A wheel rotates around a stationary axis so that the rotation angle `theta` varies with time as `theta=at^(2)` where `a=0.2rad//s^(2)`. Find the magnitude of net acceleration of the point A at the rim at the moment `t=2.5s` if the linear velocity of the point A at this moment is `v=0.65m//s`.

Explanation

Instantaneous angular velocity at time `t` is
`omega=(dtheta)/(dt)=(d)/(dt)(at^(2))`
or `omega=2at=0.4t`
(as `a=0.2rad//s^(2))`
Further, instantaneous angular acceleration is
`alpha=(domega)/(dt)=(d)/(dt)(0.4t)`
or `alpha=0.4rad//s^(2)`
Angular velocity at `t=2.5s` is
`omega=0.4xx2.5=1.0rad//s`
further, radius of the wheel
`R=(v)/(omega)`
or `R=(0.65)/(1.0)=0.65m`
Now, magnitude of total acceleration is,
`a=sqrt(a_(n)^(2)+a_(t)^(2))`
Here, `a_(n)=Romega^(2)=(0.65)(1.0)^(2)=0.65m//s^(2)`
and `a_(t)=Ralpha=(0.65)(0.4)=0.26m//s^(2)`
`thereforea=sqrt((0.65)^(2)+(0.26)^(2))`
or `a=0.7m//s^(2)`