Question

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 speed

• A. zero
• B. $v_{ 0}$
• C. $2v_{ 0}$
• D. $\sqrt{2v_{o}}$

Answer 1

Answer: (D)

$\sqrt{2v_{o}}$

Answer 2

The given solution can be shown as

Here $v_0 = R\omega$
At $P, v = r \omega$
$= \sqrt{(R^2 + R^2)\omega}$
$= \sqrt{2} R \omega = \sqrt{2} v_0$