The centre of a wheel rolling on a plane surface moves with a speed $v_0$. A particle on the rim of the wheel at the same level as the centre will be moving at a speed

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Aicharya · Accepted Answer

According to question, speed of the centre of wheel is $v_0$.

Particle is situated on the rim of wheel as shown in Fig. (ii). $\begin{aligned} \therefore \quad v=r \omega & =\sqrt{(O P)^2+(O M)^2} \cdot \omega=\sqrt{R^2+R^2} \cdot \omega \\ v & =\sqrt{2} R \omega \quad \end{aligned}$ But $v_0=R \omega$ ...(ii) From Eqs. (i) and (ii), we get $v=\sqrt{2} v_0$

The centre of a wheel rolling on a plane surface moves with a speed \(v_0\). A particle on the rim of the…

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