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Angular position theta of a particle mov...

Angular position `theta` of a particle moving on a curvilinear path varies according to the equation `theta=t^(3)-3t^(2)+4t-2`, where `theta` is in radians and time t is in seconds. What is its average angular acceleration in the time interval `t=2s` to `t=4s`?

Text Solution

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Like average linear acceleration, the average angular acceleration `alpha_(av)` equals to ratio of change in angular velocity `Deltaomega` to the concerned time interval `Deltat`.
`alpha_(av)=(Deltaomega)/(Deltat)=(omega_("final")-omega_("initial"))/(t_("final")-t_("initial"))` ...(i)
The angular velocity `omega` being rate of charge in angular position can be obtained by equation
`omega=(d theta)/(dt)`
Substituting the given expression of the angular position `theta`, we have
`omega=3t^(2)-6t+4` ...(ii)
From the above eq. (ii), angular velocities `omeha_(2)` and `omega_(4)` at the given instants `t=2s` and `4s` are
`omega_(4)=4 rad//s` and `omega_(4)=28 rad//s`
Substituting the above values in eq. (1), we have `alpha_(av)=12 rad//s^(2)`
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