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 56 m long that spin at 13 rpm .
A. At the tip of a blade, what is the speed?
B. At the tip of a blade, what is the centripetal acceleration?
C. 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 .
D. What force is required to keep a 10 mg water droplet moving in this circular arc?
E. What is the ratio of this force to the weight of a droplet?

Answer 1

A) The speed of the tip of the blade is 76.2 m/s

B) The centripetal acceleration of the tip of the blade is $103.7 m/s^2$

D) The force required to keep the droplet moving in circular motion is 0.39 N

E) The ratio of the force to the weight of the droplet is 4.0

Explanation:

A)

We know that the blade of the turbine is rotating at an angular speed of

$\omega = 13 rpm$

First, we have to convert this angular speed into radians per second. Keeping in mind that

$1 rev = 2 \pi$

$1 min = 60 s$

We get

$\omega = 13 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=1.36 rad/s$

The linear speed of a point on the blade is given by:

$v=\omega r$

where

$\omega=1.36 rad/s$ is the angular speed

r is the distance of the point from the axis of rotation

For a point at the tip of the blade,

r = 56 m

Therefore, its speed is

$v=(1.36)(56)=76.2 m/s$

B)

The centripetal acceleration of a point in uniform circular motion is given by

$a=\frac{v^2}{r}$

where

v is the linear speed

r is the distance of the point from the axis of rotation

In this problem, for the tip of the blade we have:

v = 76. 2 m/s is the speed

r = 56 m is the distance from the axis of rotation

Substituting, we find the centripetal acceleration:

$a=\frac{(76.2)^2}{56}=103.7 m/s^2$

D)

The force required to keep the 10 mg water droplet in circular motion on the dog's fur is equal to the centripetal force experienced by the droplet, therefore:

$F=m\frac{v^2}{r}$

where

m is the mass of the droplet

v is the linear speed

r is the distance from the centre of rotation

The data in this problem are

m = 10 mg = 0.010 kg is the mass of the droplet

v = 2.5 m/s is the linear speed

r = 16 cm = 0.16 m is the radius of the circular path

Substituting,

$F=(0.010)\frac{2.5^2}{0.16}=0.39 N$

E)

The weight of the droplet is given by

$F_g = mg$

where

m = 10 mg = 0.010 kg is the mass of the droplet

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$F_g = (0.010)(9.8)=0.098 N$

The force that keeps the droplet in circular motion instead is

F = 0.39 N

Therefore, the ratio between the two forces is

$\frac{F}{F_g}=\frac{0.39}{0.098}=4.0$

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