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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] .

Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Kate Reply
what is the change in momentum of a body?
Eunice Reply
what is a capacitor?
Raymond Reply
Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
Maria Reply
please solve
Sharon
8m/s²
Aishat
What is Thermodynamics
Muordit
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
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50 m/s due south east
Someone
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
Ramon Reply
I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
Someone
about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
Someone
definitely of physics
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what is field
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physics, biology and chemistry this is my Field
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field is a region of space under the influence of some physical properties
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what is ogarnic chemistry
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determine the slope giving that 3y+ 2x-14=0
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Another formula for Acceleration
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a=v/t. a=f/m a
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innocent
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pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
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Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
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No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Robert
like charges repel while unlike charges atttact
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What is specific heat capacity
Destiny Reply
Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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