A grindstone increases in angular speed from 6.00 rad/s to 12.20 rad/s in 16.00 s. Through what angle does it turn during that time if the angular acceleration is constant?

Given that the grand stone has initial angular velocity of

w (ini) = 6rad/

And it has a final angular velocity of

w (fin) = 12.20rad/sec

Time taken is t=16s

Using equation of angular motion

To get angular acceleration (α)

w (fin) = w (ini) + αt

12.20=6+16α

16α=12.20-6

16α=6.2

α=6.2/16

α=0.3875rad/sec²

The angular acceleration is 0.39rad/s²

Angle that he turn using

w (fin) ²=w (ini) ²+2αθ

12.2²=6²+2*0.3875θ

12.2²-6²=0.775θ

0.775θ=112.84

Then, θ=112.84/0.775

θ=145.6radian

The angular displacement is 145.6rad