Practice: A Ping-Pong ball goes in a horizontal circle (radius 5 cm) inside a red cup twice per second. Find its:

(a) period;

(b) speed;

(c) centripetal acceleration.

Subjects

Sections | |||
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Intro to 2D Motion | 32 mins | 0 completed | Learn |

Projectile Motion | 67 mins | 0 completed | Learn |

More Projectile Motion | 42 mins | 0 completed | Learn |

Initial Velocity in Projectile Motion | 20 mins | 0 completed | Learn |

Circular Motion | 27 mins | 0 completed | Learn |

Additional Practice |
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Motion in 2D |

Motion in 2D & 3D With Calc |

Projectile Motion: Horizontal & Negative Launch |

Projectile Motion: Positive Launch |

Projectile Motion: Launch From Moving Vehicle |

Relative Motion |

Concept #1: Intro to Circular Motion

**Transcript**

Hey guys. So in this video I'm going to introduce uniform circular motion and show you the variables in the equation that you're going to use. Let's check it out. So in uniform circular motion an object moves with constant speed in a circular path. That's the definition. OK. So uniform the uniform part of the term means constant speed. You can think of it it has uniform speed because it's going to move with constant speed. This thing will be a uniform motion around a circle. So that's why that's what that where it goes. And then circular obviously circular path. Right. So it's a particular type is very special type of circular motion. It's the simplest one and that's the one we're going to look at now. So imagine you're going around a circle like this. Let's say this way. At every point in the circle I've selected three different points here. But really at every point of the circle you're going to have what's called the tangential velocity . You're going to have a tangential velocity. And it looks like this. If you go in this way it's going to look like this at this point at this point it would look like this. And at this point you would look like this is called tangential velocity because it's pointed it's directed Tangent from your path. OK. Tangent just means one way to think about tangent is this is if you're making a turn here and then you miss the turn then you would or you stop turning then you would just go in a straight line. That's what tangent means. OK. Another way that you could have thought of that you can think about this is for example the tangent at this point is just a straight line that barely touches the circle right there. OK. That's the same thing. So the tangent velocity sometimes referred to as a tangential speed. It's abbreviated Vt or Vtan or just V, okay. You also have a centripetal acceleration Ac centripetal acceleration. Now centripetal means Center seeking. And it just means it's point to the. This is also sometimes referred to as a radio acceleration radio. So instead of Ac you might see a rad and the two are the same thing. OK. And it's points towards the center so looks like this. Wherever you are. Notice that these two make an angle of 90 degrees. At every point. This is an equation that relates the and centripetal acceleration tangential velocity square over r and this little r here is distance to the center. Distance to center. You can also think of this as the radius of your rotation. So the idea is that if you moving around a circle that has a radius r. And you're at the edge you are at a distance little r equals big R radius ok, big R is a radius little r is a distance. But in many cases they're effectively the same thing. OK. All right let's move on here. So when an object completes one lap which we can also refer to as one lap is the same thing as one revolution that's just a different name for one rotation one spin one cycle. So those are some of the key words. One lap one revolution, one cycle. It covers a distance of one circumference. OK. So if you go around the entire circle you covered one circumference that for distance. Your displacement is technically zero. OK if you go around the entire point back to the same point but the distance is the circumference which is 2 pi r. Where r is a circle is the radius of the circle that you made. You could think of this as 2 pi or as well because if you go round a circle of radius r they a distance from the center. Little r equals big R. OK so a couple of things when going on circle the distance is 2 pi r the time that it takes to make one cycle one spin one revolution. It's called the period ok period and time to make one cycle it is represented by the letter T and it isn't measured in seconds. OK. So if you think about the word period probably the first thing that comes to mind is women's periods. And I know it's awkward but it's a cycle right. The time that it takes for a cycle is a period so you know 28 days or whatever. But in physics its measured in seconds. So it's the time for one cycle should make sense. The inverse of a period is frequency. And by the way inverse mathematically means this that the frequency is the inverse of a period. So it's one over T. OK. And I can say that T is just one over f, they are inverses of each other. Frequency we use the letter little f not to be confused with friction which is also little f and it's measured in Hertz ,Hertz frequency could also be measured in 1 over seconds. Here Seconds is your unit. So the units for frequency is the inverse of that which is going over seconds and we call that Hertz. There's another equation you should know here and this equation says that the velocity is 2 Pi r that's my that's my distance ok over the total time so distance for one circle 2 pi r time for one circle period. ok so this is just the velocity equals displacement over time or actually speed equals distance over time.

I can rewrite this in terms of f. So if t is in the bottom and f is the inverse of t it shows up appear as a 2 pi r f. So you can use either one of these versions of this equation. OK. This Vt right here. Is the same the Vt here. So we're going to see later and pretty soon actually that these two equations kind of work together. Alright. Those are the two main equations for uniform circular motion. One thing that people struggle with sometimes is how to figure out whether a piece of information that you're given is a period or a frequency. And one way to think about is a period is seconds per cycle. How many seconds does it take to do something and frequency is cycles per second. How many times can you get it done in one second the frequency of something. How often it happens. And then here is how long does it take for it to happen. OK. And one last point before we do some examples is sometimes you might get something in terms of RPM you'll be told something spins up you know 30 revolutions per minute RPM stands for revolutions per minute. The problem with revolutions per minute is that you can't use minutes as a unit in physics. So Rev's per minute has to be converted. I convert this into seconds by doing this. And when you do this you get revs per second, revs per second is the definition of frequency. OK. So this is actually frequency what's frequency, revolutions divided by 60 are actually the number of revolutions you have so your RPM divided by 60m, ok.. So long story short you have this equation here as well. If I give you 60 RPM you plug in a 60 right here and then your frequency will be 60 over 60, one. OK. So you're always going to want to get, you're always going to want to know your frequency. So if I give you RPM you have to convert that frequency because my equations are in terms of frequency and not in terms of RPM. OK. All right. So let's do this, calculate the period frequency and speed of an object moving in uniform circular motion of radius 10. So again my radius is 10 and you're going around a circle of radius 10 at the edge of that circle. So that's also your little r. And it completes 100 cycles in 60 seconds. What's your period. What's your frequency. So what I recommend you do here whenever you have something like this you do this in this month's time or you do this many things or whatever. You can just write these as a fraction. Now there's two ways you can write this one hundred cycles in 60 seconds means you're going to divide one by the other. Or you could write this backwards you could have written 60 seconds for 100 cycles. Once you're here. It's pretty easy to figure out which one is which one of these is the period the other one is going to the frequency. Here seconds is in the bottom and when seconds is in the bottom I have frequency. Here I have seconds at the top. So this is my period. So this this fraction right here 60, 100 over 60 is my frequency. And it gives us 1.67 Hertz. If I wrote it this way this fraction here 60 over 100 would have been my period. Which is point six seconds. OK. And then the last thing is I want speed. Speed is the magnitude of the velocity. So it's just the velocity without the direction basically. So it is 2 pi r Over t or times doesn't matter, I'm gonna pick the times I've said I have to do is divide here and then I'm gonna get 2 pi r is 10 frequency is one point six seven. And if you're plugging just a second. I get one of 4.9 around this one of five meters per second. All right. So what I recommend to do right now is pause the video. Try to do this next one see if you get it. I'm going to keep going but hopefully try it. It's pretty straightforward. Again I have a radius of 10. Three minutes in one cycle. Three minutes. First of all you can use minutes. So I have to convert three minutes one minute goes in the bottom 60 seconds goes up here. Or maybe you just knew that you just had to multiply 180 seconds. OK. And what is this. This is my period. So right away you know that once you convert that has to be your period. OK. Because it's 180 in one cycle. If you weren't sure you could have stack them up as a fraction 180 seconds for one cycle, seconds at the top. So period is 180 divided by one or simply 180. Frequency is 100 over t, so it's 1 over 180. I got it here and it is 0.00556 hertz. And then we can find the velocity equals 2 pi r, I could do times this number or divide it. So it's a nicer number. I got point thirty five so I'm rounding it a little bit but point thirty five meters per second. OK.

That's it for this one. Hopefully you agree. Pretty straightforward. That's just showing how to use information how to figure out if information is period or frequency. Let's do this example very quick, the car below takes 10 seconds to go from A to B. So I can write that the time delta-T from A to B is 10 seconds at a constant speed. If it's not immediately obvious just to confirm this is circular motion, circular motion doesnÕt mean necessarily you have to go around an entire circle as long as a tiny little piece of your circle of your motion is circular. This is circular motion. It tells it has a semi-circle has a radius of five. This is potentially confusing that's why put it here, the semi-circle having a radius of five. It just means that if you want to complete the circle here the circle would have a radius of five. So that's just a number you're going to use the fact that it's a radius of a semi-circle doesn't matter. In fact any kind of slightly curved path has a radius let's say and you shouldn't draw this but I'm trying to sketch something here. Let's move in this way and then you're moving this way. Nor to to switch from one to the other. Let's say you do slights. You have a slight turn. Something like this. Well this is a piece of a bigger circle. Right. And then this would be the radius of that circle so any kind of curvature and the kind of curve has a radius. Ok that's the radius and you're moving at the edge of this radius. So that is your distance from the center. Those are just most of the time the same thing. And I want to find a period. So, and some other stuff. So what is the period. So it takes 10 seconds to go from here to here. So my period is going to be 20, ok. This is again potentially tricky which is why I put this here and I want you to know the definition of period is the time that it takes to make a full rotation, even if you don't make a full rotation for example let's say it took you 10 seconds to go from here to here. If this took 10 seconds this is one quarter of a circle. Then the period would have been 40 seconds. So you can just multiply it to get a full circle. But that's not the case but hopeful that makes sense the period here is 20 not 10 not anything else, ok. Find tangential velocity, now this is just plug and chug, V equals two pi r over t, 2 pi r is 5, t is 20. And if you do this you get a 1.57 meters per second and then for part C the centripetal acceleration. It's just v squared over r. So 1.57 squared over r which is 5. This gives you point forty nine meters per second squared. OK. So these are the answers. For this question. All right. And then one last point that I want to make here one note I want to add is that even though it says here even though this is the really important by the way conceptual point if you have any conceptual, your professor gives any conceptual test questions you might see the show up it's a very classic one, in a lot of people get confused here. But even though the object speed is constant right usually constant speed tells us the acceleration is zero. You're used to seeing that a lot. Even though the sweetest cars some of the direction changes therefore the velocity changes and the acceleration is not zero. Now let me show you. Velocity is a vector. OK. So as a vector it's made up of two things. It has a magnitude which is speed. And it has a direction which is usually given by your theta. So here's the thing. If you go around a circle with constant speed it means that this is a five here. This is a five here etc. So your speed stays the same. But your direction changes. And because the direction changes your v changes as well. If either one of the parts that make up a vector change then the vector itself changes. That's what happens here because direction changes velocity changes and remember acceleration is the fun in terms of velocity, acceleration is change of velocity of a change in time. So since the velocity changes. Right. It's you can make sure you know its a velocity by putting a vector hat there. So this is the velocity changes there is an acceleration, another way that you can remember this is just by having a mental picture that this is always the same number. So let's say five, five, five. Yet there is an acceleration this way. OK. So this is the only instance circular motion is the only time where a constant speed doesn't mean acceleration zero. If you're moving in a straight line and your speed is constant then acceleration certain is zero. OK. That's it for this one. Hope it helps. Let me know if you guys have any questions

Concept #2: More Circular Motion (Practice Intro)

**Transcript**

Hey guys so now that I've introduced uniform circular motion I want to do a few more problems the first one I actually want you to try yourself I put the equations of uniform circular motion to know they're all here and I think you should be able to figure this out from the previous page let's try.

Practice: A Ping-Pong ball goes in a horizontal circle (radius 5 cm) inside a red cup twice per second. Find its:

(a) period;

(b) speed;

(c) centripetal acceleration.

Example #1: More Circular Motion

**Transcript**

All right. So this next question I want to solve is an example. I've seen this in a lot of textbooks like professors like to put this in particular in homework questions. It's a little tricky at first but it's not that bad. So let me show you. It says your one way to simulate gravity or create artificial gravity in a space station is to spin it, so if you have a space station and you spin it generates, it simulates gravity. Ok now it's as if a cylindrical space station. If that means that it's basically shaped like a cylinder like a big soda can has diameter 500 so diameter equals 500. Now in physics you're never going to use diameter. So whenever you have diameter immediately change it for a radius and radius is half of the diameter. So this is 250 meters. It spun about its central axis. So if you have a cylinder the central axis is this right, it's like the obvious way you would spin a cylinder the least awkward anyway. Anyway the question here is if it spun on central axis so looks like this. How many and how many revolutions per minute must that turn. So I want to know what is the RPM that I need, so that the outermost points. So basically if you are obviously inside but over here. Have an acceleration equal to the acceleration due to gravity at the surface of the earth.

This here the acceleration due to gravity at the surface of the earth this whole phrase just means little g. And at the surface of the earth it's g on Earth which is 9.8 meters per second square or nine point one. So here's the idea this is sort of a view at an angle. But if I want to have a top view it would look like this right if I want to look from here. That's what I mean my top view I guess if I wanna look from here I'm going to see the sky here. This thing let's say spinning like that and you're being pulled towards the middle and that creates the centripetal acceleration that then simulates gravity. So the idea is how how fast do I have to spin so that the outermost points the blue dot right there has an acceleration so that my centripetal acceleration this way equals g on earth. That's the idea. If Ac equals g what must my RPM be, RPM be so that I can accomplish that. OK. So you might remember I mentioned that too. You're usually going to convert rpm into frequency. Those things are very well connected. Frequency period and RPM are all linked up together. And when I get RPM I usually gonna, I'm usually gonna replace it with frequency using this equation frequency equals rpm by sixty. And this problem is going to be kind of backwards. If I want RPM I'm going to have to find frequency and then turn it into RPM. OK. Now looking at the information that I have here how can I try to extract frequency from here. Right. If you look at it doesn't look like frequency is anywhere here. But if you look at the equations we have which is V equals 2 pi r f that has frequency in it divided by actually that's it. And Ac is V squared over r. I hope you see how A which I'm given here and I know A is 9.8 is connected to V and V is connected to f. So I'm going to go from A to V to f too rpm by using basically all the equations we have almost for circular motion. OK. So Ac is 9.8. Ac is also V squared over r. And that equals 9.8.

So I'm going to be able to solve for V. V equals the square root of 9.8 times r, r is 250. When I do this I get forty nine point five meters per second. Now that I have V, I can try to find f. So V equals 2 pi r f so f equals V over 2 pi r, so forty nine point five divided by 2 pi 250. And if you plug this you got a frequency of point zero thirty two Hertz. Lastly there's the equation that says frequency is r.p.m. over 60. But I want rpm. So RPM is just going to be 60 f. Here's my f. Once you multiply this you rounds up to about one point nine RPMs, RPM equals 1.9. This means that you need 1.9 RPM, ok so if these things rotates at a rate of almost twice per minute. Right. So that's RPM means right, how many times per minute. It's going to create a simulated gravity of nine point eight. So that's how that works. Basically a lot of changing of these variables by using a bunch of equations we have. But hopefully we agree that it's pretty straightforward, itÕs just kind of long and annoying, alright that's it for this one.

Practice: A 3kg rock spins horizontally at the end of a 2-m string at 90 rpm. Calculate its:

(a) speed;

(b) acceleration.

0 of 5 completed

Shown here are the velocity and acceleration vectors for an object in several different types of motion. In which case is the object's velocity changing while its speed is not changing?

You are taking a turn at 30.0 m/s on a ramp of radius 25.0 m. What is your acceleration?
A) 20.8 m/s2
B) 0.833 m/s2
C) 1.20 m/s2
D) 36.0 m/s2

In a game of tetherball, the ball is hit such that it spins around the pole at 500 rpm, spinning with a 0.5 m orbit. What is the centripetal acceleration of the tetherball during this motion?

An object is moving with a constant speed of 18 m/s and an acceleration of 4.7 m/s2. Which statement is true?
A) This could be a leaf falling from a tree.
B) Nonsense. with speed constant, the acceleration must be zero.
C) This could be an arrow at the highest point of its trajectory.
D) The object moves in a circle.

You spin in a circle with a ball attached to the end of a 1 m string. If the maximum centripetal acceleration the string can withstand is 15 m/s2, what is the maximum number of revolutions you could complete in 1 minute without breaking the string.

An object is moving with a constant speed of 11 m/s and an acceleration of 2.6 m/s2. Which statement is true?A) Impossible. With acceleration, the speed has to change.B) This could be a feather undergoing freefall.C) This is circular motion.D) This could be an arrow at the highest point of its trajectory.

Which of the following is an accurate statement?A) All points on a rotating disk experience the same radial acceleration.B) The vector sum of the tangential acceleration and the centripetal acceleration can be zero for a point on a rotating disk.C) All points on a car tire have zero acceleration if the car is moving with constant linear velocity.D) All points on a rotating disk have the same linear speed.E) All points on a rotating disk have the same angular velocity.

If the acceleration of an object is always directed perpendicular to its velocity,A) the object is slowing down.B) the object is speeding up.C) this situation would not be physically possible.D) the object is turning.

It is common to see birds of prey rising upward on thermals. The path they take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume a bird completes a circle of radius 5.00 m every 5.60 s and rises vertically at a rate of 2.80 m/s. Find the magnitude of the birds’ acceleration.

An object is in a uniform circular motion. Which of the following statements must be true?A) The net force acting on the object is zero.B) The velocity of the object is constant.C) The speed of the object is constant.D) The acceleration of the object is constant.

Four children are playing in a field. The children form a line, holding hands. The player at the front of the line starts to spin around faster and faster, causing the others to run around her in a circle, as shown in (Figure 1).Part 1While the line of children is rotating, which of the following statements are correct? Check all that apply.a) The player at the front of the line has the smallest angular acceleration.b) The player at the front of the line has the smallest linear velocity.c) All the children have the same angular acceleration.d) All the children have the same linear velocity.Part 2Now consider the children's linear accelerations. Which of the following statements are correct? Check all that apply.a) The last child in the line has the greatest tangential acceleration.b) The last child in the line has the greatest radial acceleration.c) All the children have the same tangential acceleration.d) All the children have the same radial acceleration.

The Nardo ring is a circular test track for cars. It has a circumference of 12.5km. Cars travel around the track at a constant speed of 100km/h. A car starts at the easternmost point of the ring and drives for 15 minutes at this speed.(a) What distance, in km, does the car travel?(b) What is the magnitude of the car's displacement, in km, from its initial position?(c) What is the speed of the car in m/s?

A big dog has a torso that is approximately circular, with a radius of 16cm. At the midpoint of a shake, the dog's fur is moving at a remarkable 2.5m/s.a) What force is required to keep a 10 mg water droplet moving in this circular arc?b) What is the ratio of this force to the weight of a droplet?

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceleration . The flywheel is assumed to be at rest at timea) Find the time it takes to accelerate the flywheel to if the angular acceleration is .Express your answer in terms of and .b)Find the angle through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity.Express your answer in terms of some or all of the following: , , and .Find the angle through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity.Express your answer in terms of some or all of the following: , , and .

A bobsled turn banked at 78° is taken at 24.65 m/s.a. What is the radius of the bobsled turn in n, assuming it is ideally banked and there is no friction between the ice and the bobsled? b. Calculate the centripetal acceleration in m/s2.c. Calculate the centripetal acceleration as multiple of g.

A disk at rest experiences a constant angular acceleration for t=15 s, at the end of which it is spinning with a frequency of f = 75rpm. Please answer the following questions. Randomized Variables t = 5s f=75 rpm Part (a) Write an expression for the magnitude of the angular velocity of the disk after the acceleration in terms of the frequency.Part (b) Calculate the magnitude of the angular velocity of the disk in rad/s at time t = 15 s. Part (c) Write an expression for the average angular acceleration of the disk aave in terms of angular velocity.Part (d) Calculate the magnitude of the average angular acceleration of the disk aave from rest to t = 15 s in rad/s2.Part (e) Write an expression for the period T of the disk in terms of frequency f.Part (f) Calculate the magnitude of the perios of the disk T at time r = 15 s in seconds.

A penny is placed a distance r from the center of a record spinning at ω = 90π rad/min. The coefficient of static friction between the penny and the record is μs.= 0.11 on the horizontal plane. Part (a) Select an expression for the maximum distance, r, the penny can be placed from the center and not move. Part (b) What is the distance, r in meters?

A bicyclist notes that the pedal sprocket has a radius of rp = 8.5 cm while the wheel sprocket has a radius of rw = 6.5 cm.The two sprockets are connected by a chain which rotates without slipping. The bicycle wheel has a radius R- 65 cm. When pedaling the cyclist notes that the pedal rotates at one revolution every t = 1.6s. When pedaling, the wheel sprocket and the wheel move at the same angular speed.Randomized Variablesrp = 8.5 cmrw = 6.5 cmR = 65 cmt = 1.6 sPart (a) The pedal sprocket and the wheel sprocket have the same _________.a) Angular and tangential speedb) Radiusc) Tangential speed at their outer edges.d) Angular and tangential speedPart (b) Calculate the angular speed of the pedal sprocket ωp, in radians per second.Part (c) Calculate the linear speed of the outer edge of the pedal sprocket vp, in centimeters per second. Part (d) Calculate the angular speed of the wheel sprocket ωw, in radians per second.Part (e) Calculate the linear speed of the bicycle v, in meters per second, assuming the wheel does not slip across the ground. Part (f) If the cyclist wanted to travel at a speed of v2 = 5.5 m/s, how much time, in seconds, should elapse as the pedal makes one complete revolution?

Suppose a piece of dust rests on a CD. The variables to use for the problem are as follows: f = 420 rpm r = 4.1cm t = 2.8min a) If the spin rate of the CD is 420 rpm what distance is traveled by the piece of dust (in radians) in 2.8 mins? b) If the dust is 4.1 cm away from the center, what is the total distance traveled by it during the same time interval in meters?

Which is true for a car moving around a circular track with constant speed? (Check all that apply)1. It has zero acceleration2. it has an acceleration directed toward the center of its path3. it has an acceleration directed away fromt he center of its path4.it has an acceleration with a direction that cannot be determined from the information given5.it has an acceleration component in the direction of its velocity.

If an object travels at a constant speed in a circular path, the acceleration of the object is:1- larger in magnitude the smaller the radius of the circle.2- smaller in magnitude the smaller the radius of the circle.3- in the same direction as the velocity of the object.4- in the opposite direction of the velocity of the object.5- zero.

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and the centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an "artificial gravity" of 9.80 m/s2 at the rim?

A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed.What is the ratio of the normal force to the gravitational force?

A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have without flying off the road at the top of the hill?

Adjusting the test-mass riser along the platform changes the radius of the circular path. True or false?

The passengers in a roller coaster car feel 50% heavier than their true weight as the car goes through a dip with a 30.0 m radius of curvature.What is the car's speed at the bottom of the dip?

A string is wrapped around a uniform solid cylinder of radius r, as shown in the figure (Figure 1). The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass m. Note that the positive y direction is downward and counterclockwise torques are positive. Find the magnitude α of the angular acceleration of the cylinder as the block descends. Express your answer in terms of the cylinder's radius r and the magnitude of the acceleration due to gravity g.

A cyclist competes in a one-lap race around a flat, circular course of radius 140m. Starting from rest and speeding up at a constant rate throughout the race, the cyclist covers the entire course in 60s. The mass of the bicycle (including the rider) is 76kg. What is the magnitude of the net force acting on the bicycle as it crosses the finish line?

The wheel in the figure is rolling to the right without slipping.Rank in order, from fastest to slowest, the speeds of the points labeled 1 through 5.Rank items from fastest to slowest. To rank items as equivalent, overlap them.

Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass m and radius of orbit L. The satellites fire rockets that provide the force needed to maintain a circular orbit around the space station. The gravitational force is negligible.Rank the net force acting on each satellite from their rockets. Rank from largest to smallest. To rank items as equivalent, overlap them.A) m=200 kg and L= 5000mB) m=400 kg and L=2500mC) m=100kg and L=2500mD) m=100kg and L=10000mE) m=800kg and L=5000mF) m=300kg and L=7500m

Why does a satellite in a circular orbit travel at a constant speed?A. There is no component of force acting along the direction of motion of the satellite.B. The net force acting on the satellite is zero.C. There is a force acting opposite to the direction of the motion of the satellite.D. The gravitational force acting on the satellite is balanced by the centrifugal force acting on thesatellite.

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon has an arm length (hand to shoulder) of 0.60 m. We can model its motion as that of a point mass swinging at the end of a 0.60-m-long, massless rod. At the lowest point of its swing, the gibbon is moving at 3.2 m/s .What upward force must a branch provide to support the swinging gibbon?Express your answer to two significant figures and include the appropriate units.

Figure 1 shows four rotating wheels. For each, determine the signs (+ or -)of ω and α.

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