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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/

**(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/

Calculate the change in length of the upper leg bone (the femur) when a **70.0 kg** man supports **62.0 kg** of his mass on it, assuming the bone to be equivalent to a uniform rod that is **45.0 cm** long and **2.00 cm** in radius.

Strategy

The force is equal to the weight supported:

and the cross-sectional area of the upper leg bone(femur) is:

To find the change in length we use the Young’s modulus formula. The Young’s modulus reference value for a bone under compression is known to be **9×10 ^{9} N/m^{2}**. Now,all quantities except

**ΔL**are known. Thus:

Discussion

This small change in length seems reasonable, consistent with our experience that bones are rigid. In fact, even the rather large forces encountered during strenuous physical activity do not compress or bend bones by large amounts. Although bone is rigid compared with fat or muscle, several of the substances listed in Table 5.3(*see reference below*) have larger values of Young’s modulus Y . In other words, they are more rigid.

**Reference:**

This worksheet is a modified version of Example 5.4 page 188 found in :

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/

Water hammer (or, more generally, fluid hammer) is a pressure surge or wave caused when a fluid (usually a liquid but sometimes also a gas) in motion is ... more

Factor of safety (FoS) or (FS), is a term describing the structural capacity of a system beyond the expected loads or actual loads. Essentially, how much ... more

In lead climbing using a dynamic rope, the fall factor (f) is the ratio of the height (h) a climber falls before the climber’s rope begins to stretch ... more

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A typical small rescue helicopter, like the one in the Figure below, has four blades, each is

4.00 mlong and has a mass of50.0 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 of1000 kg.(a)Calculate the rotational kinetic energy in the blades when they rotate at300 rpm.(b)Calculate the translational kinetic energy of the helicopter when it flies at20.0 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?