UNIT 3: CIRCULAR MOTION, WORK AND ENERGY

Unit 3A Circular Motion – Acceleration with Changes in Direction

Acceleration has been defined as the rate of change of velocity. Velocity has been defined as a vector with a magnitude (speed) and a

direction.

So far, we have considered acceleration only for objects slowing down or speeding up, ie., changes in the speed component of velocity. But objects can be accelerating without changing speeds. When objects are going at a constant speed but continually change their direction, they are accelerating as well. The result of this acceleration is circular motion.

Period T and Frequency f

The time it takes for one complete revolution of circular motion is called the period and given a special symbol T. The inverse of this time period T is the frequency f, which gives the number of revolutions in one second. The units for frequency are Hertz Hz, where 1 Hz = 1 revolution/second.

Example 1: A 30 cm vinyl record revolves on a turntable that "turns" a record around and around at 33 1/3 RPM (revolutions per minute)

Example: What is the period and frequency of the 33 1/3 RPM record above?

Since 33 1/3 RPM refers to 33 1/3 revolutions per minute

=f

(per 1 revolution)

Example: Old vinyl records were separated into two categories: albums or LP’s (for long play) which played at 33 1/3 RPM, and “singles” with one song on an A side and another song on the B side, playing at 45 RPM.

a) What is the period of a single revolving on a turntable?

For one revolution, the needle covers the distance around the perimeter of the record's circle, d = 2 R, where R is the radius of the circle.

For circular motion, the formula for speed v of an object at the perimeter of the circle is a special application of the constant speed formula v = d/t

Going back to the record question: (45RPM for a single)If the diameter of a single is 13.0 cm, what is the speed of the turntable needle as it revolves at the outer tip of the record?

Here v represents the constant speed of an object undergoing circular motion, its direction always at right angles to a line through the centre of the circle or tangential to its speed at any point.

Acceleration in Circular Motion: Centripetal Acceleration

If the speed is not changing in circular motion, how does an object accelerate in circular motion? To understand circular motion acceleration, let us freeze the motion of an object in circular motion at several instants in its circular path.

Two things to note: 1) while the speed v is constant, the direction of v is constantly changing, and 2) the direction of v is always perpendicular (at right angles) to the line from the center of the circle (tangential to the circle). Following Newton's First Law (inertia), the object at any instant will want to keep going at its constant velocity, which includes its constant speed and constant direction. It keeps going at its constant speed, but it is constantly changing direction, bending towards the circular path, so it is accelerating.

If we could draw a vector indicating which direction v is changing, the arrow would point directly to the center of the circle.

This then is the direction of the acceleration: the direction of acceleration in circular motion is always directed towards the center of the

circle. The term for this acceleration is the centripetal ("centre-seeking") acceleration, ac, given by the formula

where v refers to the constant speed and r refers to the radius of the circular orbit.

Note that this formula only gives the magnitude of the centripetal acceleration as a function of the speed v. The direction of the acceleration is always towards the center of the circle. We can extend this formula by substituting the speed formula v = 2R/T to get

ac = (2 R/T) 2 R

= 4 2 R T2

Example 2: Calculate the magnitude and direction of the acceleration of the turntable needle at the outside of the 33 1/3 RPM (R=15 cm) record.

Derivation of Centripetal Acceleration Formula (*optional)

Consider a velocity vector v1 originating from a point P1 on a circle of radius R. The vector is of magnitude v and in the direction perpendicular to the line from the centre of the circle C. Now consider another point P2 on the circle a distance from P1. A line from C to P2 will also be of magnitude R and set at an angle from the line from C to P2.

We can draw a new vector v1 (in dashed lines above) originating from P2, parallel to the original v1 vector. By similar angles this new vector v1 will be set at an angle = 90o – from the line from C to P2.

At point P2 we can insert a second velocity vector v2, again of magnitude v and now at right angles to the line from C to P2. To get a change in velocity vector v, we can add the negative of vector v1 tail-to-tip to the second vector v2. The resultant is v = v2 + - v1.

If we denote the displacement from P1 to P2 as the change in displacement d, we get a triangle C-P1-P2. By similar angles, the angle between v1 and v2 is also . Since v1 and v2 are of the same magnitude v and angle between them, and the lines C to P and C to P are both of the same magnitude R with the same angle between them, we have two similar triangles. This allows us to state:

v = d v R

For small angle , we can approximate the displacement d = v t. Substituting in the above, and rearranging, we get

v = v t v R

v = v (v)t R

Since by definition, acceleration is the change in velocity over the change in time, then

ac = v 2 R

Centripetal Force

If there is acceleration, then, by Newton's 2nd Law (F=ma), there must be a net force accompanying centripetal acceleration in circular motion. This net force is directed towards the center of the circle like the acceleration, and appropriately named the centripetal force.

centripetal force can be caused by; the tension of the string or rope when an object is whirled around in a

circle on a string the friction of the wheels on the road when a car is rounding an

unbanked curve the electric force of attraction between an electron circling a proton. gravity toward a large central body

Without the force – tension, friction or electrostatic - the object would continue going in its straight path. The centripetal force causes the centripetal acceleration, in this case a change in direction of v towards the center of the circle. The constant change in the direction of v causes the object to follow circular motion, instead of a straight line path.

Centripetal force is the net force. The formula for centripetal force, Fc, derives directly from centripetal acceleration and Newton's 2nd Law:

Again, we can substitute the formula for v in circular motion to get the expanded formula for centripetal force

Example 3: Calculate the centripetal force on the needle of the record player in the above 33 1/3 RPM record example if the needle weighs 1.5 grams.

Horizontal Circular Motion

On the surface of the earth, when an object is undergoing circular motion along the horizontal (flat), since the force of gravity is perpendicular to the centripetal force, there is no interaction- the net force is the centripetal force causing circular motion.

Example 4: A 10 kg hammer is swung horizontally around and around on a 2.0 meter radius chain (including arm length) at a constant speed of 20 m/s.

1) What is the centripetal acceleration of the hammer?

2) What is the tension on the chain?

3) What horizontal direction will the hammer travel when the chain is released?

a) outward because of the centrifugal forceb) inward because of the centripetal forcec) in a straight line along the present v direction because of its inertiad) continue along its circular path as it was doing before

4) Assuming the hammer is released from the horizontal, at a height of 1.5 m,how far will the hammer travel horizontally before it hits the ground?

Centrifugal Force

Centripetal force is a real force. The tension in a string whirling horizontally, the force of the track on the roller coaster with or against gravity, or the friction between the road and a car as it rounds a curve: these are all examples of centripetal force.

Unfortunately, we do not feel centripetal force as a normal push or pull on us, like other forces. For example, what do you feel as you round a bend? Imagine you are driving in a car and you are turning right around a corner. The centripetal force is directed to the right. Yet, you feel yourself pushed to the left. Why? Likewise, in a merry-go-round, or like the amusement park ride where you stand back against a wall that circles around and around, until the floor drops off beneath your feet. The centripetal force is directed inward, yet you find yourself being pushed outward, to the point where you find yourself pinned against the outer wall and defying gravity in the amusement park ride. Why is the direction of the force we feel opposite to the direction of the centripetal force?

Imagine you are standing in an elevator, accelerating upward. What do you feel? While the tension on the cable lifting the elevator is directed upward, you feel as if you are being pushed down the elevator floor. This is very much like the feeling you get in circular motion. Pretend, that instead of accelerating upward, the elevator is instead circling horizontally at the end of a large rope. The elevator is still accelerating "upward", but now it is a centripetal acceleration - and the centripetal force/tension on the rope causing this elevator's acceleration is directed towards the centre of a circle. Inside the elevator it would still feel like you were being pushed downward, to the floor. This is very similar to the case of the amusement park ride where you feel as if you are being pushed outward. The floor of the elevator corresponds to the outside wall of the circle.

In actuality, what you are feeling is called a pseudo-force (pseudo = false). In both cases above, what is happening is that you are moving at constant velocity and your inertia makes you want to stay moving at whatever speed and direction you're moving. The elevator and the object, on the other hand, are accelerating. The elevator is accelerating "upwards" and you - wanting to stay still - you feel as if you are being pushed "downward", but it is the elevator that is actually pushing upwards at you. Likewise in the circling object, inside the object, you want to keep moving in a straight-line path. The object, though, is accelerating inward. The outer wall of the object is pushing inward at you, not you pushing outward at all. The feeling of pushing outward opposite to centripetal force in circular motion seems real, though, and it is given a special name, called the centrifugal force. Centrifugal force is a pseudo-force. (Newton’s 3rd Law)

Simulating Gravityhttp://www.wired.com/wiredscience/2013/05/gravity-in-the-elysium-space-station/

The force of gravity holds us to the surface of the earth. In space travel there is no gravity, but we can simulate gravity through circular motion. Speculation has space stations in the future built as giant wheels, revolving around its own central axis. If the space station was revolving at the appropriate rate, the centrifugal force would mimic the force of gravity on the outside rim of the wheel.

Example 7: Let us put the radius of a circular space station at 50 meters. What would be the speed needed in meters per second and revolutions per minute to mimic the force of gravity here on the surface of the earth? (hint: start with ac = g).

(22 m/s or 4.2 rpm)

Inside the revolving space station, an interesting situation arises in terms of the forces. To an observer inside the revolving space station, the situation is identical to the surface of the earth: she feels a force of gravity downward, balanced by a normal force the floor of the space station is exerting back at her. She feels like she is not moving at all, like standing on earth, where the forces are balanced.

Newton- The Great Thinker (*optional)

There is a story about Newton’s discovery of the universal law of gravitation while sitting in his garden one day. Back in Newton’s early days at Cambridge university, the Black Plague (bubonic plague carried by rats) swept over Europe, wiping out a large percentage of the population. In 1466, the university was forced to shut down and Newton went home, back to his mother’s farm in Lincolnshire. It was during this time that he formulated most of his important contributions to mathematics and physics including the binomial theorem, differential calculus, vector addition, the laws of motion, centripetal acceleration, optics, and universal gravitation. Newton was not even 22 at the time.

Newton’s biographer writes in his memoirs that an apple falling from a tree provided Newton his inspiration for the law of universal gravitation. As Newton thought about the apple and its fall, he began to see the relationship between the similarity between the moon and the apple’s motions. Might not the motion of the moon and the apple be caused from the same thing?

“ I began to think of gravity extending to the orb of the moon, and … from Kepler's rule I deduced that the forces which keep the Planets in their orbs must be reciprocally as the squares of their distances from the centres about which they revolve: and thereby compared to the force required to keep the moon in her orb with the force of gravity at the surface of the earth, and found them to answer pretty nearly. All this was in the two plague years of 1665 and 1666, for in those days I was in the prime of my age for invention, and minded mathematics and philosophy more than at any time since…”

We can attempt to recreate Newton’s derivation of the force relationships between the apple on earth and the moon as an example of the use of proportions rather than equations to solve a physics problem. Knowing only the radius of the earth (RE = 6.37 x 108 m) and approximating the orbital radius of the moon as 60 times RE, Newton was able to arrive at a very close approximation to the period of the moon, verifying the inverse square relationship.

For an object in circular motion due to the force of gravity, we start with the centripetal force Fc (force causing circular motion- see next unit) being due to the gravitational force Fg.

This inverse square relation gives an approximation to the force of gravity on the moon relative to the force of gravity on an apple on the surface of the earth. If the distance R to the moon is approximately 60 times the radius of the earth, then the force of gravity on the moon is approximately 1/60 squared or 1/3600 as compared to the apple on earth.

Again, starting with the first principle that the gravitational force is the centripetal force:

This equation allows Newton to estimate the period T of the moon’s orbit. Substituting the new radius of the moon as 60 times the radius on earth, and the gravitational field on the moon as 1/60 squared times the acceleration due to gravity on earth, we can obtain a very close approximation of the period of the moon’s orbit.

The true value for the moon’s orbit is 27.3 days. It should be remembered that Newton was performing these calculations in the days without computers or calculators.

Newton was a great thinker but rather eccentric, and often mean and vindictive. He did not bother to publish many of his works or ideas, often until much later or when he was goaded to by friends like Edmund Halley (of Halley’s comet). Yet when other scientists or mathematicians claimed credit for a discovery, he was merciless in denouncing them, as with his lifelong attacks of Robert Hooke (of Hooke’s Law and microscope fame) and Gottfried Liebniz (co-discoverer, with Newton, of calculus). In his later years he became quite famous and was appointed Warden of the Royal Mint, a lucrative and important post. One of his duties there was to chase down counterfeiters and Newton managed to convict many prominent ones who were then tried, executed by hanging then drawn and quartered. He is reported to have died a virgin.

Kepler's Laws of Planetary Motion

Kepler's 1st Law states that planetary orbits are ellipses.

Kepler's 2nd Law states that equal areas subtended by a string from the sun around a planet's orbital path will correspond to equal periods of time. This means that planets move at different speeds along different parts of the elliptical orbit.

Kepler's 3rd Law, which states the mathematical relationship between the periods of orbit, T, and the radius of orbit, R, is given by:

Satellites in Orbit

Newton applied his ideas of forces to orbiting bodies. He said that an object in orbit must have a centripetal force that is equal to the force causing it to orbit (force of gravity).

Fc = Fg

The orbiting body must maintain a particular speed that would make the forces equate. If the speed varies, then the orbiting body would not maintain its same orbital radius. (This explains Kepler’s law of changing speed and radius for a planet at different distances from the sun in it’s elliptical orbit.)

Geosynchronous /Geostationary Satellites

Lastly, satellites can be positioned so that they follow the rotation of the earth (T = 24 hours). These geosynchronous or geostationary satellites are fixed hovering over one point of the earth, which is advantageous for instance in tv satellites. Because of its fixed position above a point on the earth, your tv satellite dish need not move while tuned to one satellite station. Like pearls on a string, a large number of tv satellites are positioned above the equator to allow for the greatest range covering the surface of the earth.

Example 16:Calculate the height above the earth necessary to plant a satellite in geosynchronous orbit above Ecuador.

G forces

Pilots and astronauts have to withstand tremendous g-forces sometimes in their line of work. G-forces refers to multiples of the acceleration due to gravity, g. For example, 3 G's refers to 3 times the acceleration due to gravity, and your weight would then be 3 times the normal weight you would feel if you were standing on the surface of the earth.

To test and train pilots ability to withstand G-forces, a training machine using circular motion creates artificial g-forces. The pilot is seated on the outer edge of the circle, facing inwards and the entire apparatus starts revolving.

Using our formulae, we can quite readily calculate the speed needed for any particular G-force. For instance, to calculate the speed needed to produce 1G of force (FN = 1xFg ):

Since Fc = FN ( =__xFg )

mv 2 = 1x m g R

v = ( R g )1/2

Exanple 9: Calculate the speed needed to revolve a 5.0 m radius G-force trainer to produce 6.0 G's of force.

(17 m/s)

Simulated Weightlessness

Conversely, for space travel, astronauts must get used to weightlessness. To train future astronauts for weightlessness, a special aircraft is flown in a circular arc close enough to earth’s surface that g=9.81m/s2) which simulates weightlessness and provides a brief training time in each arc. Fg on the astronaut is balanced by the inertial force(centrifugal) that is created by Fc. (see workbook page 264)

(this airplane: dubbed the “Vomit Comet” was used for the "space" sequences in the movie, Apollo 13).

Example 10:What circular radius must a weightlessness(FN =0) simulator aircraft with a speed of 1000 km/h have to fly? [Fc (centrifugal) is balanced with Fg]

(7.87 km)

Vertical Circular Motion

On the surface of the earth, when an object is undergoing circular motion along the vertical, because the force of gravity is acting along the vertical, we have to add the force of gravity Fg to any tension T or FN, their vector sum being the net or centripetal force Fc.

Example 5: A 1000 N man rides a roller coaster with a vertical loop of radius 10 m. A radar gun clocks the roller coaster at a constant speed of 35.66 km/h around the loop.

1) Sketch a free-body diagram of the roller coaster at the top of the loop. What is the apparent weight of the man if he stands on a Newton

scale?

At the top, since the apparent weight and gravity are in the same direction as the net force Fc, then

Fc =

2) Sketch a free-body diagram of the roller coaster at the bottom of the loop. What is the apparent weight of the man if he stands on a Newton

scale?

At the bottom, since the tension is the same direction as the (centripetal) net force Fc, but the force of gravity is in the opposite (negative) direction, then

Fc =

Example 6: What is the tension at the top in a 1.0 m string with a 10 g eraser circling vertically at 10.0 m/s?

What speed would T = 0?

What would happen to the object if you swing it at a speed below that speed?

Banked Curves (*optional)

Friction between the road and car tires is what supplies the centripetal force when cars round a curve. If the friction is not great enough, the cars would continue in their straight line paths according to their current inertia. This is a real problem road engineers must worry about, especially here in Canada when the roads get icy. Besides getting cars to slow down before the curve (reducing the centripetal force needed), road engineers can use gravity to their favour and bank curves sloping down towards the center of the giant circle that the curve represents.

From the above, the formula for banked curves is:

Example 11:The Indianapolis Motor Speedway is a 2.5 mile oval track, home to the world famous Indianapolis 500, where cars can reach speeds around 240 mph (400 km/h). Supposing turn No. 3 is banked 9.20 to the horizontal to handle the tremendous speed. Assuming there is no friction and no slowing down in the turn, what is the radius of the circle turn No. 3 inscribes at 400 km/h?

(7.8 km)

Unit 3B- Work and Energy

Energy is Work

Historically, the concept of energy arose from the industrial revolution and the study of machines doing work for us (mechanical energy). Energy is defined strictly in physics as the ability to do work.

Energy = Work

E = W

The unit for energy and work is the same: the Joule (J).

The formula for work, like all previous formulae so far, derives from a graph: here the force-displacement graph. Work is always equal to the area under the force-displacement (F vs d) graph. For a constant or average force, the area under the F vs d graph is the area of a rectangle of length F and width d.

W = area of a rectangle = length x width = Fconstant x d

This gives the general formula for work with an average or constant force.

W = F d work with a constant or average force

It is important to realize that work – and energy – although derived from vectors like force and displacement is a scalar quantity (no directions). The units for work or energy is the Joule, equivalent to the force of 1 Newton across a distance of 1 metre.

1 J = 1 N m

Usually, the force under consideration is the man-made or applied force, T (FT).

Work Needs a Distance

Example 1. I hold a 100 g coffee cup in the air without moving. How much work am I doing on the coffee cup?

Start with a free-body diagram. The work being done is from the applied force T opposing the force of gravity Fg. By Newton’s 2nd Law:

But since displacement is zero (no motion)

W = T x d= Fg x 0= 0

In general, if there is no displacement, the net work is zero.

Work Done Against a Force (of Gravity)

Example 2: I lift a 100 g coffee cup in the air 1.00 m at constant velocity. How muchwork am I doing on the coffee cup? Again, by Newton’s 2nd Law:

Note that the same forces for the coffee cup at rest apply for the constant velocity case as well. The difference is that the constant velocity case has a displacement which then yields the work done.

W = T x d= Fg x d= (mg) h= (0.100 kg) (9.81 m/s2) 1.00m= 0.981 J(almost 1 joule of work! Phew!)

Work Done to Accelerate

Example 3: I move this 100 g coffee cup from rest a distance of 1.00 m across this table with an average force of 1.00 N.

a) What is the work being done on the coffee cup?

Note that in this case there is no work being done to overcome a force like the force of gravity, only work being done to accelerate the object.

W = = =

(1.00 J)

b) Assuming the table is frictionless, what would be the final speed of the coffee cup after 1.00 m?

(4.47 m/s)

In general, there are two basic types of work:

1) Work done to overcome a force like the force of gravity Fg

2) Work done to accelerate an object (F).

Which work, if any or both, applies to any particular problem depends on what the relevant force - usually the applied force FA – does.

Work Done by a Force at an Angle to the Displacement

Remembering that work is a product of force and displacement – both vectors - the force and displacement must be in the same direction to count for work. Since the component of force in the same direction as the displacement is given as F cos, the general formula for work includes the angle between the force and displacement.

W = F d cos

Example 4: A 20.0 kg wagon is pulled with a force of 40.0 N by its handle at an angle of 30.00 to the horizontal. Calculate the work done in pulling it 10.0 m..

(346 J)

For the case when force and displacement are in the same direction, = 0o and cos = 1, and we get back the original equation W = F d.

Example 5: If the wagon above starts from rest, what is its speed after 10.0 m?(Remember that the acceleration is due to the net force.)

(5.88 m/s)

Work Done Against Friction

The usual cases of motion are assumed ideal – to not involve friction. If friction is considered, its direction opposite to motion means it is usually subtracted from the forces in the direction of motion.

The work done by brakes on a vehicle is an example of a force in the opposite direction of the displacement, eg the brakes of a car. This produces a decrease in energy or a negative work (W= -F d).

Example 6: A 1200 kg car is traveling down the road at 60.0 km/h.

a) If it takes 33.0 m to stop the car, calculate the work done by the brakes in stopping the car.

(-1.67 x 105 J)

b) What is the average force of friction applied by the brakes during the car’s stopping?

(-5.05 x 103 N)

Example 7: A 20.0 kg wagon is pulled with a force of 40.0 N by its handle at an angle of 30.00 to the horizontal against a force of friction of 20.0 N.

a) Determine the net force in the direction of motion.

(14.6 N)

a) Calculate the work done in pulling the cart 10.0 m.

(346 J)

b) If the wagon starts from rest, what is its speed after 10.0 m?(Remember that the acceleration is due to the net force.)

(3.83 m/s)Pg 294 do # 1-3

Work on an Incline

For an incline, the forces parallel to the surface determine the motion of the object up or down the incline. This means the force of gravity resolved into its parallel component Fg// is added or subtracted to the applied force FA (and/or friction Ff as need be).

Example 8: A girl pushes a 20.0 kg box, initially at rest, up a frictionless 10.0 m long ramp inclined 30.00 to the horizontal. If the final velocity of the box at the top of the ramp is 1.00 m/s, what was the work done on the box by the girl?

Start by drawing the free-body diagram to determine the forces acting on the box.

Determine, by Newton’s 2nd Law, the applied force FA. Here, work is being done by the applied force FA to overcome the force of gravity Fg and also accelerate the object with a net force F.

Power

Power, P, is the rate of doing work, the work or energy per second or time.

P = W = E= Fd = F vave

t t t

The unit for power is the Watt (W), unfortunately the same symbol as for work. To differentiate between Watt W and work W, two bars can be placed above the Watt symbol W.

1 Watt = 1 Joule 1 second

1 W = 1 J 1 s

Example 9: A 50.0 kg student runs up three flight of stairs at constant velocity, a vertical distance of 10.0 m. The first time it takes her 9.81 s. The second time it takes her 19.62 s. What was the power she exerted for both times up the stairs? (Hint: first find the work done)

Note that for both times up the stairs, she is doing the same work (against the force of gravity). But because it takes her half as long for the first time as the second time, she is exerting twice the power the first time as for the second time up the stairs.

Mechanical Energy

Energy has many forms: heat, light, sound, chemical, electrical, nuclear. Any one form readily converts to any other. So, for instance, chemical energy can be converted to electrical energy, nuclear energy to heat energy, and so on.

Historically, the concept of energy arose from the Industrial Revolution with the idea of machines able to do work for us. All forms of energy can transform into the work of machines, or have the potential to be transformed into the work that machines can do for us. Notably, one of the first machines in the Industrial Revolution was the steam engine, used in the coal mines of England to hoist water out of the mine. There, heat in the form of steam (from boiling water) was used to power a giant pump, replacing the work being done by horses drawing load after load of water out from underground in the mine (from which came the term “horsepower” to rate machines).

The first steam engines were very inefficient, burning up a lot of coal (luckily there was lots nearby). It was James Watt who developed a vastly improved steam engine, one much more efficient that generated a lot more work for any given amount of heat. Watt’s version of the steam engine essentially powered the beginning of the Industrial Revolution – used from factories to trains.

For physicists then, the important forms of energy – historically – were mechanical energy (energy of machines). Mechanical energy has two parts:

1) Kinetic energy Ek or energy of motion, Eg. Water moving or being moved

2) Potential energy Ep, the energy of position or state with the potential to produce motion (or more correctly cause changes in motion).

Eg. A boulder perched on top of a hill, a stretched spring, a chemical bond.

In any isolated system where energy is not allowed to come in or out the total mechanical energy ET is the sum of the individual kinetic and potential energies.

ET= Ek + Ep

The law of conservation of energy states that energy is neither created nor destroyed, or the total energy of any isolated system is a constant value. In a perfect (frictionless) system, the total mechanical energy is a constant value.

Example 10:Describe the mechanical energy transformations in a pendulum brought up to a height of 0.100 metres above its lowest point, and then releaseda) in a frictionless system

b) in a system with friction.

Gravitational Energy Potential Ep

The potential energy of a system is related to the amount of work required to put that system into its position or state (from a position where the energy potential is zero).

Example 11: A 100 g coffee cup is lifted 1.00 m straight up at constant velocity. What is the potential energy of the coffee cup?

Ep = W= T d= Fg d= (mg) h= (0.100 kg) (9.81 m/s2) (1.00 m)= 0.981 J

The potential energy from gravity is equal to the work done against gravity. The general formula for potential energy due to gravity is

Ep = m g h

Note that the energy potential due to gravity is zero when the height is zero. The general position of zero potential energy is the point where there is no further potential to move, here the ground level zero. However, potential energy can be made relative to any given position.The change in potential energy (from gravity) exactly equals the work done from that position against the force of gravity.

Example 12:A 250 g pendulum of length 0.500 m is initially at rest 1.00 m from the floor. The pendulum bob is pulled sideways a distance of 0.300 m. What is the potential energy of the pendulum bob in its raised position?

Kinetic Energy Ek

Kinetic energy is energy of motion. Any object with mass m and speed v has kinetic energy Ek given by

Ek = ½ mv2

Example 13:Which has more kinetic energy, a 100 g baseball thrown at 150 km/h, or a 10.0 g bullet speeding at 1500 km/h?

Like potential energy, the formula for kinetic energy is derived from the work done, in this case the change in kinetic energy exactly equals the work done to accelerate an object.

Example 14:Calculate the work done to accelerate a 5.0 g bullet to a speed of 1100 km/h from a gun with a 10.0 cm barrel.

W = Fgun d

Elastic Potential energy and Hooke’s Law:

Elastic potential can be stored in any object which when deformed by a force soon returns to its original shape undamaged. The work that was done to change the object’s shape is returned by the object as work. There are many common examples of elastic potential energy – diving boards, trampolines, sling shot, springs in a car, etc.

A British scientist, Robert Hooke, stated that ‘The amount of deformation of an elastic object is proportional to the force applied to deform it’. Up to a certain point the amount of stretch or compression (x) in a spring varies directly as the force (F).

Fnet = kx Since the spring responds in an opposite direction to the applied force it is often written as Fr = -kx and called a ‘restoring force’. Fnet = -Fr

The slope of the graph is called the spring constant (k) in N/m. Beyond a certain applied force, the graph will no longer be linear and damage can be done to the spring by applying that amount of force. This is called the non-elastic region and at the end of this region is the breaking point for the spring. Because force is a vector quantity we can give different signs for a stretch than a compression. In either case ‘k’ is constant and +.

Note: As the x gets larger, the force gets larger. This is not a constant force!! And therefore the acceleration of a spring is not constant either!

Eg. A spring which has a force constant of 15.0 N/m has a 1.50 x 102 g mass attached to it. How far will the spring stretch?

When a spring is compressed/stretched, the work done on the spring to compress/extend it is stored in the spring as POTENTIAL ENERGY. The amount of potential energy is given by:

Ep = ½ kx2 (since the force changes, you can not use the work formula unless average force is given – show formula!)

The Law of Conservation of Mechanical Energy

The sum of potential and kinetic energy is called the mechanical energy.

Total Mechanical Energy = Energy Potential + Energy Kinetic

ET = EP + EK

Energy forms are readily converted into other energy forms. The first law of thermodynamics states that energy is conserved in these energy transformations. This means that the total amount of energy will always remain the same before and after any energy conversion, or any form of energy will convert completely to any other form of energy.

In ideal (frictionless) system this means that work can convert completely (100%) to mechanical energy, or that one form of mechanical energy like potential energy can convert completely to another form like kinetic energy, and vice versa.

Example 19: An archer draws back a bowstring with an average force of 500 N to a distance of 20.0 cm from equilibrium, outline the energy conversions. What is the maximum speed an 80.0 g arrow could be shot from this drawn bowstring?

In an isolated frictionless system where energy cannot go in or out, the total amount of energy is the total mechanical energy, which stays a constant value. This means that in an isolated system, any system’s loss in potential energy is equal to its gain in kinetic energy (or a system’s gain in potential energy is equal to its loss in kinetic energy). If an isolated system

starts with only potential energy, it can convert any part or all of that potential energy to kinetic energy. At the end it may have converted anywhere from 0 to 100 per cent of that potential energy to kinetic energy. Irrespective of the conversions, the total energy – here the total mechanical energy – remains a constant.

ET(position 1) = ET(position 2) or 0(energy added)= Ek + Ep

Example 20:A coffee cup falls off a1.00 m high table. Assuming no friction, complete the following table of the mechanical energy of the coffee cup as it is

a) at the table top just before it starts fallingb) at the middle of its fall c) at the bottom just before it hits the ground.

Note that the change in energy potential is exactly equal to the opposite change in kinetic energy.

The law of conservation of energy allows the calculation of a speed at any point along the energy conversion, knowing only the initial energy of a system. By equating the total mechanical energy from any one point to any other point, once the height h is known the speed v can be determined. Note that mass m is not required as it cancels out from both sides of the equation.

E Top = E Middle= E Bottom

mgh = (mgh + ½ mv2)Middle = (½ mv2)Bottom

Example 21.Calculate the maximum speed of the coffee cup just before it hits the floor.

E Top = E Bottom

Example 22. Calculate the speed of the coffee cup at the midpoint of its drop to the floor.

Example 23:A weightlifter lifts a 250 kg barbell to a height of 2.00 m from the floor.

When he releases it, what is the maximum speed before it hits the ground?

Example 24: A 100 g coffee cup is lifted 1.00 m straight up from the floor at constant velocity, then dropped. Assuming no friction,

a) What is the work done on the coffee cup?

b) What is the potential energy of the coffee cup at the top?

c) What is the kinetic energy at the bottom just before it hits the ground?

d) What is the maximum final speed just before it hits the

ground?

e) What is the speed of the coffee cup at the middle of its fall?

Roller Coaster Problem:

Example 20 Initially the roller coaster is at rest at A (see diagram below). It startsrolling down the track. Assuming a frictionless surface, what is the speed of the roller coaster at B, C and D:

Pendulum:

Falling object (no resistance):

What would the graph look like if there is friction?

t

The Work-Energy Theorem

The equations for kinetic and potential energy derive from Newtonian mechanics and the work done in either accelerating (kinetic) or overcoming the force of gravity (potential). By first considering the (changes in) potential or kinetic energies, most motion problems can often be solved much simpler than by Newtonian mechanics and its consideration of forces across distances.

To analyze a motion problem from an energy standpoint, first consider: are there any changes in kinetic energy Ek from changes in speed, or potential energy Ep from changes in height or position? If so, then the total work WT (Fd) involved will be a sum of these two types of energies.

WT= Ek + Ep (ideal systems-all energy is accounted for)

Let us work through a few examples from above to illustrate this point.

Example 15. What is the work done in holding a 100 g coffee cup stationary on a table?

(Here there is no change in speed or position)

(0)

Example 16. What is the work done in lifting the 100 g coffee cup straight up from the table 1.00 m at constant velocity? (Here there is a change in position)

Example 17. What is the work done in moving a 100 g coffee cup straight across a frictionless table 1.00 m from rest to a final speed of 1.00 m/s?(Here there is a change in speed)

Example 18:Calculate the work done to lift a 100 g coffee cup 1.00 m straight up from the table while it accelerates to1.00 m/s from rest. (Here there is both)

Work can be added to the system by an applied force, or taken away from the system by friction. Work theorem is useful in both cases. Signs are used to show an increase or decrease in the work. The change in energies (Ek, Ep) need to be final – initial to reflect the increase or decrease with the appropriate sign.

Eg. Pg 308 #1,2

Lab 7- Conservation of Energy – Friction Down an Incline

Objective: To use the law of conservation of energy to determine what percentage of energy is lost to friction by a cart rolling down an incline.

Background:The law of conservation of mechanical energy predicts that for ideal systems,

potential energy can be completely (100%) converted to kinetic energy. For a cart rolling down a ramp, this means the potential energy of the cart at rest at the top of a frictionless incline will be exactly equal to the final kinetic energy of the cart at the bottom of the incline.

Ep, top = Ek, bottom

Real systems on earth include friction. As a result, some energy will be lost to friction. By the law of conservation of energy, the total energy of the system should still remain constant. This means that the potential energy at the top is not converted 100% to kinetic energy, but that some percentage of energy will be lost to friction.

Ep, top = Ek, bottom + Wfriction

Eg. 100% = 80% + 20% (20% of the energy is lost to friction here)

Rearranging, we can determine the energy lost to friction by taking the potential energy and subtracting from it the kinetic energy at the bottom. The potential energy is calculated from the mass m and height g. The kinetic energy is calculated from the mass m and speed v at the bottom. The difference is the energy lost to friction Wf or Efriction: Wf = Ep top - Ek bottom WT= Ek + Ep (Theorem)

Wfriction = Ep, top - Ek, bottom

= mgh - ½ mv2

This energy lost to friction is usually in the form of heat or sound. Since the total energy of the system to begin with is the energy potential at the top of the ramp (Ep,top=100%), to calculate the percentage of energy lost to friction, we divide the energy lost to friction Efriction (Wf) by this total energy and multiply by 100.

% Energy lost = Efriction x 100 Ep, top

Materials: Ticker tape timer and tape (2.0 m) 80-100 cm ramp or inclineCart (paper clip or tape to attach ticker tape to end) metre stick

Lab Write-up: Design an experiment using the above materials to determine the percent of energy lost to friction on an incline. Perform the experiment collecting your own values (work in groups of two) and determining the percent of energy lost to friction. Your write-up should include sufficient detail and information for the experiment to be reproducable by one of your classmates.