Question

Circular Motion, Kinetic Energy, Work and Power

Wind energy is the fastest growing renewable energy source, the US has a goal of producing 20 percent of it’s electricity from wind by 2030. All of our energy options come with a set of pros and cons of varying degrees. One concern, which is minor unless it lands on your house, with wind turbines is ice throw off from the tips of the blade. Consider a 50 meter radius wind turbine (about 3 MW) rotating clockwise 15 times a minute with a hub height of 150 meters. Calculate the following for a 100 kg piece of ice at the tip of the turbine blade (r=50m):

a) The angular velocity of the rotating blades in radians per second, the period, or time it takes for a rotor blade to make one revolution and the frequency, or number of rotations made in one second.

b) The magnitude of the force required at the top circular path to keep the piece of ice in circular motion.

c) The magnitude of the force required at the bottom of the circular path to keep the piece of ice in circular motion.

d) The range or distance the piece of ice would travel if it was released at the highest point in the rotation. Consider clockwise rotation so the ice lands in the positive x direction.

e) The kinetic energy of the piece of ice upon release.

f) The kinetic energy of the piece of ice upon landing.

g) The work done by the gravitational force from release to landing.

h) Is this the maximum range? Is this the most likely point of release? Briefly explain your reasoning.

Answer 1

a) angular veloicty=angular veloicty of turbine=15 rpm

=15*2*pi radians/60 seconds=1.57 rad/sec

period=2*pi/angular veloicty=4 seconds

frequency=1/period=1/4=0.25 Hz

b)magnitude of force=cetripetal force=mass*angular velicity^2*radius=12337 N

c)at the bottom , extra force has to be provided in order to support the weight of the ice

hence net force=12337+100*9.8=13317 N

d)when ever it leaves the leaves the turbine at the highest point,

it will travel with a horizontal speed =angular speed*radius=78.54 m/s

height before it reaches ground=50*2=100 m

time taken to reach ground=sqrt(2*height/g)=4.517 seconds
then distance travelled horizontally=78.54*4.517=354.64 m

e)kinetic energy=0.5*mass*speed^2=308426.58 J

f)upon landing, horizontal speed=78.54 m/s (as horizontal acceleration is 0)

vertical speed=sqrt(2*vertical acceleration*distance travelled)=44.27 m/s

then net speed=sqrt(78.54^2+44.27^2)=90.158 m/s

then kinetic energy=0.5*mass*speed^2=406426.58 J

g)work done by gravity=change in potential energy=mass*g*height=98000 J