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alukav5142 [94]
3 years ago
9

Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a

great amount of angular momentum when rotating. A wind turbine has a total of 3 blades. Each blade has a mass of m = 5500 kg distributed uniformly along its length and extends a distance r = 46 m from the center of rotation. The turbine rotates with a frequency of f = 11 rpm.
Required:
a. Calculate the total moment of inertia of the wind turbine about its axis, in units of kilogram meters squared.
b. Calculate the angular momentum of the wind turbine, in units of kilogram meters squared per second.
Physics
1 answer:
Anton [14]3 years ago
4 0

Answer:

Explanation:

moment of inertia of each blade which is similar to rod rotating about its one end

= 1/3 ml²

moment of inertia of 3 blades = ml²

= 5500 x 46²

I = 11638 x 10³ kg m²

angular velocity = 2πn where n is rotation per second

n = 11 / 60

angular velocity = 2π x 11/60

= 1.1513 rad /s

angular momentum

= moment of inertia x angular velocity

=  11638 x 10³ x 1.1513

= 13399 x 10³ kg m² per second.

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A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule
ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

p_i = 0

The final total momentum is instead:

p_f = m_a v_a + m_c v_c

where

m_a = 125 kg is the mass of the astronaut

v_a = 2.50 m/s is the velocity of the astronaut

m_c = 1900 kg is the mass of the capsule

v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

so

m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

F \Delta t = \Delta p

The change in momentum of the astronaut is

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F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N

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(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J

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