Question

A wheel is rolling along the ground with a speed of 2 m/s .The magnitude of the velocity of the points at the points at the extremities of the horizontal diameter of the wheelis equal to

Answer 1

Speed of the wheel = 2 m/s

The extreme points on the horizontal will have two components of 2 m/s, i.e. one in vertical direction and one in the direction of the wheel's motion.

Hence, the angle between the two components is 90°.

Now, we need to find the resultant of two speed components using Pythagoras Theorem.

$v_x = \sqrt{2^2 + 2^2}$

$v = \sqrt{4 +4}$

$v =\sqrt{8}$

$v = 2\times\sqrt{2}$ m/s

Hence, the speed of the extreme points on the horizontal diameter is $2\sqrt{2}$ m/s.

Answer 2

v= umder root 2^2+2^2

=umder root 4+4

=2umder root 2