A particle is projected with a speed of `10 m//s` at an angle `37^(@)` from horizontal, anglular speed to particle with respect to point of projection at `t = 0.5 sec`. Is `((16X)/(61))` rad/sec then calculate `X` :

Answer

2

Explanation

`overset(vec)(r) = 4hat(i) + [6(1)/(2) - (1)/(2) 10 (1)/(4)hat(j)]`
`= 4hat(i) + (3 - 1.25)hat(j)`

`= 4hat(i) + 1.75 hat(j)`
`overset(vec)(v) = 8hat(i) + hat(j)`
`omega = (|overset(vec)(v) xx overset(vec)(r)|)/(|r|^(2)) = (1.75 xx 8 - 4)/(16 + (49)/(16))`
`= (10 xx 16)/(256 + 49) = (160)/(305) = (32)/(61) rad//sec`.