Rotational Motion

JEE Main level, numerical answer, medium

162 of 280

Question 162 of 280

The centre of a wheel rolling on a plane surface moves with a speed v0.{v}_{0}. A particle on the rim of the wheel at the same level as the centre will be moving at a speed xv0.\sqrt{x}{v}_{0}. Then the value of xx is ..

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Answer: 2


For no slipping, v0=ωR{v}_{0}=\omega R
Now, vA=vB=v02+(ωR)2{v}_{A}={v}_{B}=\sqrt{{v}_{0}^{2}+(\omega R{)}^{2}}
=2v0=\sqrt{2}{v}_{0}
⇒x=2\Rightarrow x=2

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