is inclined at 45∘45^\circ45∘ to the xyxyxy-plane.

Again, the entire rigid system rotates about the zzz-axis with angular speed ω\omegaω. Therefore the angular velocity vector is still

ω⃗inst=ω k^.\vec\omega_{\text{inst}} = \omega \, \hat{k}.ωinst​=ωk^.

So the instantaneous axis through the CM is again vertical.

  1. Checking options
  • A: Vertical for both (a) and (b) — Correct
  • B: Wrong, because in case (b) the axis is not in the plane of the disc; it is vertical.
  • C: Wrong, because case (a) is not horizontal, and case (b) is also not normal to the disc.
  • D: Wrong, because case (b) is not along the normal to the disc.
  1. Final answer

The instantaneous axis through the centre of mass is vertical in both cases.

Therefore, the correct option is:

A\boxed{\text{A}}A​
PreviousNext

More from Rotational Motion

  • The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be… Includes diagram2012 · MCQ
  • Four solid spheres each of diameter 5​ cm and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is N × 10−4 kg-m2, then N is Includes diagram2011 · Numerical
  • A metal rod of length L and mass m is pivoted at one end. A thin disk of mass M and radius R ( < L) is attached at its centre to the free end of the rod. Consider two ways the disc is attached : (case A). The disc is not free to rotate… Includes diagram2011 · Multiple correct
  • A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.2 m/s2. The coefficient of friction between the… Includes diagram2011 · Numerical
  • If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that2009 · Multiple correct
  • A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure below, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then, Includes diagram2009 · Multiple correct
  • STATEMENT - 1 : Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder…2008 · MCQ
  • A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically… Includes diagram2008 · MCQ