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AIIMS 201926 May
Exam
AIIMS
Year
2019
Date
26 May
Subject
Physics
Chapter
Motion in Two Dimensions

AIIMS 2019 · 26 May · Physics · Motion in Two Dimensions

A disc of radius 2 ~m is rotating with angular is to be put on the...

A disc of radius 2 m2 \mathrm{~m} is rotating with angular is to be put on the disc. Friction coefficient between disc and block is μK=0.3\mu_{\mathrm{K}}=0.3, then find the maximum distance from axis where the block can be placed without sliding

Options

  1. A2 cm2 \mathrm{~cm}
  2. B3 cm3 \mathrm{~cm}
  3. C4 cm4 \mathrm{~cm}
  4. D6 cm6 \mathrm{~cm}

Correct answer

B. 3 cm3 \mathrm{~cm}

Step-by-step solution

 Friction force = centripetal force μK(mg)=mω2xμK⋅g=ω2x0.3×10=(10)2x⇒x=3100m=3cm\begin{aligned} & \text{ Friction force }=\text{ centripetal force } \mu_{\mathrm{K}}(m g) \\ & =m \omega^2 x \\ & \mu_{\mathrm{K}} \cdot g=\omega^2 x \\ & 0.3 \times 10=(10)^2 x \Rightarrow x=\frac{3}{100} \mathrm{ m}=3 \mathrm{ cm}\end{aligned}
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