Q. A wheel is rolling along the ground with a speed of $2\ ms^{- 1}$. The magnitude of the velocity of the points at the extremities of the horizontal diameter of the wheel is equal to
A) $2\sqrt{10}\ ms^{- 1}$
B) $2\sqrt{3}\ ms^{- 1}$
C) $2\sqrt{2}\ ms^{- 1}$
D) $2\ ms^{- 1}$
Explanation
(c) $v_{t} =$ velocity due to translator motion
$v_{R} =$ velocity due to rotational motion
$v_{N} = \sqrt{v_{t}^{2} + v_{R}^{2}} = \sqrt{v^{2} + v^{2}} = \sqrt{2v} = 2\sqrt{2}m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.