Given: α=6t2−2t
Using relation α=dtdω=6t2−2t
Integrating the above,
∫10ωdω=∫0t(6t2−2t)dt
⇒ω−10=2t3−t2
Now, ω=dtdθ=10+2t3−t2
∫4θdθ=∫0t(10+2t3−t2)dt
Integrating the above relation,
θ−4=10t+2t4−3t3
Thus, angular position of the ball is θ=2t4−3t3+10t+4.