Question

A machine part rotates at an angular speed of 0.39 rad/s; its speed is then increased to 2.5 rad/s at an angul?
A machine part rotates at an angular speed of 0.39 rad/s; its speed is then increased to 2.5 rad/s at an angular acceleration of 0.84 rad/s2.
(a) Find the angle through which the part rotates before reaching this final speed.
b)In general, if both the initial and final angular speed are doubled at the same angular acceleration, by what factor is the angular displacement changed? Why?

Answer 1

a) \[\omega^2=\omega_{0}^2+2\alpha \Delta \theta \]everything is known here except delta theta, which is what you are asked to solve for. Inputting,\[(2.50)^2=(0.39)^2+2(0.84)\Delta \theta \]Solving,\[\Delta \theta=3.63\]radians.

Answer 2

b) \[\Delta \theta =\frac{(2\omega)^2-(2\omega_{0})^2}{2\alpha}\]\[\Delta \theta=4\frac{\omega^2-\omega_{0}^2}{2\alpha}\]Comparing with the equation in part a we see that doubling both the initial and final angular speed while keeping the angular acceleration the same results in a change in the angular displacement of a factor of four.