Question

The center 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 center will be moving at speed:

• A. zero
• B. $v_0$
• C. $\sqrt{2}{v}_0$
• D. $2v_0$

Answer 1

Answer: (C)

$\sqrt{2}{v}_0$

Assuming that the radius of the wheel be r

We know that the angular speed of the wheel will be given by the relation-

$\omega =\frac{v_0}{r}$

We know that the particle on the rim of the wheel at the same level as the center will have a vertical speed=$\omega$r

=$\frac{v_0}{r}\times r$

=$v_0$

And horizontal speed also equals to the $v_0$

Therefore the resultant speed will be given by,

$R=\sqrt{\left(v_0^2\right)+\left(v_0^2\right)}$

$R=\sqrt{2\left(v_0^2\right)}$

$R=v_0\sqrt{2}$