Question

Modern wind turbines are larger than they appear, and despite their apparently lazy motion, the speed of the blades tips can be quite high-many times higher than the wind speed. A turbine has blades 59m long that spin at13rpm .

At the tip of a blade, what is the speed?
At the tip of a blade, what is the centripetal acceleration?
A big dog has a torso that is approximately circular, with a radius of 16cm . At the midpoint of a shake, the dog's fur is moving at a remarkable 2.5m/s .
What force is required to keep a 10 mg water droplet moving in this circular arc?
What is the ratio of this force to the weight of a droplet?

Answer 1

speed and acceleration of the turbine V = 80.24 m/s and a = 109.1 m/s²

Step-by-step explanation:

The wind turbine describes a circular motion, with constant angular velocity, we can find the speed with the equation

V = x / t

Where x is the distance traveled which is the length of the circle (X = 2πr) and time is the time in a revolution, which in this case is called the period

t = 60/13

t = 4.62 s

x = 2 p1 59

x = 370.71 m

V = 370.1 / 4.62

V = 80.24 m / s

The centripetal acceleration is given by the equation

a = V² / r

a = 80.24²/59

a = 109.1 m / s²

The second part of the exercise seems totally independent

The dog r = 16cm with a speed of 2.5 m/s, ask us the centripetal force

F = m a

F = m (v² / r)

m = 10 mg (1g / 1000mg) 1 Kg / 1000 g) = 1 10⁻⁵ Kg

r = 16 cm (1m / 100cm) = 16 10⁻² m

F = 1 10⁻⁵ 2.5²/16 10⁻²

F = 3.91 10⁻² N

Let's calculate the relationship between the force and the weight of the drop

F / W = 3.91 10⁻² (1 10⁻⁵ 9.8)

F / W = 3.99 10 ²

See that the force of the shake is approximately 400 times greater than the weight of the drop