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1st Equation of Motion - Linear Velocity

In mathematical physics, equations of motion are equations that describe the behaviour of a physical system in terms of its motion as a function of ... more

1st Equation of Motion for Rotation - Angular Velocity

In mathematical physics, equations of motion are equations that describe the behaviour of a physical system in terms of its motion as a function of ... more

2nd Equation of Motion - Final Position

In mathematical physics, equations of motion are equations that describe the behaviour of a physical system in terms of its motion as a function of ... more

2nd Equation of Motion for Rotation - Final Angle

In mathematical physics, equations of motion are equations that describe the behaviour of a physical system in terms of its motion as a function of ... more

Acceleration

Acceleration, in physics, is the rate at which the velocity of an object changes over time. Mathematically, instantaneous acceleration—acceleration over an ... more

Angular Acceleration

Torque, moment, or moment of force is the tendency of a force to rotate an object about an axis, fulcrum, or pivot.
Moment of inertia is the mass ... more

Angular Frequency

In physics, angular frequency ω (also referred to by the terms angular speed, radial frequency, circular frequency, orbital frequency, radian frequency, ... more

Angular Momentum

In physics, angular momentum, moment of momentum, or rotational momentum is a measure of the amount of rotation an object has, taking into account its ... more

Area of a Cylinder

A cylinder is a geometric shape, that its surface is forming by the points at a fixed distance from a given line segment, the axis of the cylinder. The ... more

Area of a Sphere

A sphere is a perfectly round geometrical and circular object in three-dimensional space that resembles the shape of a completely round ball. A sphere is ... more

Centripetal Force

Centripetal force (from Latin centrum “center” and petere “to seek”) is a force that makes a body follow a curved path: its ... more

Cosine function

The trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its ... more

Coulomb's law

Coulomb’s law, or Coulomb’s inverse-square law, is a law of physics describing the electrostatic interaction between electrically charged ... more

Density

The density of a material is defined as its mass per unit volume. For a pure substance the density has the same numerical value as its mass concentration. ... more

Disk area

Calculation of the area of a disc. The area of a disk is half its circumference times its radius or the product of the constant π (the constant ratio of ... more

Elastic Potential Energy

According to Hooke’s Law, Elastic potential energy is stored in a simple harmonic oscillator at position x,for example, the energy saved in an object ... more

Electric Current

An electric current is a flow of electric charge. In electric circuits this charge is often carried by moving electrons in a wire. It can also be carried ... more

Electric power

The electric power produced by an electric current “ I “ passing through an electric potential (voltage) difference of “ V “

... more

Electrical resistance

Electrical resistance of an electrical conductor is the opposition to the passage of an electric current through that conductor and depends on: What ... more

Worksheet 289

Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust T , for the four-rocket propulsion system shown in the Figure below. The sled’s initial acceleration is 49 m/s 2, the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N.

A sled experiences a rocket thrust that accelerates it to the right.Each rocket creates an identical thrust T . As in other situations where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force N on the system that is equal in magnitude and opposite in direction to its weight,w.The system here is the sled, its rockets, and rider, so none of the forces between these objects are considered. The arrow representing friction ( f ) is drawn larger than scale.
Assumptions: The mass of the Sled remains steady throughout the operation

Strategy

Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure.

Solution

Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with

Force (Newton's second law)

Fnet is the net force along the horizontal direction, m is the rocket’s mass and a the acceleration.

We can see from the Figure at the top, that the engine thrusts add, while friction opposes the thrust.

Subtraction

Tt is the total thrust from the 4 rockets, Fnet the net force along the horizontal direction and Ff the force of friction.

Finally, since there are 4 rockets, we calculate the thrust that each one provides:

Division

T is the individual Thrust of each engine, b is the number of rocket engines

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/

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/

Worksheet 334

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:

Pythagorean theorem (right triangle)

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

Cosine function
Subtraction

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:

Cosine function

For y axis:

Sine function

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/
Creative Commons License : http://creativecommons.org/licenses/by-sa/3.0/us/

Worksheet 290

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

Terminal Velocity (without considering buoyancy)
Rectangle area

where Vt is the terminal velocity, m is the mass of the skydiver, g is the acceleration due to gravity, Cd 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

Worksheet 296

(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 :

Density

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:

Density

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

Force (Newton's second law)

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:

Density

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

Force (Newton's second law)

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/

Worksheet 302

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.

Force (Newton's second law)
Mechanical Advantage - Law of Lever
Subtraction

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

Division

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/

Worksheet 306

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 FB 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 FB, that of the elbow joint is FE, that of the weights of the forearm is wa , and its load is wb. Two of these are unknown FB, so that the first condition for equilibrium cannot by itself yield FB . But if we use the second condition and choose the pivot to be at the elbow, then the torque due to FE is zero, and the only unknown becomes FB .

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

Force (Newton's second law)
Torque
Force (Newton's second law)
Torque

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

Mechanical equilibrium - 3=3 Torque example

which yields

Torque
Addition

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

Division

Discussion

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

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/

Worksheet 308

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:

Newton's law of universal gravitation

Jupiter’s gravitational force on the baby is:

Newton's law of universal gravitation
Division

(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

Newton's law of universal gravitation

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.)

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/

Dedicated to little Konstantinos

Worksheet 316

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:

Force (Newton's second law)

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

Disk area

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 9×109 N/m2. Now,all quantities except ΔL are known. Thus:

Young's Modulus

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/

Force (Newton's second law)

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

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

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