Question

At t = 0, particle starts and moves in a circular path of radius ‘R’. If tangential acceleration of the particle is ‘a’ then find the time when the total acceleration vector makes an angle 45° with tangential acceleration vector.

Answer 1

 

Relation between at and t is 

\(a_t=tan\,45°t+0\)

\(a_t=t\)

\(\frac{du}{dt}=t\)

\(\int dv=\int t\,dt\)

\(v=\frac{t^2}{2}\)

centripetal acceleration 

\(a_c=\frac{v^2}{r}\)

\(a_c=\frac{t^4}{4R}\)

\(a_{net}=\sqrt{a_t ^2+a_c^2}\)

\(a_{net}=\sqrt{t^2+\frac{t^3}{16R^2}}\)

The velocity will always be in direction of the tangential acceleration.

\(tan\theta=\frac{a_t}{a_c}\)

\(\theta=45°\)

\(a_t=a_c\)

\(t=\frac{t^4}{4R}\)

\(4R=t^3\)

\(t=(4R)^{\frac{1}{3}}\,sec\)