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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

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In a video game design, a map shows the location of other characters relative to the player, who is situated at the origin, and the direction they are facing. A character currently shows on the map at coordinates **(-3, 5)**. If the player rotates counterclockwise by **20 degrees**, then the objects in the map will correspondingly rotate **20 degrees** clockwise. Find the new coordinates of the character.

To rotate the position of the character, we can imagine it as a point on a circle, and we will change the angle of the point by **20 degrees**. To do so, we first need to find the radius of this circle and the original angle.

Drawing a right triangle inside the circle, we can find the radius using the Pythagorean Theorem:

To find the angle, we need to decide first if we are going to find the acute angle of the triangle, the reference angle, or if we are going to find the angle measured in standard position. While either approach will work, in this case we will do the latter. By applying the cosine function and using our given information we get

While there are two angles that have this cosine value, the angle of **120.964** degrees is in the second quadrant as desired, so it is the angle we were looking for.

Rotating the point clockwise by **20 degrees**, the angle of the point will decrease to **100.964 degrees**. We can then evaluate the coordinates of the rotated point

For **x** axis:

For **y** axis:

The coordinates of the character on the rotated map will be **(-1.109, 5.725)**

Reference : PreCalculus: An Investigation of Functions,Edition 1.4 © 2014 David Lippman and Melonie Rasmussen

http://www.opentextbookstore.com/precalc/

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Find the terminal velocity of an **85-kg** skydiver falling in a spread-eagle position.

where **Vt** is the terminal velocity,
**m** is the mass of the skydiver,
**g** is the acceleration due to gravity,
**C _{d}** is the drag coefficient,

**ρ**is the density of the fluid through which the object is falling, and

**A**is the projected area of the object.

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/

where h is skydiver height and w the width at “spread-eagle” position

**(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/

In the wheelbarrow of the following figure the load has a perpendicular lever arm of **7.50 cm**, while the hands have a perpendicular lever arm of **1.02 m**.**(a)** What upward force must you exert to support the wheelbarrow and its load if their combined mass is **45.0 kg**? **(b)** What force does the wheelbarrow exert on the ground?

**(a)** In the case of the wheelbarrow, the output force or load is between the pivot and the input force. The pivot is the wheel’s axle. Here, the output force is greater than the input force. Thus, a wheelbarrow enables you to lift much heavier loads than you could with your body alone. **(b)** In the case of the shovel, the input force is between the pivot and the load, but the input lever arm is shorter than the output lever arm. The pivot is at the handle held by the right hand. Here, the output force (supporting the shovel’s load) is less than the input force (from the hand nearest the load), because the input is exerted closer to the pivot than is the output.

Strategy

Here, we use the concept of mechanical advantage.

Discussion

An even longer handle would reduce the force needed to lift the load. The MA here is:

http://openstaxcollege.org/textbooks/college-physics

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Calculate the force the biceps muscle must exert to hold the forearm and its load as shown in the figure below, and compare this force with the weight of the forearm plus its load. You may take the data in the figure to be accurate to three significant figures.

**(a)** The figure shows the forearm of a person holding a book. The biceps exert a force **F _{B}** to support the weight of the forearm and the book. The triceps are assumed to be relaxed.

**(b)**Here, you can view an approximately equivalent mechanical system with the pivot at the elbow joint

Strategy

There are four forces acting on the forearm and its load (the system of interest). The magnitude of the force of the biceps is **F _{B}**, that of the elbow joint is

**F**, that of the weights of the forearm is

_{E}**w**, and its load is

_{a}**w**. Two of these are unknown

_{b}**F**, so that the first condition for equilibrium cannot by itself yield

_{B}**F**. But if we use the second condition and choose the pivot to be at the elbow, then the torque due to

_{B}**F**is zero, and the only unknown becomes

_{E}**F**.

_{B}Solution

The torques created by the weights are clockwise relative to the pivot, while the torque created by the biceps is counterclockwise; thus, the second condition for equilibrium (net **τ = 0**) becomes

Note that **sin θ = 1** for all forces, since **θ = 90º** for all forces. This equation can easily be solved for **F _{B}** in terms of known quantities,yielding. Entering the known values gives

which yields

Now, the combined weight of the arm and its load is known, so that the ratio of the force exerted by the biceps to the total weight is

Discussion

This means that the biceps muscle is exerting a force **7.38** times the weight supported.

http://openstaxcollege.org/textbooks/college-physics

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Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The only known force a planet exerts on Earth is gravitational.

**(a)** Calculate the gravitational force exerted on a **4.20 kg** baby by a **100 kg** father **0.200 m** away at birth (he is assisting, so he is close to the child).

**(b)** Calculate the force on the baby due to Jupiter if it is at its closest distance to Earth, some **6.29e+11 m** away. How does the force of Jupiter on the baby compare to the force of the father on the baby?

Father’s gravitational force on the baby is:

Jupiter’s gravitational force on the baby is:

**(c)** What should be the father’s weight, so that he exerts the same force on the baby as that of Jupiter? *****this section is not included in the Reference material*

Discussion

Other objects in the room and the hospital building also exert similar gravitational forces. (Of course, there could be an unknown force acting, but scientists first need to be convinced that there is even an effect, much less that an unknown force causes it.)

http://openstaxcollege.org/textbooks/college-physics

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*Dedicated to little Konstantinos*

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/

In physics, a force is any influence which tends to change the motion of an object.In other words, a force can cause an object with mass to change its ... more

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.

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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?