## College Physics (4th Edition)

The drill rotates through $(1.80\times 10^6)^{\circ}$
We can find the angular displacement in $1.00~s$: $\Delta \theta = \omega ~t$ $\Delta \theta = (3.14\times 10^4~rad/s)(1.00~s)$ $\Delta \theta = 3.14\times 10^4~rad$ We can express the angular displacement in degrees: $\Delta \theta = (3.14\times 10^4~rad) \times \frac{180^{\circ}}{\pi~rad}$ $\Delta \theta = (1.80\times 10^6)^{\circ}$ The drill rotates through $(1.80\times 10^6)^{\circ}$.