A particle is moving in a circular orbit when a constant tangential acceleration. After `2s` from the beginning of motion, angle between the total acceleration vector and the radius E becomes `45^(@)`. What is the angular acceleration of the particle?

Explanation

In the adjoining figure are shown the total acceleration vector `vec(a)` and its components the tangential accelerations `vec(a)_(tau)` and normal acceleration `vec(a)_(n)` are shown. These two components are always mutually perpendicular to each other and act along the tangent to the circle and radius respectively. Therefore, if the total acceleration vector makes an angle of `45^(@)` with the radius, both the tangential and the normal components must be equal in magnitudr.
Now from equation and, we have
`a_(tau)=a_(n) rarr alpha R=omega^(2)R rArr alpha=omega^(2)` ...(i)
Since angular acceleration is uniform, form equation, we have `omega=omega_(o)+alphat`
Substituting `omega_(0)=0` and `t=2 s`, we have `omega=2alpha` ...(ii)
From equation (i) and (ii), we have `alpha=0.25 rad//s^(2)`