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A solid body rotates about a stationary ...

A solid body rotates about a stationary axis accordig to the law `theta=6t-2t^(3)`. Here `theta`, is in radian and `t` in seconds. Find
(a). The mean values of thhe angular velocity and angular acceleration averaged over the time interval between `t=0` and the complete stop.
(b). The angular acceleration at the moment when the body stops.
Hint: if `y=y(t)`. then mean/average value of `y` between `t_(1)` and `t_(2)` is `ltygt=(int_(t_(1))^(t_(2))y(t)dt))/(t_(2)-t_(1))`

Text Solution

Verified by Experts

`omega=(dtheta)/(dt)=6-6t^(2)`
`alpha=(domega)/(dt)=-12t`
`omega=0`
at `t=1s`
(a) `(omega)_(0-1)=((int_(0)^(1)omegadt)/(1)=int_(0)^(1)(6-6t^(2))dt`
`=4rad//s`
`(alpha)_(0-1)=((int_(0)^(1)(-12t)dt)/(1))`
`=-6rad//s^(2)`
(b). At `t=1s`
`alpha=-12rad//s^(2)`
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