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Physics · Rotational Dynamics

JEE Advanced 2020 — Paper 1 — Question 3

A small roller of diameter 20 cm has an axle of diameter 10 cm (see figure below on the left). It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle (see figure on the right). The scale is now pushed slowly on the axle so that it moves without slipping on the axle, and the roller starts rolling without slipping. After the roller has moved 50 cm , the position of the scale will look like (figures are schematic and not drawn to scale)

Question figure
  1. Option A:
    Option A figure
  2. Option B:
    Option B figure
    Correct
  3. Option C:
    Option C figure
  4. Option D:
    Option D figure

Answer: B

Step-by-step solution

v−ωR=0v-\omega R=0

vp=v+ωr=v+vRr=v(1+rR)v_{p}=v+\omega r=v+\frac{v}{R} r=v\left(1+\frac{r}{R}\right)

So, Δxp=50(1+510)=75 cm\Delta x_{p}=50\left(1+\frac{5}{10}\right)=75 \mathrm{~cm}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 1
Subject
Physics
Chapter
Rotational Dynamics
Topic
Rolling Motion

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