A grindstone rotates at constant angular acceleration `alpha=0.35rad//s^(2)`. At time t = 0, it has an angular velocity of `omega_(0)=-4.6rad//s` and a reference line on it is horizontal, at the angular position `theta_(0)=0`.
(a) At what time after t = 0 is the reference line at the angular position `theta=5.0 ` rev ?
(b) Describe the grindstone's rotation between t = 0 and t = 32 s.
(c) At what time t does the grindstone momentarily stop?

Explanation

(a) The angular acceleration is constant, so we can use the rotation equations of. We choose Eq.
`theta-theta_(0)=omega_(0)t+1/2alphat^(2)`,
because the only unknown variable it contains is the desired time t.
Calculations: Substituting known values and setting `theta_(0)=0andtheta=5.0rev=10pirad` given us
`10pirad=(-4.6rad//s)t+1/2(0.35rad//s^(2))t^(2)`.
(We converted 5.0 rev to 10 `pirad` to keep the units consistent.) Solving this quadratic equation for t, we find
t = 32 s.

(b) Reasoning: The wheel is initially rotating in the negative (clockwise) direction with angular velocity `omega_(0)=-4.6rad//s`, but its angular acceleration `alpha` is positive. This initial opposition of the signs of angular velocity and angular acceleration means that the wheel slows in its rotation in the negative direction, stops, and then reverses to rotation in the positive direction. After the reference line comes back through its initial orientation of `theta=0`, the wheel turns an additional 5.0 rev by time t = 32 s.
( c) Calculation: We again go to the table of equations for constant angular acceleration, and again we need an equation that contains only the desired unknown variable t. However, now the equation must also contain the variable `omega`, so that we can set it to 0 and then solve for the corresponding time t. We choose, which yields
`t=(omega-omega_(0))/(alpha)=(0-(-4.6rad//s))/(0.35rad//s^(2))=13s`.