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Solution for (a)

The rotational kinetic energy is

KE rot = 1 2 2 . size 12{"KE" rSub { size 8{"rot"} } = { {1} over {2} } Iω rSup { size 8{2} } } {}

We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find KE rot . The angular velocity ω size 12{ω} {} is

ω = 300 rev 1.00 min 2π rad 1 rev 1.00 min 60.0 s = 31.4 rad s . size 12{ω= { {"300"" rev"} over {1 "." "00 min"} } cdot { {2π" rad"} over {"1 rev"} } cdot { {1 "." "00"" min"} over {"60" "." 0" s"} } ="31" "." 4 { {"rad"} over {s} } } {}

The moment of inertia of one blade will be that of a thin rod rotated about its end, found in [link] . The total I size 12{I} {} is four times this moment of inertia, because there are four blades. Thus,

I = 4 Mℓ 2 3 = 4 × 50.0 kg 4.00 m 2 3 = 1067 kg m 2 . size 12{I=4 { {Mℓ rSup { size 8{2} } } over {3} } =4 times { { left ("50" "." 0" kg" right ) left (4 "." "00"" m" right ) rSup { size 8{2} } } over {3} } ="1067"" kg" cdot m rSup { size 8{2} } } {}

Entering ω size 12{ω} {} and I size 12{I} {} into the expression for rotational kinetic energy gives

KE rot = 0.5 ( 1067 kg m 2 ) 31.4 rad/s 2 = 5.26 × 10 5 J alignl { stack { size 12{"KE" rSub { size 8{"rot"} } =0 "." 5 left ("1067"" kg" cdot m rSup { size 8{2} } right ) left ("31" "." 4" rad/s" right ) rSup { size 8{2} } } {} #" "=5 "." "26" times "10" rSup { size 8{5} } " J" {} } } {}

Solution for (b)

Translational kinetic energy was defined in Uniform Circular Motion and Gravitation . Entering the given values of mass and velocity, we obtain

KE trans = 1 2 mv 2 = 0.5 1000 kg 20.0 m/s 2 = 2 . 00 × 10 5 J . size 12{"KE" rSub { size 8{"trans"} } = { {1} over {2} } ital "mv" rSup { size 8{2} } =0 "." 5 left ("1000"" kg" right ) left ("20" "." 0" m/s" right ) rSup { size 8{2} } =2 "." "00" times "10" rSup { size 8{5} } " J"} {}

To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is

2 . 00 × 10 5 J 5 . 26 × 10 5 J = 0.380 . size 12{ { {2 "." "00" times "10" rSup { size 8{5} } " J"} over {5 "." "26" times "10" rSup { size 8{5} } " J"} } =0 "." "380"} {}

Solution for (c)

At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies:

KE rot = PE grav size 12{"KE" rSub { size 8{"rot"} } ="PE" rSub { size 8{"grav"} } } {}

or

1 2 2 = mgh . size 12{ { {1} over {2} } Iω rSup { size 8{2} } = ital "mgh"} {}

We now solve for h size 12{h} {} and substitute known values into the resulting equation

h = 1 2 2 mg = 5.26 × 10 5 J 1000 kg 9.80 m/s 2 = 53.7 m . size 12{h= { { { size 8{1} } wideslash { size 8{2} } Iω rSup { size 8{2} } } over { ital "mg"} } = { {5 "." "26" times "10" rSup { size 8{5} } " J"} over { left ("1000"" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right )} } ="53" "." 7" m"} {}

Discussion

The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades—something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades.

The given figure here shows a helicopter from the Auckland Westpac Rescue Helicopter Service over a sea. A rescue diver is shown holding the iron stand bar at the bottom of the helicopter, clutching a person. In the other image just above this, the blades of the helicopter are shown with their anti-clockwise rotation direction shown with an arrow and the length of one blade is given as four meters.
The first image shows how helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades. The second image shows a helicopter from the Auckland Westpac Rescue Helicopter Service. Over 50,000 lives have been saved since its operations beginning in 1973. Here, a water rescue operation is shown. (credit: 111 Emergency, Flickr)

Making connections

Conservation of energy includes rotational motion, because rotational kinetic energy is another form of KE size 12{"KE"} {} . Uniform Circular Motion and Gravitation has a detailed treatment of conservation of energy.

How thick is the soup? or why don’t all objects roll downhill at the same rate?

One of the quality controls in a tomato soup factory consists of rolling filled cans down a ramp. If they roll too fast, the soup is too thin. Why should cans of identical size and mass roll down an incline at different rates? And why should the thickest soup roll the slowest?

The easiest way to answer these questions is to consider energy. Suppose each can starts down the ramp from rest. Each can starting from rest means each starts with the same gravitational potential energy PE grav size 12{ ital "PE" rSub { size 8{ ital "grav"} } } {} , which is converted entirely to KE , provided each rolls without slipping. KE , however, can take the form of KE trans size 12{ ital "KE" rSub { size 8{ ital "trans"} } } {} or KE rot size 12{ ital "KE" rSub { size 8{ ital "rot"} } } {} , and total KE is the sum of the two. If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. Thus, the can goes slower than it would if it slid down. Furthermore, the thin soup does not rotate, whereas the thick soup does, because it sticks to the can. The thick soup thus puts more of the can’s original gravitational potential energy into rotation than the thin soup, and the can rolls more slowly, as seen in [link] .

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Source:  OpenStax, Mechanics. OpenStax CNX. Apr 15, 2013 Download for free at http://legacy.cnx.org/content/col11506/1.2
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