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Laws Of Motion

Question
CBSEENPH11016903

Derive the relation, 

straight theta equals straight theta subscript 0 plus straight omega subscript 0 straight t plus 1 half αt squared

for uniform angular accelerated motion.

Solution

We know, 
Angular velocity, straight omega equals dθ over dt
or                     dθ equals straight omega space dt
For uniform angular accelerated motion, 
space space space space space space space space space straight omega space equals space straight omega subscript straight o space plus space αt space
therefore space space space dθ space equals space left parenthesis straight omega subscript straight o space plus space αt space right parenthesis space dt space

Integrating space both space sides comma space we space get

integral subscript straight theta subscript straight o end subscript superscript straight theta dθ space equals space integral subscript 0 superscript straight t left parenthesis straight omega subscript straight o space plus space αt space right parenthesis space dt space equals straight omega subscript straight o space integral subscript 0 superscript straight t dt space plus space straight alpha space integral subscript 0 superscript straight t straight t space dt space

rightwards double arrow space space space right enclose straight theta space subscript straight theta subscript straight o end subscript superscript straight theta space equals right enclose space straight omega subscript straight o straight t end enclose subscript 0 superscript straight t space plus space 1 half right enclose αt squared end enclose subscript 0 superscript straight t space

rightwards double arrow space space straight theta space minus space straight theta subscript 0 space equals ωt space plus space 1 half αt squared
rightwards double arrow space straight theta space equals space space straight theta subscript 0 space plus ωt space plus space 1 half αt squared
Hence, proved.