JEE Challenger
More from System of Particles and Rotational Motion
PhyMedAdv 2020 • Paper 1

Position of Meter Scale Resting on Axle of Rolling Cylinder

A small roller having a diameter of 20 cm20\text{ cm} features an axle with a diameter of 10 cm10\text{ cm} (as shown in the left figure). It rests on a flat horizontal floor while a meter scale is placed horizontally atop its axle (as shown in the right figure). The scale is pushed gradually over the axle such that it moves without slipping relative to the axle, causing the roller to roll without slipping on the ground. After the roller travels a distance of 50 cm50\text{ cm}, which of the following diagrams correctly represents the final position of the scale? (Schematics are not to scale)

Question Diagram 1

Options

A
Option A Diagram 1
B
Option B Diagram 1
Correct
C
Option C Diagram 1
D
Option D Diagram 1

Topics & Concepts

System of Particles and Rotational MotionRolling Motion

Step-by-Step Solution

To determine the final position of the meter scale, we analyze the rolling motion of the cylinder and the relative motion of the scale resting on its axle.

1. System Parameters

  • Radius of the outer cylinder (roller): R=20 cm2=10 cmR = \frac{20\text{ cm}}{2} = 10\text{ cm}
  • Radius of the inner axle: r=10 cm2=5 cmr = \frac{10\text{ cm}}{2} = 5\text{ cm}
  • Total length of the scale: L=100 cmL = 100\text{ cm}

2. Kinematics of Pure Rolling Motion

Let vcv_c be the forward velocity of the center of mass of the roller. Since the outer cylinder rolls without slipping on the horizontal ground, its angular velocity ω\omega is related to vcv_c by: ω=vcR\omega = \frac{v_c}{R}

The top point of the axle, where the meter scale makes contact, moves forward with a linear velocity vscalev_{\text{scale}} given by: vscale=vc+ωr=vc(1+rR)v_{\text{scale}} = v_c + \omega r = v_c \left(1 + \frac{r}{R}\right)

Substituting the given values of R=10 cmR = 10\text{ cm} and r=5 cmr = 5\text{ cm}: vscale=vc(1+510)=1.5vcv_{\text{scale}} = v_c \left(1 + \frac{5}{10}\right) = 1.5 v_c

3. Displacement Calculations

Integrating the velocities with respect to time gives the relationship between the displacement of the scale (Δxscale\Delta x_{\text{scale}}) and the displacement of the roller's center (Δxc\Delta x_c): Δxscale=1.5Δxc\Delta x_{\text{scale}} = 1.5 \Delta x_c

Given that the roller moves a distance of Δxc=50 cm\Delta x_c = 50\text{ cm}: Δxscale=1.5×50 cm=75 cm\Delta x_{\text{scale}} = 1.5 \times 50\text{ cm} = 75\text{ cm}

The displacement of the scale relative to the axle is: Δxrel=Δxscale−Δxc=75 cm−50 cm=25 cm\Delta x_{\text{rel}} = \Delta x_{\text{scale}} - \Delta x_c = 75\text{ cm} - 50\text{ cm} = 25\text{ cm}

4. Final Position on the Scale

Initially, the right edge of the meter scale (at x=100 cmx = 100\text{ cm}) rests on top of the axle, and the rest of the scale (towards x=0x = 0) extends to the left.

Since the scale moves to the right relative to the axle by 25 cm25\text{ cm}, the contact point on the scale shifts towards the left end (towards x=0x = 0) by 25 cm25\text{ cm}.

Thus, the mark on the scale directly above the axle after the motion is: x=100 cm−25 cm=75 cmx = 100\text{ cm} - 25\text{ cm} = 75\text{ cm}

This matches the diagram shown in Option (B).

Correct Option: (B)

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