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Question
The centre of a wheel rolling on a plane surface moves with a speed \[\nu_0\] A particle on the rim of the wheel at the same level as the centre will be moving at speed ___________ .
Options
zero
\[\nu_0\]
\[\sqrt{2} \nu_0\]
\[2 \nu_0\]
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Solution
\[\sqrt{2} \nu_0\]
For pure rolling,
\[\omega r = v_0\]
As shown in the figure, the velocity of the particle will be the resultant of v0 and ωr.
Therefore, we have
\[v_{net} = \sqrt{{v_0}^2 + \left( \omega r \right)^2}\]
\[ v_{net} = \sqrt{{v_0}^2 + {v_0}^2}\]
\[ v_{net} = \sqrt{2} v_0\]
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