JEE Challenger
More from System of Particles and Rotational Motion
PhyHardAdv 2016 • Paper 2

Net Reaction Force on Block in Radial Slot of Rotating Disc

Comprehension Passage

A frame of reference that accelerates relative to an inertial reference frame is defined as a non-inertial frame. For example, a coordinate frame attached to a circular disc rotating about a fixed vertical axis with constant angular velocity ω\omega serves as a non-inertial frame.\n\nThe force F⃗rot\vec{F}_{\text{rot}} observed in the rotating frame and the force F⃗in\vec{F}_{\text{in}} in an inertial frame on a particle of mass mm are connected by:\nF⃗rot=F⃗in+2m(v⃗rot×ω⃗)+m(ω⃗×r⃗)×ω⃗,\vec{F}_{\text{rot}} = \vec{F}_{\text{in}} + 2m(\vec{v}_{\text{rot}} \times \vec{\omega}) + m(\vec{\omega} \times \vec{r}) \times \vec{\omega},\nwhere v⃗rot\vec{v}_{\text{rot}} represents the velocity of the particle within the rotating frame, and r⃗\vec{r} is its position vector measured from the centre of the disc.\n\nConsider a smooth slot along a diameter of a disc of radius RR that rotates counter-clockwise with a constant angular speed ω\omega around its central vertical axis. We set up a coordinate system with its origin at the disc's center, where the xx-axis lies along the slot, the yy-axis is perpendicular to it, and the zz-axis coincides with the rotation axis (so ω⃗=ωk^\vec{\omega} = \omega \hat{k}). A small object of mass mm is gently placed inside the slot at position r⃗=(R/2)i^\vec{r} = (R/2)\hat{i} at time t=0t = 0 and is restricted to move purely along the slot.

Passage Diagram 1

Find the total normal reaction vector exerted by the rotating disc on the block.

Options

A

12mω2R(e2ωt−e−2ωt)j^+mgk^\frac{1}{2} m \omega^2 R (e^{2\omega t} - e^{-2\omega t}) \hat{j} + mg \hat{k}

B

12mω2R(eωt−e−ωt)j^+mgk^\frac{1}{2} m \omega^2 R (e^{\omega t} - e^{-\omega t}) \hat{j} + mg \hat{k}

Correct
C

−mω2Rcos⁡ωtj^−mgk^-m \omega^2 R \cos \omega t \hat{j} - mg \hat{k}

D

mω2Rsin⁡ωtj^−mgk^m \omega^2 R \sin \omega t \hat{j} - mg \hat{k}

Topics & Concepts

System of Particles and Rotational MotionNewton's LawsRelative Motion

Step-by-Step Solution

To find the total normal reaction force vector exerted by the rotating disc on the block, we analyze the motion of the block of mass mm in the rotating frame of reference attached to the disc.

1. Coordinate System and Forces in the Rotating Frame

The coordinate system is set up as follows:

  • xx-axis lies along the slot.
  • yy-axis is perpendicular to the slot in the plane of the disc.
  • zz-axis is along the axis of rotation, pointing vertically upwards (ω⃗=ωk^\vec{\omega} = \omega \hat{k}).

The formula given for the apparent force in the rotating frame is: F⃗rot=F⃗in+2m(v⃗rot×ω⃗)+m(ω⃗×r⃗)×ω⃗\vec{F}_{\text{rot}} = \vec{F}_{\text{in}} + 2m(\vec{v}_{\text{rot}} \times \vec{\omega}) + m(\vec{\omega} \times \vec{r}) \times \vec{\omega}

Let the position of the block along the slot at any time tt be r⃗=x(t)i^\vec{r} = x(t) \hat{i}, so its velocity in the rotating frame is v⃗rot=x˙i^\vec{v}_{\text{rot}} = \dot{x} \hat{i}.

  • Centrifugal Force: m(ω⃗×r⃗)×ω⃗=m(ωk^×xi^)×ωk^=m(ωxj^)×ωk^=mω2xi^m(\vec{\omega} \times \vec{r}) \times \vec{\omega} = m(\omega \hat{k} \times x \hat{i}) \times \omega \hat{k} = m(\omega x \hat{j}) \times \omega \hat{k} = m \omega^2 x \hat{i}

  • Coriolis Force: 2m(v⃗rot×ω⃗)=2m(x˙i^×ωk^)=−2mωx˙j^2m(\vec{v}_{\text{rot}} \times \vec{\omega}) = 2m(\dot{x} \hat{i} \times \omega \hat{k}) = -2m \omega \dot{x} \hat{j}

  • Real Forces (F⃗in\vec{F}_{\text{in}}): The real forces acting on the block are the gravitational force −mgk^-mg \hat{k}, the normal force from the bottom of the slot Nzk^N_z \hat{k}, and the normal force from the side wall of the slot Nyj^N_y \hat{j}: F⃗in=Nyj^+Nzk^−mgk^\vec{F}_{\text{in}} = N_y \hat{j} + N_z \hat{k} - mg \hat{k}

2. Motion Along the Slot (xx-axis)

In the rotating frame, the equation of motion along the smooth radial slot (xx-direction) is: mx¨=mω2x  ⟹  x¨−ω2x=0m \ddot{x} = m \omega^2 x \implies \ddot{x} - \omega^2 x = 0

The general solution to this differential equation is: x(t)=C1eωt+C2e−ωtx(t) = C_1 e^{\omega t} + C_2 e^{-\omega t}

Using the initial conditions at t=0t = 0:

  1. x(0)=R2  ⟹  C1+C2=R2x(0) = \frac{R}{2} \implies C_1 + C_2 = \frac{R}{2}
  2. x˙(0)=0  ⟹  ω(C1−C2)=0  ⟹  C1=C2\dot{x}(0) = 0 \implies \omega(C_1 - C_2) = 0 \implies C_1 = C_2

Solving these equations gives: C1=C2=R4C_1 = C_2 = \frac{R}{4}

Thus, the position and velocity along the slot at time tt are: x(t)=R4(eωt+e−ωt)x(t) = \frac{R}{4}\left(e^{\omega t} + e^{-\omega t}\right) x˙(t)=ωR4(eωt−e−ωt)\dot{x}(t) = \frac{\omega R}{4}\left(e^{\omega t} - e^{-\omega t}\right)

3. Reaction Forces in the yy and zz Directions

Since the block is constrained to move purely along the xx-axis, its acceleration components in the yy and zz directions relative to the rotating frame are zero (F⃗rot⋅j^=0\vec{F}_{\text{rot}} \cdot \hat{j} = 0 and F⃗rot⋅k^=0\vec{F}_{\text{rot}} \cdot \hat{k} = 0).

  1. Along the yy-axis: 0=Ny−2mωx˙0 = N_y - 2m \omega \dot{x} Ny=2mωx˙=2mω[ωR4(eωt−e−ωt)]=12mω2R(eωt−e−ωt)N_y = 2m \omega \dot{x} = 2m \omega \left[ \frac{\omega R}{4}\left(e^{\omega t} - e^{-\omega t}\right) \right] = \frac{1}{2} m \omega^2 R \left(e^{\omega t} - e^{-\omega t}\right)

  2. Along the zz-axis: 0=Nz−mg  ⟹  Nz=mg0 = N_z - mg \implies N_z = mg

4. Total Normal Reaction Vector

Combining the components, the total normal reaction force vector exerted by the disc on the block is: N⃗=Nyj^+Nzk^=12mω2R(eωt−e−ωt)j^+mgk^\vec{N} = N_y \hat{j} + N_z \hat{k} = \frac{1}{2} m \omega^2 R \left(e^{\omega t} - e^{-\omega t}\right) \hat{j} + mg \hat{k}

This corresponds to Option B.

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