A particle is moving in a circular orbit with a constant tangential acceleration. After a certain time t has elapsed after the beginning of motion, the between the total acceleration a and the direction along the radius r becomes equal to `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)`