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JEE Main 2021 (22 Jul, Shift 2) · Rolling Motion

Rolling Motion question from JEE Main 2021 (22 Jul, Shift 2)

Q54JEE Main 2021 · 22 JUL · SHIFT 2Medium

Rolling Motion

The centre of a wheel rolling on a plane surface moves with a speed v0. A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt{x}\,v_0$ . Then the value of x is _____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 02

Rim point at centre level: translational $v_0$ (horizontal) plus rotational $R\omega = v_0$ (vertical).
These are perpendicular, so speed $= \sqrt{v_0^2 + v_0^2} = \sqrt{2}\,v_0$ .
So $x = 2$ .

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