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Laws Of Motion

Question
CBSEENPH11016907

Derive the relation straight omega squared minus straight omega subscript straight o superscript 2 equals 2 straight alpha left parenthesis straight theta minus straight theta subscript straight o right parenthesis for uniform angular accelerated motion.

Solution

We know, 
Angular acceleration,
space space straight alpha equals dω over dt equals dω over dθ dθ over dt equals dω over dθ straight omega
rightwards double arrow space straight omega space dω space equals space straight alpha space dθ
Now, on integrating both sides, we get
space space space space space space integral subscript straight omega subscript 0 end subscript superscript straight omega omega space d omega space equals space integral subscript theta subscript o end subscript superscript theta alpha space d theta

rightwards double arrow space 1 half right enclose omega to the power of 2 space end exponent end enclose subscript omega subscript 0 end subscript superscript omega space equals space alpha space right enclose theta subscript theta subscript o end subscript superscript theta space

rightwards double arrow space 1 half left parenthesis omega squared space minus space omega subscript o squared right parenthesis space equals space alpha space left parenthesis theta space minus space theta subscript o right parenthesis

rightwards double arrow space space space space space omega squared space minus space omega subscript o squared space equals space 2 space alpha space left parenthesis theta space minus space theta subscript o right parenthesis