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The total acceleration of a particle in circular motion is 5 m/s2. If the speed of the particle decreases at a rate of 3 m/s2, assuming r=2 m, the angular speed ω of the particle ( in rad/s) relative to the centre of the circle is 


Solution

Let the tangential acceleration be at 
and centripetal acceleration be ac
Given, a=5 m/s2,at=3 m/s2,r=2 ma=a2t+a2ca2=a2t+a2c52=32+a2c
ac=4 m/s2

We know ac=v2r
i.ev2r=ω2r2r=4ω2=42=2
ω=2  rad/s

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