Question

A cyclist is moving on a circular path with constant speed v. What is the change in its velocity after it has described at an angle of ${30^ \circ }$
A) $v\sqrt 2 $
B) $\dfrac{v}{2}$
C) $v\sqrt 3 $
D) None of these

Answer 1

Hint:When an object is moving along the circumference of a circle or rotating along a circular path it is known to be in circular motion. If the speed of the object is moving in a circular path with constant speed, then the motion is uniform circular motion.

Complete step by step solution:
Step I:
The direction of the velocity vector of the cyclist moving will be the same as the direction of the motion of the object. Therefore, the change in velocity is given by using$\Delta v = \overrightarrow {{v_2}} - \overrightarrow {{v_1}} $
Where ${v_1}$is the initial velocity
And ${v_2}$is the final velocity of the cyclist moving on a circular path
\[\left| {\Delta v} \right| = \left| {\overrightarrow {{v_2}} - \overrightarrow {{v_1}} } \right|\]
\[ = \sqrt {{{\left| {\overrightarrow {{v_2}} } \right|}^2} + {{\left| {\overrightarrow {{v_1}} } \right|}^2} - 2\left| {\overrightarrow {{v_2}} } \right|\left| {\overrightarrow {{v_1}} } \right|\cos \theta } \]
Since it is given that the cyclist is moving with constant speed v, therefore, ${v_1} = {v_2} = v$. So the above equation becomes,
$ = \sqrt {{v^2} + {v^2} - 2{v^2}\cos \theta } $
$ = \sqrt {2{v^2} - 2{v^2}\cos \theta } $
$ = \sqrt 2 v\sqrt {(1 - \cos \theta )} $
$ = \sqrt 2 v(\sqrt 2 \sin \dfrac{\theta }{2})$
$ = 2v\sin \dfrac{\theta }{2}$
Given that the angle is ${30^ \circ }$, so the above equation can be written as
$ = 2v\sin \dfrac{{30}}{2}$
$ = 2v\sin {15^ \circ }$
$ = 2v \times 0.25$
$ = 0.5v$

$ \Rightarrow $Option B is the right answer.

Note:It is to be noted that in case the object moves in uniform circular motion, it changes its direction continuously. It moves in a direction tangent to the circle. An object that moves in a circle will always be accelerating and its velocity also changes continuously. The acceleration will be perpendicular to the velocity. The force acting on the object will always be in an inward direction and is called centripetal force. If this force is not there, the object will move in a straight line.