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Worksheet 341

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.

Division
Potential energy

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.

Multiplication

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

Multiplication

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

Multiplication
Multiplication

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

Division
Worksheet 292

What is the cost of running a 0.600-kW computer 6.00 h per day for 30.0 d if the cost of electricity is \$0.120 per kW ⋅ h ?

Multiplication

where t is the total consumption time, td is the days of consumption and th the hours of consumption per day

Power-average (related to Work)

where P is Power consumption rate, E is the energy supplied by the electricity company and t is consumption time

keywords:
ballistics

Multiplication

where C is the total cost and CkW is the cost per kilowatt hour

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/

Software equation

In the study of software project estimation, the Software Equation is a model with multiple variables based on assumptions of a specific distribution of ... more

Worksheet 333

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 Er . The angular velocity ω for 1 r.p.m is

Angular velocity

and for 300 r.p.m

Multiplication

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

Moment of Inertia - Rod end

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

Multiplication

and so The rotational kinetic energy is

Rotational energy

Solution for (b)

Translational kinetic energy is defined as

Kinetic energy ( related to the object 's velocity )

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

Division

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:

Potential energy

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/

Richardson's law

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Gyromagnetic ratio for a nucleus

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Bend allowance

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Reserve factor as function of the ultimate strength and load

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Darcy friction factor

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Darcy friction factor - Blasius correlation with correction for curved or helically coiled tubes

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Nγ bearing capacity factor (simplified)

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