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Laws Of Motion

Question
CBSEENPH11016938

A particle moves in a circle of radius r with constant angular velocity straight omega. What is the change in velocity when the particle describes an angle of 90°?

Solution
Let the particle revolve in circular orbit of radius r with constant angular velocity straight omega.
The speed of particle will be, 
                      nu equals omega r
Let  stack straight v subscript 1 with rightwards arrow on top be the velocity of particle at P and space space space stack straight v subscript 2 with rightwards arrow on top  be the velocity of particle at Q after revolving by angle 90 degree.
Therefore, angle between stack straight v subscript 1 with rightwards arrow on top space and space stack straight v subscript 2 with rightwards arrow on top is 90 degree.
∴           Change in velocity equals left parenthesis stack straight v subscript 2 with rightwards arrow on top minus stack straight v subscript 1 with rightwards arrow on top right parenthesis
i.e.         space space open vertical bar stack straight v subscript 2 with rightwards arrow on top minus stack straight v subscript 1 with rightwards arrow on top close vertical bar space equals space square root of straight v subscript 2 superscript 2 plus straight v subscript 1 superscript 2 minus 2 straight v subscript 1 straight v subscript 2 cos 90 degree end root 
                       space space space space space space equals square root of straight omega squared straight r squared plus straight omega squared straight r squared minus 0 end root         
                            equals square root of 2 space space ωr