NEET-XII-Physics
10: Rotational Mechanics
- #20The centre of a wheel rolling on a plane surface moves with a speed
ν0. A particle on the rim of the wheel at the same level as the centre will be moving at speed
(a) zero
(b)
ν0
(c)
2ν0
(d)
2ν0.digAnsr: cAns : (c) `` \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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