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AIIMS 201925 May
Exam
AIIMS
Year
2019
Date
25 May
Subject
Physics
Chapter
Rotational Motion

AIIMS 2019 · 25 May · Physics · Rotational Motion

A disc of radius 20 ~cm and mass 1 / 2 ~kg is rolling on an...

A disc of radius 20 cm20 \mathrm{~cm} and mass 1/2 kg1 / 2 \mathrm{~kg} is rolling on an inclined plane. Find out the frictional force such that disc performs pure rolling. A disc of radius 20 ~cm and mass 1 / 2 ~kg is rolling on an inclined plane. Find out the frictional force such that

Options

  1. A523N\frac{5 \sqrt{2}}{3} N
  2. B532 N\frac{5}{3 \sqrt{2}} \mathrm{~N}
  3. C52 N\frac{5}{\sqrt{2}} \mathrm{~N}
  4. D523 N\frac{5}{2 \sqrt{3}} \mathrm{~N}

Correct answer

B. 532 N\frac{5}{3 \sqrt{2}} \mathrm{~N}

Step-by-step solution

Let mm be the mass and RR be the radius of the disc. mgsin⁡θ−f=macmm g \sin \theta-f=m a_{c m}τcm=IcmαfR=mR22(acmR)\begin{aligned} & \tau_{c m}=I_{c m^\alpha} \\ & f R=\frac{m R^2}{2}\left(\frac{a_{c m}}{R}\right)\end{aligned}f=macm2oracm=2fmf=\frac{m a_{c m}}{2} or a_{c m}=\frac{2 f}{m} So, mgsin⁡θ−f=m(2fm)f=mgsin⁡θ3=12×103×12=532 Nm g \sin \theta-f=m\left(\frac{2 f}{m}\right) f=\frac{m g \sin \theta}{3}=\frac{1}{2} \times \frac{10}{3} \times \frac{1}{\sqrt{2}}=\frac{5}{3 \sqrt{2}} \mathrm{~N} A disc of radius 20 ~cm and mass 1 / 2 ~kg is rolling on an inclined plane. Find out the frictional force such that
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