Search results

A typical small rescue helicopter, like the one in the Figure below, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it?

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)

Strategy

Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy.

Solution for (a)

We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find E_r . The angular velocity ω for 1 r.p.m is

and for 300 r.p.m

The moment of inertia of one blade will be that of a thin rod rotated about its end.

The total I is four times this moment of inertia, because there are four blades. Thus,

and so The rotational kinetic energy is

Solution for (b)

Translational kinetic energy is defined as

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

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:

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.

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/

The awe‐inspiring Great Pyramid of Cheops was built more than 4500 years ago. Its square base, originally 230 m on a side, covered 13.1 acres, and it was 146 m high (H), with a mass of about 7×10^9 kg. (The pyramid’s dimensions are slightly different today due to quarrying and some sagging). Historians estimate that 20,000 workers spent 20 years to construct it, working 12-hour days, 330 days per year.

a) Calculate the gravitational potential energy stored in the pyramid, given its center of mass is at one-fourth its height.

b) Only a fraction of the workers lifted blocks; most were involved in support services such as building ramps, bringing food and water, and hauling blocks to the site. Calculate the efficiency of the workers who did the lifting, assuming there were 1000 of them and they consumed food energy at the rate of 300 Kcal/hour.

first we calculate the number of hours worked per year.

then we calculate the number of hours worked in the 20 years.

Then we calculate the energy consumed in 20 years knowing the energy consumed per hour and the total hours worked in 20 years.

The efficiency is the resulting potential energy divided by the consumed energy.

(a) Calculate the buoyant force on 10,000 metric tons (1.00×10 7 kg) of solid steel completely submerged in water, and compare this with the steel’s weight.

(b) What is the maximum buoyant force that water could exert on this same steel if it were shaped into a boat that could displace 1.00×10 5 m ³ of water?

Strategy for (a)

To find the buoyant force, we must find the weight of water displaced. We can do this by using the densities of water and steel given in Table [insert table #] We note that, since the steel is completely submerged, its volume and the water’s volume are the same. Once we know the volume of water, we can find its mass and weight

First, we use the definition of density to find the steel’s volume, and then we substitute values for mass and density. This gives :

Because the steel is completely submerged, this is also the volume of water displaced, Vw. We can now find the mass of water displaced from the relationship between its volume and density, both of which are known. This gives:

By Archimedes’ principle, the weight of water displaced is m w g , so the buoyant force is:

The steel’s weight is 9.80×10 7 N , which is much greater than the buoyant force, so the steel will remain submerged.

Strategy for (b)

Here we are given the maximum volume of water the steel boat can displace. The buoyant force is the weight of this volume of water.

The mass of water displaced is found from its relationship to density and volume, both of which are known. That is:

The maximum buoyant force is the weight of this much water, or

Discussion

The maximum buoyant force is ten times the weight of the steel, meaning the ship can carry a load nine times its own weight without sinking.

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/