Answer to Question #172492 in Physics for Diocos

Question

A wind turbine is rotating counterclockwise at 0.5 rev/s and slows to a stop in 10 s. Its blades are 20 m in length

1)The angular acceleration of the turbine is

2)The centripetal acceleration of the tip of the blades at t=0s is

3)The magnitude of the total linear acceleration of the tip of the blades at t=0s is

4)The direction of the total linear acceleration of the tip of the blades at t=0s is Blank ____° counterclockwise

Answer 1

1) We can find the angular acceleration of the turbine as follows:

$$\omega=\omega_0+\alpha t,$$ $$\alpha=\dfrac{\omega-\omega_0}{t},$$ $$\alpha=\dfrac{0-0.5\ \dfrac{rev}{s}\cdot\dfrac{2\pi\ rad}{1\ rev}}{10\ s}=-0.314\ \dfrac{rad}{s^2}.$$

The sign minus means that the turbine decelerates.

2) We can find centripetal acceleration of the tip of the blades as follows:

$$a_c=\omega_0^2R=(0.5\ \dfrac{rev}{s}\cdot\dfrac{2\pi\ rad}{1\ rev})^2\cdot20\ m=197.2\ \dfrac{m}{s^2}.$$

3) Let's first find the tangential acceleration of the tip of the blades:

$$a_t=\alpha R=(-0.314\ \dfrac{rad}{s^2})\cdot20\ m=-6.28\ \dfrac{m}{s^2}.$$

Finally, we can find the magnitude of the total linear acceleration of the tip of the blades from the Pythagorean theorem:

$$a=\sqrt{a_c^2+a_t^2}=\sqrt{(197.2\ \dfrac{m}{s^2})^2+(-6.28\ \dfrac{m}{s^2})^2}=197.3\ \dfrac{m}{s^2}.$$

4) We can find the direction of the total linear acceleration of the tip of the blades from the geometry:

$$tan\theta=\dfrac{|a_t|}{|a_c|},$$ $$\theta=tan^{-1}(\dfrac{|a_t|}{|a_c|}),$$ $$\theta=tan^{-1}(\dfrac{|-6.28\ \dfrac{m}{s^2}|}{|197.3\ \dfrac{m}{s^2}|})=1.82^{\circ}.$$