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A particle is at rest, It starts rotatin...

A particle is at rest, It starts rotating about a fixed point. Its angle of rotation `(theta)` with time (t) is given by the relation :
`theta = (6t^3)/(15) - (t^2)/(2)`
where `theta` is in radian and t is seconds. Find the angular velocity and angular acceleration of a particle at the end of 6 second.

Text Solution

Verified by Experts

Here, `theta = (6t^3)/(15) - (t^2)/(2)`
Angular velocity, `Omega = (d theta)/(dt) = (d)/(dt) [(6t^3)/(15) - (t^2)/(2)] = (6)/(5)t^2 - t`
Angular acceleration, `alpha = (domega)/(dt) = (d)/(dt) ((6)/(5)t^2- t) = (12)/(5) t -1`
When t =6s, `omega = (6)/(5)xx6^2 -6 =43.2-6 = 37.2 rad//s`
`alpha = (12)/(5)x6-1 =13.4 rad//s^2`
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