Q146 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Physics
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
- A0
- B\(v_0\)
- Correct answerC\(\sqrt{2} v_0\)
- D\(2 v_0\)
Solution
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 \(v=\sqrt{2} v_0\)
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 \(v=\sqrt{2} v_0\)
Hence, the correct answer is option: 'C'.