Question

A particle is moving in a circular motion with constant speed "v". The magnitude of change in velocity for the following angular displacements
90°
180°
360°​

Answer 1

Explanation:

See attachment

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Answer 2

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put (0,0){\circle {14}}\put(8.3,0){\scriptsizeA}\put(-1,8.9){\scriptsizeB}\put(-10.5,0){\scriptsizeC}\put(30,0){\vector (1,0){10}}\put(30,0){\vector (0,1){10}}\put(41,0){\scriptsizeX}\put(30,11){\scriptsizeY}\end{picture}$

Let the particle start its circular motion from A with constant speed $v.$

→ The velocity of the particle at A is $\bf {v_A}=v\ \bf{\hat j}.$

→ The velocity of the particle at B is $\bf{v_B}=-v\ \bf{\hat i}.$

→ The velocity of the particle at C is $\bf{v_C}=-v\ \bf{\hat j}.$

On travelling from A to B, the angular displacement is 90°. Then the magnitude of the change in velocities for travelling AB is,

$|\bf{v_B-v_A}|=|- v\ \bf{\hat i}\ -\ v\ \bf{\hat j}|\\\\\\|\bf{v_B-v_A}|=|-v(\bf{\hat i} + \bf{\hat j})|\\\\\\|\bf{v_B-v_A}|=|\text { -v\sqrt {i^2+j^2+2ij\cos 90^{\circ}} }|\\\\\\|\bf{v_B-v_A}|=|-v\sqrt {1+1+0}|\\\\\\|\bf{v_B-v_A}|=|-v\sqrt {2}|\\\\\\\underline {\underline {|\bf{v_B-v_A}|=v\sqrt {2}\ ms^{-1}}}$

On travelling from A to C, the angular displacement is 180°. Then the magnitude of the change in velocities for travelling AC is,

$|\bf{v_C-v_A}|=|-v\ \bf{\hat j}-v\ \bf{\hat j}|\\\\\\|\bf{v_C-v_A}|=|-2v\ \bf{\hat j}|\\\\\\\underline {\underline {|\bf{v_C-v_A}|=2v\ ms^{-1}}}$

For getting angular displacement of 360° the particle has to reach the starting point A. Then the magnitude of the change in velocities for reaching the initial point is,

$\underline {\underline {|\bf{v_A-v_A}|=0\ ms^{-1}}}$