5.0 Central Attraction: centripetal acceleration When we see a body moving in a circular path, we know that there must be force acting on it because without a force, things travel in a straight line. We likely want to look at this through a dynamics lens. From our experiences we can also see that the force is radially inward:

a) Gravity pulls the moon into a circular orbit. The force of gravity on the moon pulls the moon directly inward toward the earth.

b) If you attach a rock to the end of a string and spin it around your head in a horizontal circle, the string pulls the rock radially inward.

c) If you roll a marble around the bottom of a cylindrical container, the normal force of the wall on the marble is inward.

We know that ∑ �⃗� = 𝑚�⃗�, so if the force on the object is inward, we can be sure the resulting acceleration is inward too. At right, you see that the velocity of an object in circular motion is tangential to the circle (and would continue in this direction without some force accelerating it), and the acceleration is radially inward. We can show that the acceleration of an object in uniform circular motion scales like the square of the velocity and inversely with the radius of the circular path. Consider that at some velocity, v, you go around a half turn of radius R, and come back with velocity –v. Remembering �⃗� = ∆()⃗

∆*:

a) If you double your speed the change in velocity (-2v) also doubles. However, at twice the speed, you take half the time to make the turn, so the acceleration should increase by a factor of 4, or the square of the speed.

b) If you double the radius of the turn, the change in velocity is still the same (-2v), but you take twice as long to do the turn, so the acceleration should be half as much

Thus, we are not surprised that centripetal acceleration is 𝑎+ =(,

-.

Forces cause acceleration (dynamics lens) not the other way around. I often hear that centripetal acceleration means there’s centripetal force. There’s no such thing as centripetal force, just like there’s no such thing as linear force. A force is an interaction between two bodies whereby they exchange momentum. Force can cause linear acceleration or centripetal acceleration. When we see centripetal acceleration (or any other acceleration), we know that there is some force pushing or pulling on the object to cause this acceleration. So, if we see circular motion, we should consider the dynamics lens and look for the force or net force toward the center of the circular path.

Exercise 1: Please first read the previous paragraph. You see a 10 kg rock in space moving with constant speed of 10 m/s in a circle of radius 20 m. You wonder about the rock, and look at it through different lenses.

a) Do you think there’s a force acting on it? Why? b) Find the acceleration of the rock, including direction of the acceleration. c) Calculate the force necessary to accelerate this rock. d) What kind of force is this? – if you say, “it is centripetal force!” I will be sad. I will be

pleased if you say, “I have no idea what force is acting on it, because I can’t see anything that the rock is interacting with, so I have to look around at what object must be applying a force of_____(put answer from c) on the rock to make it accelerate at______(put answer from b).”

e) Then you see a string attached to my arm as I spin the rock in a circle. What kind of force is it? Find the tension in the string.

f) Then the string breaks – what happens to the rock? Please draw a picture. g) ….instead of a string, you see a large sphere at the center of the rock’s circular path. What

kind of force might be acting on the rock now? h) ….instead, you notice that the 10 kg rock is actually a small 10 kg toy car driving around in a

20 m circle on a flat parking lot at 10 m/s. Now what force is acting on the car? Please find the coefficient of friction necessary to keep the car moving in this circle. Is this static friction or dynamic friction?

Exercise 2: You’re taking a turn with a car or bike on flat pavement and there is a coefficient of static friction of 1.2 between the rubber and the road. To execute a turn of a radius of 10 m,

• what force causes the centripetal acceleration allowing you to make this turn? • How fast can you go without “sliding out?” • State your lens; set up the drawings and logic; and reflect on your answers in light of your

experiences. • Why did we use static friction rather than dynamic friction?

Exercise 3: Prove to yourself that we can also write centripetal acceleration as 𝑎+ = 𝜔/𝑅, and 𝑎+ =

12,-3,

, where 𝜔 is angular velocity, and T is the period of revolution. Astronomers like the second expression because T is the period or time it takes for a full revolution, or length of year (of a planet for instance). Exercise 4: A concrete flywhere of uniform thickness has a mass of 50 kg and a radius of 40 cm. The string is wound around a pulley of radius 16 cm. I pull on the string with a force of 100 N. After some time, the wheel is spinning such that it takes two seconds for a full revolution. At this time, you notice a fly standing on the rim of the wheel at the extreme left (red dot):

a) What’s the fly’s tangential acceleration? Include direction b) What’s the fly’s centripetal acceleration? Include direction

Exercise 4:

PROVE IT! Proving the formula for centripetal acceleration is a classic proof I expect you to be able to do. Please learn it. The drawing at right shows a picture of something moving in circular trajectory. The diagram to the right shows how to find the change in velocity using the velocities at the two points in the cicular trajectory. Please prove to yourself that the two isocelese trianges are similar, and thus their parts are in proportion. You can then start with the definition of acceleration as the rate of change of velocity and express ∆𝑣 and ∆𝑡 in terms of radius and speed. In this proof it is also important to take the limit of ∆𝑡 = ∆𝑣 = ∆𝜃 ⟹ 0. Why? Do we want the average acceleration, or the instantaneous acceleration, 𝑎)))⃗ = :()⃗

:*? The average acceleration over one

revolution is zero because the change in velocity is zero. The acceleration, like the velocity is constantly changing, so it’s the instantaneous accelertion that is of interest. When we take the limit of short time or small ∆𝜃, we see that ∆𝑟 (the change in position) becomes the same as the distance traveled along the perimeter of the circle and ∆<

∆*⇒ 𝑣, the tangential speed. Please carry out this proof

with a good drawing. 5.1 Ubiquitous Inverse Square Relationship: It’s everywhere When things spread out in three dimensions from their source (like light, sound, or shrapnel from a cluster bomb) intensity decreases like the inverse of the distance squared: 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 ∝ E

<,= 𝑟F/.

Why is that? When things come from a point source, they spread out with spherical symmetry, so they are spread out over the surface of a sphere that gets bigger as it expands in time. Because the surface area of a sphere = 4𝜋𝑟/, the area is proportional to the square of the radius. So, if you double your distance to a lightbulb, the light from the bulb is spread out over 4 times the surface area, and the intensity of light at the new location is ¼ as great as before: If 𝑅 ⇒ 𝟐𝑅J, 𝐼 ⇒

𝟏𝟒𝐼J.

This may be better seen from the diagram at right, where if light travels through a window of a sphere at distance 𝑅J, the same light would need 4 windows to pass through a sphere at distance 2𝑅J: 2 windows wide, and 2 windows high. In the next section, we describe how Cavendish measured the relationship between mass, distance and gravitational attraction. However, even before he experimentally determined this relationship, one could imagine the gravitational force (from a planet) spreading outward into space in all

∆"#⃗%

&⃗'

#⃗'&⃗%∆#⃗

&⃗%&⃗'

∆&⃗

Light propagates equally in all directions from a point source in the middle. The flux of power that escapes through the smaller sphere must also pass through the larger sphere. Because the same flux is spread over a larger surface at the larger sphere, the intensity of light is weaker at the larger distance.

R

2R

directions. Thus, we might have expected the gravitational force to drop off according to the inverse square relationship. Exercise 1: The distance between the centers of the earth and moon is about 385,000 km and the radius of the earth is about 6,400 km.

a) How many earth radii is the distance between the center of the earth and moon? b) If the moon were to stop moving, estimate the moon’s acceleration toward the earth. It may

help to know that the acceleration from gravity one earth radius away from the earth’s center is 10 m/s2.

c) But the moon is moving! In this case, should its acceleration be the same as in “a” above? Why or why not?

d) From your answers above, estimate the speed of the moon in its orbit around the earth. e) From your answer above, estimate the period of the moon… is it close to a month?

Exercise 2 In 1930, it was discovered that a beta decay: n => p+ + e- didn’t conserve energy, momentum or angular momentum. Wolfgang Pauli imagined the existence of a new particle, the neutrino with the correct angular momentum to conserve angular momentum in this process. We now estimate that 65 billion neutrinos from the sun pass through each square centimeter on earth, per second. The earth is about 150 billion meters from the sun and Venus is about 108 billion, or about 0.72 (or ~ E

√/) the earth-solar distance.

• How fast does the sun produce neutrinos? • How many pass through you during a one hour class? • How about if you were on Venus?

It’s interesting to note that neutrinos interact with matter very very weakly making them very difficult to detect. In fact, the flux of solar neutrinos is negligibly attenuated by the earth, so the flux coming out of the earth at night is about the same as that entering the earth on the other side facing the sun. This was astutely noted by John Updike in 1960: Neutrinos they are very small. They have no charge and have no mass And do not interact at all. The earth is just a silly ball To them, through which they simply pass, Like dustmaids down a drafty hall Or photons through a sheet of glass. They snub the most exquisite gas, Ignore the most substantial wall, Cold-shoulder steel and sounding brass, Insult the stallion in his stall, And, scorning barriers of class, Infiltrate you and me! Like tall And painless guillotines, they fall Down through our heads into the grass. At night, they enter at Nepal And pierce the lover and his lass

From underneath the bed – you call It wonderful; I call it crass. The poem, Cosmic Gall was published before the 1998 discovery that neutrinos actually do have mass. 5.2 Universal Gravity Gravity attracts any two bodies with masses m and M: 𝐹QR = QR

<,𝐺, where r is the distance between the centers of

the two bodies, and 𝐺 = 6.67𝑥10FEE YZ

,

[\,,

Cavendish measured gravitational force with an apparatus like that at right. 1 The two small masses m are attracted to the larger masses, M, causing a rotational deflection of the hanging barbell. The force of gravity is very weak compared to the force of electric charges. If two electrons are close to each other, the repulsion from the electric charges will be greater than the gravitational attraction by more than a factor of 1043. The only way we are able to feel the earth’s gravity is because the earth is very massive, and both the earth and our bodies have almost no charge since we have roughly equal numbers of protons and electrons. The force of gravity being so weak made it difficult to measure the gravitational constant. Cavendish’s torsional pendulum was very sensitive to the tiny torque produced by the weak gravitational attraction between m and M. Additionally, the light reflecting from the mirror allowed him to measure very small rotational displacements. Exercise 1: We know about gravitational force on the earth’s surface, and that the distance from the North Pole to the equator along the earth’s surface is 10 million meters. With this information, please calculate the mass of the earth. Check your answer against a known value. Before the Cavendish Experiment, people likely expected the formula to have the general form of the product of the two masses divided by the square of the separation of the centers:

• Our last chapter discussed the inverse square relationship that we expect for field interactions (that is, light, electric fields, gravitational fields, etc.)

• We already knew that the force of gravity on an object is proportional to the object’s mass, AND that a force is an interaction between two bodies affecting each with equal strength in opposite directions. Thus, if we double the mass of one of two gravitationally attracting bodies, the force of gravity on both the bodies must double.

The Satellite Equation: In the vacuum of space, the only force acting on a body is often gravity. For circular orbits, centripetal acceleration results from this force. Thus, for circular orbits, the Satellite Equation can be written:

1 From OpenStax.org: https://cnx.org/contents/jQSmhtXo@14.60:OMxfrjQL@3/Newtons-Universal-Law-of-Gravi

m formeters

𝐹] = 𝑚𝑎+ Exercise 2, Lower Earth Orbit (LEO): Above an altitude of 100 miles (160 km), the atmosphere is thin enough to allow near frictionless orbiting.

• Make an argument that at this elevation, the acceleration of a satellite is reasonably close to that at sea level, g.

• What is the speed, and orbital period, T, of a satellite in LEO? • If you were in LEO, about how many times a day would you expect to see the sun rise? • Read up on Lower Earth Orbit by looking it up on Wikipedia (or somewhere else).

Exercise 3, Geosynchronous Orbit (GSO): Communication satellites are used to bounce signals around the curve of the earth. It is nice for them to stay over the same place on the earth by orbiting the earth in a single day.

• LEO orbits have a period of way less than a day (see above exercise). Please explain why we might think that as we get further from the earth the orbital period would increase.

• With a full drawing and reasoning, calculate how far from the earth’s center GSO is. Please put your answer in terms of the number of earth radii.

o What lens do you use? o Did you execute it well? o Does your answer make sense to you? o Should GSO be closer or further than the moon? How can you be sure?

• If you were in GSO and stopped moving, you would start falling toward the earth. What would your acceleration be?

Force is a vector! We know that gravity is an attractive force, so the force is always in the direction of −�̂�. The vector notation of the force of gravity then becomes: �⃗�QR = −QR

<,𝐺�̂�.

5.3 Loop-the-Loop: Circular motion in the vertical plane When you see something moving in a circle, it’s a good bet that dynamics is involved. Make a free body diagram and note acceleration is towards the center of the circle. A review from section 2.5 (just this page): If we were in outer space, or in free fall inside the space station as it orbits the earth, there is no apparent force of gravity. How would we keep water inside a bucket, so that the water is touching the bottom of the bucket? “Touching” the bottom of the bucket means that there is a nonzero normal force; that is 𝑁 > 0. Given that this is the only force, it would mean that the bucket would be accelerating in the same direction. So, you could imagine accelerating the bucket back and forth or up and down while rotating it to maintain a positive normal force.

N

Please recall the elevator problem noting that ∑ �⃗� = 𝑚�⃗�. Consider keeping a man, a ball, or water in a bucket as shown at right. Exercise 1: If the object is kept in the bucket, what do we know? Identify all correct statements. Answers at chapter end.

a) 𝑁 > 𝐹b. b) 𝐹b > 𝑁 > 0. c) 𝑁 > 0. d) acceleration must be upward or zero only. e) acceleration can be zero or any value upward or downward. f) acceleration can cannot be downward with a magnitude more than gravity.

Exercise 2: How about if we see someone in a bucket upside down, what do we know? Identify all correct statements. Answers at chapter end.

a) 𝑁 > 𝐹b. b) 𝐹b > 𝑁 > 0. c) 𝑁 > 0. d) acceleration must be downward with a magnitude more than gravity. e) acceleration can be any value, but must be downward. f) acceleration can be zero or any value upward or downward. g) acceleration can cannot be downward with a magnitude more than gravity.

When we see something moving in a circle, we only know that it is accelerating radially inward. We then look for the forces that could provide that acceleration according to ∑ �⃗� = 𝑚�⃗�. So, for instance with a “Loop-the-Loop” carnival ride, the forces include gravity and the normal force that cause centripetal acceleration inwards. So, at the top the centripetal acceleration is downward. If we want the cart to stay in contact with the track, we know that the normal force is downward. What does this say about the centripetal acceleration necessary to keep the cart on the track? Exercise 3: Please make a good free body diagram of the cart at the top of the loop and the bottom of the loop. Please identify what you know for both the acceleration, and the normal force:

• at the top of the loop • at the bottom of the loop

be sure to support your answer with a diagram and lens identification.

Fg=mg

N

Exercise 4: You go on a R = 10 m, loop-the-loop ride where the cart is let go on a low friction track and is pulled downhill by gravity. You have to choose how high to start the cart. Say you have a mass of 70 kg, like your instructor and you are sitting on a scale that reads in kg. Don’t use this drawing… please make your own. a) If you start from a vertical height of 40 m, what does the

scale under you read as you are at the top of the loop? What does it read at the bottom of the loop? Is this a good ride for pregnant women? How does it feel as you round the bottom of the loop?

b) What would happen if you decide to start the cart at the same height as the top of the loop? Why would this happen?

c) Please find the minimum vertical height, above the ground that you must start the ride to stay on the track at the top.

The low friction, circular loop-the-loop ride above has a higher speed at the bottom than at the top. Thus, the centripetal acceleration to keep a circular path is also greater at the bottom. Additionally, please show that because of the way the track faces, in order to keep the cart on the track at the top, there must be a very great normal force at the bottom. This could make for a dangerous carnival ride! Is there some way we could increase the centripetal acceleration at the top (to keep the cart on the track), and lower the centripetal acceleration at the bottom (in order to not smoosh the people)? Exercise 5: Look at the carnival ride at right. Notice that the radius of curvature at the top is not the same as the radius of curvature at the bottom. What effect does this have on the centripetal acceleration at the top and at the bottom? Why would they build it this way? Please support your answers with a good free-body diagram, and identify the lenses you use. Exercise 6: Below, you see 3 pictures of a car you are driving. You are sitting on a common “bathroom scale” that is situated between your body and the seat.

a) Do you “feel” the same at the three different places? If not, how do you feel differently?

b) Does the scale read the same at all three locations? If so how do you know? If not, where is the reading the highest and where is the reading the lowest?

c) Please back up your answers above with a good free body diagram and indication of direction of acceleration.

d) In each of three places, how would the reading on the scale compare to the force of gravity on your body?

Exercise 7: Do you weigh more at the North Pole? You stand on a scale at the north pole and then on the same scale at the equator to see where the scale reads higher. Assume that the earth is a perfect rotating sphere. You find that at the equator, the scale reads: a) More because the normal force compensates for both

gravity and centripetal acceleration. b) Less because the normal force at the equator is not

enough to keep you in equilibrium. c) The same because the normal force is always the same

as gravity. d) The same because you are in equilibrium in both

places e) None of these. f) Not enough information is given If you get confused… as with any mechanics problem, please draw a picture and do a lens analysis. There is a story that gold was mined in Alaska, weighed, and sent to Fort Knox in Washington DC where they weighed it again. What did they think in DC when they weighed the gold they’d just purchased?

Please don’t read these answers until you have made a free body diagram and discussed the answers yourself. Exercise 1 (c,f); Exercise 2 (c,d) 5.4 Gravitational Potential Energy in Outer Space We learned in section 1.7 and 2.7: W = ∆E = F∆x, and dW = 𝑑E = F𝑑x. So, if you push a car on flat ground with a force of 200 N for a distance of 10 m, you do 2000 J of work, which increases the car’s kinetic energy by 2000 J. If you change the magnitude (strength) of the force on the car half way through, you’d have to calculate the work done in each section and add it to calculate your total work, or the total change in energy of the car.

𝑊 =e 𝑊JJ

= e 𝐹J∆𝑥J

= 𝐹E∆𝑥E + 𝐹/∆𝑥/

What do we do if the force changes smoothly over time rather than in discreet steps? Then the sum becomes an integral of force over the distance, something not covered in this text. Apply to gravitational potential energy: When you raise an object some distance ∆y, the work you do on it is equal to the gain of gravitational potential energy, ∆𝐸] : 𝑊 = 𝐹b∆𝑦 = 𝑚𝑔∆𝑦 = 𝑚𝑔𝑦i − 𝑚𝑔𝑦j This works as long as the force of gravity is a constant, 𝑚𝑔. What do we do for changes in elevation so great that the force of gravity changes? The force of gravity drops off as the inverse square of the distance to the center of a planet, as described in sections 5.1 and 5.2: 𝐹QR = QR

<,𝐺.

The work and ∆𝐸] is found by integrating this force over the change in height, yielding: 𝐸]= −QR

<𝐺, so

∆𝐸]= QR<k𝐺 −QR

<l𝐺 = 𝑚𝑀𝐺 nE

<k− E

<lo

Exercise 1: With a good drawing, consider moving a mass from the earth’s surface, at 𝑟 = 𝑟j, to some greater elevation at 𝑟 = 2𝑟j. Calculate the amount of work you do using both formulas above: the one assuming the force of gravity is constant, and the one resulting from integrating over the inverse square, ever changing force of gravity. Which one yields a greater amount of work? Why?

Exercise 2: Let’s use the different equations for gravitational potential energy

• You drop a rock from a 10 m cliff. Why can we use ∆𝐸] = 𝑚𝑔∆𝑦for to find the final speed? What lens do you use to find the final speed? Please find the final speed when the rock hits the ground.

• You drop a rock from 6.4 million meters (about one earth radius) above the earth’s surface. Why is it no longer reasonable to use ∆𝐸] = 𝑚𝑔∆𝑦? Would the correct speed be less than or greater that that calculated with ∆𝐸] = 𝑚𝑔∆𝑦? Why? Use the correct formula for ∆𝐸] to calculate the speed of the rock when it hits the ground.

Exercise 3: Escape speed is the speed you need on a planet’s surface to take you “infinitely far away.” This is not infinite because as you get far from the planet, the force of gravity gets weaker.

a) Calculate the escape speed from the earth’s surface: how fast do you need to throw a rock in order for it to not to come back to earth?

b) 14 km/s is faster that the escape speed from the earth. So, if you could throw a rock at 14 km/s, it would still have kinetic energy when it got very far from the earth. What would the rock’s speed be infinitely far from the earth?

c) You really couldn’t get infinitely far from the earth with this speed because you would also need to escape from the solar system and then the galaxy requiring much more kinetic energy than just escaping from the earth alone.

Is there an absolute value of 𝐸]? It is arbitrary where we assign 𝐸] = 0 because the physical behavior is determined from ∆𝐸] . For instance, if you take a physical system like a mass on a spring and increase its elevation by 1 m, it still behaves the same. On earth, we can assign 𝐸] = 0 to sea level, but it can also be the desk top in your laboratory. But what if we want to compare the potential energy of the earth’s surface with the surface of Mars? For this, we assign 𝐸] = 0 at 𝑟 =∞, when you are infinitely far from any massive body. This can be considered a “free” mass. By integrating back from 𝑟 = ∞ to some distance 𝑟i, to a planet of mass 𝑀, we have: 𝐸](𝑟) = −ZR

<𝐺.

Our potential energy is negative everywhere! Yes, this would be true, because we do negative work bringing a mass down to a planet from 𝑟 = ∞. The force we apply to the rock is in the opposite direction we are moving. We say that on a planet, we are “stuck in a gravitational well”. And we are! Just jump up off the earth’s surface and see what happens. 6.0 Systems of Masses: What do you do when you have more than one mass such as the system at right connected with a light, low friction string? If we let this system go on a low friction surface with a low friction pulley, can we find the acceleration? the tension in the string? The speed when the 250 g mass hits the ground below? There are many ways to solve this. At first glance, we think of the dynamics lens because we see that the force of gravity accelerates the masses. However, we can also see that there is an energy transformation. There are two simplifications to assert for our system:

- If the pulley is frictionless, it exerts no force on the string. So, the tension in the string is the same on both sides of the pulley.

- If the string has no mass, then there is no force on it even if it is accelerating. Thus, for a massless string, the tension is the same everywhere, including at both ends.

Example 1: Consider the system of masses above. Here are some important questions:

a) Is the acceleration of the falling mass zero, between zero and gravity, or equal to gravity? How can you be sure?

b) Is the tension in the string >, =, or < the force of gravity on the 250 g mass? What would it mean if the tension was = to the force of gravity on the 250 g mass?

c) What is the energy transformation after the system is released? Which block(s) lose potential energy? How much? Which block(s) gain kinetic energy?

d) Say that the system is released with the hanging mass 1 m from the ground. Can you calculate the speed of the blocks in this system just before the falling block hits the ground?

e) If you can find the final speed, then you can also find the average speed, the time it until it hits the bottom, the acceleration of the system and the tension in the string. Do it!

Example 2: Consider the system above through a dynamics lens of forces on the whole system. Make a good Free Body Diagram.

a) Consider assigning the positive direction to this system. It’s not up or down or left or right but determined by one direction of the curved arrow as shown at right. This is the only way the system can move, so we need consider only forces in this direction. For instance, there is a force of gravity on the 1 kg mass. Is this in the positive or negative direction of the system, or does it not pull on the system at all?

b) Does the tension act on the system? Which way does tension pull the 1 kg mass? The 250 g mass? When we add these forces, what is the total force? Here, we can call tension an internal force. Thus, the tension doesn’t accelerate the system.

c) We should see that there is only one external force that acts on the system. We can see that if gravity didn’t pull downward on the 250 g mass, that there would be no force accelerating the system. How much mass does this force accelerate? With this information, you can find the acceleration!

d) Once you have the acceleration, we can find the tension in the string. But in order to find the tension, we have to consider a body that the tension is acting on (not the whole system). Pick and look at a single mass through the dynamics lens. Can you find the tension?

Example 3: The Standard Approach is to look at the two masses separately, setting up the dynamics equations, ∑�⃗� = 𝑚�⃗�, for each mass, leading to simultaneous equations. For this system, the two equations are used to solve for the two unknowns (the acceleration and the tension in the string). …energy of the system works well

…or we can solve it using system dynamics

… or we can use the standard dynamics on each of the masses

Which is the best method? This is for you to decide. Note that the energy lens yields final speed, and dynamics lens yields acceleration. You can use one to calculate the other:

Also, we can use dynamics of one of the masses to find the tension in the string.

Why can we assert that the tension is uniform across the length of the string? If the tension at one end were different from the tension at the other end, there would be a net force on the string. This force would accelerate the string. Thus, if the string has very little mass, and there is very little friction between the string and the pulley, then there should be very little difference in tension between the two ends of the string. Example 4: What if there is a coefficient of friction(𝜇: = 0.1) on the 1 kg mass as is slides across the horizontal surface?

a) How would this change the energy considerations in Example 1? Find the new speed of the system as it hits the ground 1 m below. Then find the time to fall and the acceleration.

b) How would this change the dynamics considerations of the system? Find the new acceleration directly (do you get the same answer as above?) and tension in the string.

Example 5: In a tug-o-war, each team pulls as hard as they can on the rope. However, the tension on the rope must be the same on both sides, pulling each team in opposite directions with the same force. So then, how can a team win?

a) Please look at a single team through a dynamics lens and do a good free-body diagram. What can you say about the forces necessary for this team to win?

b) Then look at the two connected teams as a system and do a good free-body diagram. What must be true for one team to win?

Example 6: A 2-ton tow truck pulls a 1-ton car on a smooth level road, with a rope that has a tension of 3000 N on it. If the wheels of the car are free to roll, what coefficient of friction is necessary between the tow truck’s wheels and the ground? This is a multi-step problem that will require some thought and some drawing.

Rotating Systems: At right, there is a falling block with a string wrapped around a massive pulley. As the block falls, the pulley turns. In understanding how to solve this problem, please consider and answer the following: Example 1: The system at right is released from rest:

a) Is the downward acceleration of the falling block <, =, or > g? How do you know?

b) Is the tension in the string <, =, > the force of gravity on the hanging block? How do you know?

c) Let the hanging block drop 1 m, and watch through an energy lens. Please identify the energy transformation happening. Which of the objects lose energy, and which of the objects gain energy? What are the forms of energy?

d) In order to solve this question, it is crucial to know the relationship between the speed of the falling block and the rotational velocity of the pulley. How does this relationship involve the radius of the wheel? Please check this with units to see if you have it.

At right, a string is wrapped around the circumference of a wheel. With the string anchored to the ceiling, the wheel is allowed to fall as it unwraps. Although this is only one object, we can treat it as a system because it is both falling and rotating. Example 2: The wheel at right unwraps like a “yo-yo”.

a) What energy did it start with? b) What energies does it gain as it falls? c) What is the relationship between the vertical and rotational

velocities? Example 3: Consider two identical massive wheels cemented to a smaller, low mass inner pulley. The wheel/pulley on a low friction bearing rotates from rest, pulled by identical hanging blocks on the same length of string. The only difference is that the strings are wrapped around different parts of the wheel: A around the outer circumference and B around the smaller circumference of the pulley. The systems are let go at the same time until the blocks hit the floor below.

a) In which case is the hanging block moving the fastest when it hits the ground? Or is it the same? Use a lens discussion to support your answer.

b) In which case is the tension in the string the greatest or are they the same? Explain. c) Which block took the longest time to fall to the ground or are the times the same? Explain.

Exercise 4: At right a disk of radius 10 cm and mass ½ kg falls while unwinding 4 m of string. In any order, answer the following, providing good conceptual support:

• If I double the mass of the disk, what effect, if any would it have on the acceleration of the disk? How do you know?

• How would this disk’s acceleration compare to that of one with twice the radius? How do you know?

• I drill out the center of the disk (the black circle in the center of the circle), and fill it with iron, increasing the mass of the disk considerably. What effect, if any would this have on the acceleration of the disk? How do you know?

Exercise 5: At right a disk of radius 10 cm and mass ½ kg falls while unwinding 4 m of string. In any order, answer the following, providing good conceptual support:

• Find the speed and angular velocity of the disk when it reaches the end of the string.

• Find the downward acceleration and angular acceleration of the disk as it falls.

• Find the tension in the string as the disk falls.

6.2 Center of Mass: Remember the problem with the 90 kg friend and the massless board? What if the board was not massless? Say it had a mass of 50 kg? We might think that we could take the man and the board separately and that the 50 kg board alone would have a 500 N force of gravity on it, symmetrically distributed, so both my finger and pylon A would each have an additional 250 N added to the forces we had with the man alone. And you’d be right if you guessed this. In fact, we can show that for any object, we can model the force of gravity as if all the mass were at a point called the center of mass, or the average position of the mass. For any symmetrical object of uniform density, this is the geometric center of the object. Thus, the above statics problem with a 50 kg board could be modeled as the system at right, with the mass of the board adding another force and another torque to the equations. Example 1: Please draw a good free body diagram for the 90 kg friend on a 50 kg board and solve the rotational dynamics (statics) problem. Please find the force on my finger. Center of Mass and Stability In order for an object such as a board to be stable on the ground, there must be a positive normal force pushing on both sides. Thus, the center of mass must be located between the supports. For instance, we can see that a uniform plank (with a center of mass in the middle) is stable resting on

the wider supports at left. At right the torque from gravity would rotationally accelerate the board into the page (clockwise) pivoting the board about pylon B. In order for this board to be stable, the force from pylon A would have to be downward, requiring the board to be bolted to pylon A, providing a torque about pylon B equal and opposite to the gravitational torque.

6.3 Parallel Axis We know that if a solid, uniform disk of radius R is rotated about its axis, the moment of inertia, I = ½mR2. Also, a point mass rotating around a point at a radius d will have a moment of inertia of Ipt = ½md2. So, if a disk is rotated about an axis that is parallel to its central axis, but is a distance d away from the disk’s center, the object moves around the center of rotation and rotates itself. So, the moment of inertia must include rotation of the object itself about its geometric center plus the moment of inertia of the object as a point mass around the center of rotation. We can prove this by integrating 𝑑𝑚 ∗ 𝑟/ over the entire mass of the object where r is the distance of each little bit of mass to the external center of rotation, but we’ll save that for another day. However, we can look at it through the energy lens. What if we saw a disk moving with velocity v as well as rotating with angular velocity 𝜔, we would hopefully recognize that it has both linear and rotational kinetic energy and the total kinetic energy is the sum of the two: 𝐸3 = 𝐸u + 𝐸v =

E/𝑚𝑣/ + E

/𝐼𝜔/.

Now further consider that the linear velocity is coming from the fact that the object is rotating around an external center such as the two objects above at right. Then the linear speed is 𝑣 = 𝜔𝑑. Substituting this for the speed in the above equation yields: 𝐸3 = 𝐸u + 𝐸v =

E/𝑚𝑑/𝜔/ + E

/𝐼𝜔/ = E

/(𝑚𝑑/ + 𝐼)𝜔/ = E

/𝐼wx<xuuyuxzJ{𝜔/.

The center of mass and parallel axis theorem provide a nice way to solve the yo-yo problem in just a single line of math. Consider the disk at right with a string wound around the circumference that is falling while rotating. At any given moment the disk is actually rotating about the point of contact of the string. Therefore, we could solve a simple dynamics problem with the torque from the force of gravity (acting on the center of mass), while remembering to use the parallel axis theorem to find the moment of inertia about the point of rotation.

A BFg

A BFg

Figure 1 A disk rotates about the large back dot. At left, the center of rotation is the center of the disk. At right the center of rotation is above that of the disk.

Example 1: Consider the disk at right that is falling while rolling down the length of string wrapped around the circumference. Please consider the disk to be rotating about the point of contact with the string to find the angular acceleration, and the linear acceleration and the tension in the string. Are these the same values you calculated for this system in chapter 6.1?

Example 2: A disk of uniform mass distribution, total mass mo, and radius R, is secured to a wall with a low friction pivot that allows rotation as shown at right. The disk is started in the higher position where its center is the same height as the pivot and is allowed to drop and swing.

• The moment I let the disk go from the upper position the center of mass accelerates downward. Is a <, =, or > g?

• Please find the angular acceleration of the disk the moment I let it go and from this information find the downward acceleration of the center of mass of the disk.

• Please find the force the pivot point provides to the disk the moment I let go of the disk.

• Does the angular acceleration of the disk remain constant as the disk falls to the lower position? How do you know?

We want to find the force on the pin when the disk is at the bottom location. In order to solve this complicated, multidimensional problem, please consider:

• What is the complete energy transformation happening as the disk rotates from top to bottom?

• What is the complete dynamics going on when the disk is at the bottom of the swing? Is the force on the pivot just equal to mg? Why or why not?

• Find the force on the pivot when the disk is in full swing at the bottom. Include direction.