For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

### Rotational Kinematics

- Once around a circle is 360 degrees, or two pi radians. A radian measures a distance around an arc equal to the length of the arc’s radius.
- Linear position is given by the r vector. The position around a curved path is represented by s. Angular positions / displacements are given by theta (θ).
- The linear displacement can be found by multiplying the angular displacement by the radius.
- Linear speed / velocity is given by the v vector. Angular speed and velocity are represented by omega (ω).
- Linear acceleration is given by the a vector. Angular acceleration is given by alpha (α).
- Translational (linear) kinematics parallels rotational kinematics. Equations for one mirror the equations for the other.

### Rotational Kinematics

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Objectives
- Radians and Degrees
- Example I: Radian and Degrees
- Linear vs. Angular Displacement
- Linear vs. Angular Velocity
- Direction of Angular Velocity
- Converting Linear to Angular Velocity
- Example II: Angular Velocity of Earth
- Linear vs. Angular Acceleration
- Example III: Angular Acceleration
- Kinematic Variable Parallels
- Variable Translations
- Kinematic Equation Parallels
- Example IV: Deriving Centripetal Acceleration
- Example V: Angular Velocity
- Example VI: Wheel in Motion
- Example VII: AP-C 2003 FR3
- Example VIII: AP-C 2014 FR2

- Intro 0:00
- Objectives 0:07
- Radians and Degrees 0:35
- Once Around a Circle: In Degrees
- Once Around a Circle: In Radians
- Measurement of Radian
- Example I: Radian and Degrees 1:08
- Example I: Convert 90° to Radians
- Example I: Convert 6 Radians to Degree
- Linear vs. Angular Displacement 1:43
- Linear Displacement
- Angular Displacement
- Linear vs. Angular Velocity 2:04
- Linear Velocity
- Angular Velocity
- Direction of Angular Velocity 2:28
- Direction of Angular Velocity
- Converting Linear to Angular Velocity 2:58
- Converting Linear to Angular Velocity
- Example II: Angular Velocity of Earth 3:51
- Linear vs. Angular Acceleration 4:35
- Linear Acceleration
- Angular Acceleration
- Example III: Angular Acceleration 5:09
- Kinematic Variable Parallels 6:30
- Kinematic Variable Parallels: Translational & Angular
- Variable Translations 7:00
- Variable Translations: Translational & Angular
- Kinematic Equation Parallels 7:38
- Kinematic Equation Parallels: Translational & Rotational
- Example IV: Deriving Centripetal Acceleration 8:29
- Example V: Angular Velocity 13:24
- Example V: Part A
- Example V: Part B
- Example VI: Wheel in Motion 14:39
- Example VII: AP-C 2003 FR3 16:23
- Example VII: Part A
- Example VII: Part B
- Example VII: Part C
- Example VIII: AP-C 2014 FR2 25:35
- Example VIII: Part A
- Example VIII: Part B
- Example VIII: Part C
- Example VIII: Part D
- Example VIII: Part E

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### Transcription: Rotational Kinematics

*Hello, everyone, and welcome back to www.educator.com.*0000

*I am Dan Fullerton and in this lesson we are going to talk about rotational kinematics.*0003

*To begin with our objective, understand and apply relationships between translational and rotational kinematics.*0008

*Write and apply relations among the angular acceleration, angular velocity, *0016

*and angular displacement of an object rotating about a fixed axis with constant angular acceleration.*0021

*Use the right hand rule to determine the direction of the angular velocity vector.*0027

*Let us start by talking about radians in degrees.*0033

*I know a couple of these are going to be a bit of review but as we get back in the rotation in more depth, *0036

*it will probably take a minute or 2 to make sure we have got these fundamentals down.*0041

*In degrees, once around the circle is 360° and once around the circle is 2π in radians.*0045

*The radians measures the distance around an arc equivalent to the length of the arc’s radius.*0052

*That distance around δ S is circumference or 2π r or if you are measuring diameter it would just be π × d.*0058

*Let us do a couple conversions again very quickly, 90° radians, 90° if we want that in radians, 2π radians is equal to 360°.*0068

*We would get π /2 radians or 6 radians to degrees.*0079

*If we start off with 6 radians, 2π radians is 360° and those cancel out 360 ° × 6/2π is 344°.*0086

*We have got our conversions, angular vs. linear displacement.*0103

*Linear position displacement we have talked about as δ r and δ s.*0107

*Angular position or displacement we give by δ θ.*0111

*As you go around the circle, you have increasing amounts of θ, where S the linear distances r × θ or δ S is our δ θ.*0114

*If we talked about velocity in the same way, linear speed or velocity is given by the V vector.*0125

*Angular speed or velocity is given by the squiggly W, the ω vector where velocity is the derivative of position with respect to time.*0130

*Angular velocity is a derivative of angular position or displacement with respect to time.*0141

*We talk about these angular vectors, the direction is given by the right hand rule, something that is very non intuitive.*0148

*If we think about an object going around a path like this, the radius to the side, take the right hand *0154

*or wrap the fingers of your right hand in the direction the object is moving around that circular path *0160

*and your thumb will give you the direction of the positive angular velocity vector.*0165

*The angular velocity vector does not point in the direction the object is actually moving.*0170

*Converting linear to angular velocity, we have velocity as the rate of change of position or displacement with respect to time.*0177

*But we know that S is our r × θ, the radius × θ.*0188

*Therefore, we can write that this is equal to D / DT of the derivative of r θ.*0195

*But r is a constant, our radius is not changing so we can write this then as V = r D θ dt which is r ω.*0203

*D θ dt is our ω, so V = r ω or if we want ω, ω = V/ r.*0216

*Doing an example where we look at the angular velocity of the Earth.*0230

*Find the magnitude of Earth’s angular velocity in radians per second.*0234

*Ω is δ θ /δ T which is 2 π radians / 24 hours which is going to be π radians/ 12 hours but we know that 1 hour his 3600s to get this in more standard units.*0239

*Ω would be equal to 7.27 × 10⁻⁵ radians /s.*0261

*All the things that we have done before but useful to get just as firm foundation before we get a little bit more in depth here.*0269

*If we want to talk about linear vs. angular acceleration, if linear acceleration is given by A vector, angular acceleration is given by the Α vector,*0277

*where if A is the derivative of velocity, the angular acceleration is a derivative of angular velocity.*0284

*In this case, it is how quickly you are changing your angular velocity is what we call angular acceleration.*0293

*Just like we did with the angular velocity, as far as finding the direction, the direction of the angular acceleration vector is also given by a similar right hand rule.*0300

*Let us do an angular acceleration problem.*0310

*Our friend rides a unicycle, if the unicycle wheel begins at rest and accelerates uniformly *0313

*in a counterclockwise direction to an angular velocity of 15 rpm to the time of 6s,*0317

*Find the angular acceleration of the unicycle wheel.*0323

*First, let us convert rpm to radians/ s, 15 rpm is 15 revolutions / 60s but there are 2 π radians in each revolution.*0328

*That is going to be 1.57 radians /s.*0345

*Our angular acceleration is our change in angular velocity with respect to time which will be our final - our initial angular velocity with respect to time,*0353

*Or 1.57 radians/ s ÷ 6s which will be 0.26 radians /s² and because it is accelerating counterclockwise that is what we are going to call a positive angular acceleration.*0365

*As we go through and look at rotational kinematics, it is helpful to talk about some of these variables.*0390

*We are talking about translational motion, we have had δ S.*0395

*We are talking about angular, we have δ θ, a linear velocity translational velocity V, angular velocity ω.*0403

*Translational or linear acceleration A and angular acceleration Α and time is the same across both of these paradigms.*0409

*Where it starts to get useful is when we look at the variable translations, you will start to see a pattern.*0419

*If S = r θ, V = r ω, A = r α.*0425

*All we are doing is just multiplying the angular version by the radius to get the linear or the translational version.*0431

*Similarly, θ = s /r, ω = V /r, α = A /r.*0439

*You take the linear version divided by the radius to get the angular version.*0444

*Of course, time is time regardless of which paradigm you are doing.*0448

*When we get to kinematic problems, this makes our formulas much simpler.*0453

*Our kinetic equations that we have derived earlier V = V initial + AT, if we want to look at the rotational equivalent, *0459

*all we do is we replace any velocities with angular velocity.*0466

*We replace any displacements with angular displacements.*0470

*We replace any acceleration with angular accelerations.*0473

*This becomes ω = initial ω + Α × T or δ x = V knot t + ½ at² becomes δ θ = ω knot t + ½ α t² .*0477

*Or finally, V² = V initial² + 2a δ x, ω² = ω initial² + 2α δ θ.*0492

*You are just replacing the variables but the form of those kinetic equations and they are used that is exactly the same.*0501

*We have also talked about how we derive that centripetal acceleration.*0510

*Probably, we are taking a minute and doing it again just to make sure we have it down.*0514

*So if we look at some specific point that you get after some angular displacement θ, that sometime from T0 to T, *0518

*we could call its x position will be r cos θ and its y position would be r sin θ.*0526

*Our r vector is going to be r cos where θ = ω t, that would be ω t I hat + r sin ω t j hat.*0535

*Velocity is just going to be the derivative of that, it is the derivative of r with respect to t, *0552

*that is going to be the derivative with respect to t of r cos ω t i hat + r sin ω t j hat.*0558

*Which implies then that V is equal to, if we take the derivative of this we will get the derivative of the first + the derivative of the second.*0574

*That is going to be V equal to, we will have r I hat, derivative of cos is going to be opposite of the sin.*0582

*This will be in r I hat × ω – sin ω t.*0592

*We have our term over here, we will have + r j hat ω cos ωt.*0602

*If we rearrange them a little bit and make it look a little bit more formal, V = -ω r sin ω t I hat + ω r cos ω t j hat.*0616

*There is our velocity but we can take that a step further and let us do that.*0640

*If V (t), let us give a little room for that, if V (t)= - ω r sin ω t I hat + ω t cos ω t j hat.*0652

*And our acceleration is going to be the derivative velocity with respect to time which is going to be *0671

*the derivative of all of this is going to be - ω² r cos ω t I hat - ω² r sin ωt j hat.*0678

*Which implies then that acceleration = - ω² and then we are left with r cos ω t I hat + r sin ω t j hat.*0698

*If we recalled this is our initial r vector.*0719

*A is - ω² r or we also know ω = V /r so that means A = - V /r² × r or V² /r - V² /r.*0726

*Now the negative sign, why are we worried about that?*0751

*We are talking about the centripetal acceleration, we are defining toward the center of the circle as positive so that would V² /r.*0754

*When we are talking about the vectors, we define it this way where r is from the center out to that position point while the centripetal acceleration is opposite of that.*0762

*That is where the negative comes.*0771

*R goes from the center to the circle where as the acceleration is from the object to the center.*0775

*It is easier as we do this and stop worrying about our vector signs and directions, A is V² /r.*0792

*Let us do an example here.*0803

*An object of mass M moves in a circular path of radius r according to θ = 2 t2 + t + 4, where θ is measured in radians and t is in seconds.*0805

*Find the angular velocity of the object that equals to seconds.*0817

*As I look at this, ω = d θ dt which is going to be the derivative with respect to time of 2t³ + t + 4, which is going to be 6t² + 1,*0822

*which implies then since t = 2s, that ω T = 2s is going to be 6 × 2² + 1 or 25 radians/s.*0839

*Find the object’s speed at this time.*0857

*The velocity, the speed at T = 2s is going to be r ω at t = 2s which is just going to be 25r m/s.*0860

*Let us take a look at an example with the wheel.*0878

*The wheel of radius r and mass capital M undergoes a constant angular acceleration of magnitude α,*0881

*what is the speed of the wheel after it is completed one complete turn assuming it started from rest?*0890

*This is a kinematics problem so let us figure out what we know.*0896

*Our initial angular velocity is 0 because it starts from rest, we are trying to find its final angular velocity *0901

*or trying to find its final linear velocity but angular velocity we will get to it later.*0909

*We know our displacement is 2π once around the circle and we have some angular acceleration α.*0914

*With what I know, I would go to my kinematic equation by rotational version ω final² = ω initial² + 2 α δ θ or ω final² = ω initial² is 0 so this becomes 2 × α δ θ is 2π.*0924

*This is 4π α, ω final is just going to be √4π α.*0952

*If we want that, our speed, that is going to r ω that would be r√4π α.*0964

*Let us finish up by doing a couple of AP problems from old past AP exams.*0978

*We will take a look first at the 2003 exam Mechanics question 3.*0984

*Take a minute, pull that out, you can find it here at the link above or google it, download it and give it a shot and then come back here and see what we have got.*0989

*It looks like we first are plotting some data points.*1000

*I plotted the data points first and we are supposed to draw the best fit curve.*1003

*It is kind of a goofy, easy, initial question there.*1009

*I would draw at something like this and to draw a curve, the shape, I have this kind of like that.*1014

*It says using that best fit curve determine the distance traveled by the projectile if a 250 kg is placed in the counterweight bucket.*1023

*To do that, all I do is go up here to wherever it happens to be 250 kg and come here and read off on the graph and I get an x of about 33 m.*1034

*All right, going to part B.*1051

*For part B, says students are assuming that the mass of the arm, the cup, and the counterweight bucket can be neglected and then *1060

*they develop a model for x as a function of mass using x = Vxt, where Vx is the horizontal velocity of the projectile as it flies off the top of the cup and T is the time.*1069

*First off, how many seconds after leaving the cup the projectile strike the ground?*1081

*That sounds like a kinematics question to me, so B1 we have an initial, if we look vertically, initial vertical velocity is 0.*1086

*We do not know our final vertical velocity, δ y is 15m, our acceleration is 10 m /s².*1098

*We are calling down the positive Y direction and T that is what we are trying to find.*1106

*I would use δ y = V initial t + ½ Ay T², where V initial is 0 so T is going to be 2 δ y / √A which is 2 × 15m / 10m/s², 30/10² that is about 1.73s.*1113

*Let us go onto part B2, derive the equation that describes the gravitational potential energy of the system *1145

*relative to the ground assuming the mass in the counterweight bucket is M.*1153

*For B2, as I look at that, our initial potential energy is going to be equal to the potential energy in the bucket + the potential energy of our projectile *1158

*which is going to be mass B, G × the height of B + mass of the projectile G × the height of the projectile, which is, let us see what we have, 10 m/s² for G × 3m × M + 10 kg.*1169

*And all of that is going to be equal to 300 K + 30 M.*1192

*Now let us take a look at part 3, derive the equation for the velocity as it leaves the cup.*1204

*We are getting a little bit more involved here.*1210

*Part 3, our final potential energy + our final kinetic energy must equal V initial and we are looking at when it leaves the cup there.*1214

*We can say that our final potential is going to have to equal, we have got 1 × 10 × 110 M, 1 × 10 × M + 15 × 10 m/s² × 10 =1500 + 10 M.*1226

*Our kinetic final, we know is ½, which implies there we are just coming up with it again, we are just coming up with the different pieces.*1251

*Our kinetic finals is going to be ½ × mass × square root of our velocity V x² + ½ mass of our bucket × the velocity of our bucket².*1262

*Our initial potential, we said was 300 + 30 M from part B2 up above.*1278

*Putting all of this together, we have 300 + 30 M must be equal to that 10 M + 1500 from up here + we have got 5 Vx² + ½ M VB².*1286

*The key to solving this problem at this point is realizing that both ends of that catapult, they are swinging with the same angular velocity.*1313

*If that is the case, ω B must equal ω A and since V = ω r and ω = V /r, we can write that VB / 2 must equal Vx / 12 *1322

*and then we get our relationship between the velocity of bucket and our Vx.*1337

*VB must equal Vx/6 so now we can go and we can put that back in our blue equation there to solve for the velocity as it leaves that Vx.*1343

*300 + 30 M = 10 M + 1500 + 5 Vx² + ½ M and VB² now is just going to be Vx² / 36.*1356

*It is an algebra exercise, 20 M = we take that 300 out, 20 M = 1200 + 5 Vx² + M Vx² / 72 which implies that 20 M - 1200 is going to be equal to Vx² × 5 + M / 72.*1383

*Or getting vxx all by itself, Vx is going to be equal to 20 M - 1200 ÷ 5 + M / √72.*1420

* And there is probably a way to simplify that further but that looks like plenty to me.*1434

*Onto part C, complete the theoretical model by writing a relationship for x as a function of the counterweight mass using the results from B1 and B3.*1444

*That is just x = velocity × time from our horizontal kinematics which is just going to be, time was 1.73.*1457

*This is 1.73 × √20 M - 1200/5 +√M / 72.*1467

*Part C2, compare the experimental and theoretical values of x for a counterweight bucket mass of 300 kg.*1482

*Theoretically, when we plug in for M = 300kg in our formula, I come up with about 39.6m and we write it off the graph, *1492

*if we read it off the graph actual M at 300kg was about 37m.*1507

*Where did that difference come from?*1514

*The difference could be from a lot of things, you could have had friction at the axis, you could have air resistance.*1517

*Remember, how we neglected the masses of the arm, the bucket, and the cup, any of those can all contribute to this difference there.*1522

*That covers that free response, let us take a look at one more here.*1530

*Let us go to the 2014 Mechanics exam free response number 2.*1536

*Again there is the link there, we can google it, take a minute and print it out, check it out, and try it, and come back here and see how you do.*1540

*It is kind of an interesting set up on this one.*1552

*The first thing I have to do is find an expression for the height of the ramp in terms of the V knot M and fundamental constants.*1554

*I would use conservation of energy where the final kinetic energy must be equal to the initial gravitational potential energy *1561

*or ½ M × what they call V initial² = MGH.*1570

*Therefore, just solving for H that is going to be V knot² / 2G, pretty straightforward for part A.*1579

*Move on to B, short time after passing point T, the block is in contact with the wall and moves with the speed of V.*1591

*Is the vertical component of the net force on the block upward, downward, or 0?*1599

*There is no vertical acceleration so Ay = 0, which means you can say that the net force in the y direction must equal 0 so 0.*1604

*In part 2, it says on that figure, draw an arrow starting on the block to indicate *1620

*the direction of the horizontal component of the net force on the moving block when it is that position shown.*1625

*We have got it moving in a circular path and it is right here at that position.*1631

*When it is at that position, we have the normal force acting directly to the left so that is the direction of the normal force *1641

*but we also have some amount of frictional force that is opposing the motion.*1649

*When I put those 2 together, we will add them up tip to tail normal friction, I get something that must be down into the left for my net force.*1655

*I would draw something like that for my net force.*1664

*That leads us to part C, determine an expression for the magnitude of the normal force exerted on the block by the wall as a function of velocity.*1673

*The net centripetal force is MV² /r and the only centripetal component is provided by the normal force that is M.*1685

*Therefore, I would say that N = MV² /r.*1695

*The frictional force is perpendicular to that is not going to come into play when we are talking about the centripetal force.*1699

*We have here part D, derive an expression for the magnitude of the tangential acceleration of the block at the instantaneous speed V.*1708

*Tangential acceleration has to do with our frictional force.*1721

*Our frictional force is μ × the normal force and we just found the normal force so that means that our frictional force is going to be me MV² /r and *1725

*that must be equal to mass × the tangential acceleration or the magnitude of the tangential acceleration is just going to be μ V² /R.*1739

*It looks like we got one more part of the question.*1757

*Let us give ourselves a little more room here.*1760

*For part E, derive an expression for the velocity as a function of time after passing point T.*1763

*To start with, we know that the magnitude of our tangential acceleration is – μ V² /r, that negative because the speed is in the opposite direction of the acceleration.*1771

*We also know that acceleration is the derivative of velocity with respect to time, dv dt must equal – μ V² / R or dv / V² is going to be equal –μ/R dt.*1785

*If we want to get our velocity then it looks like we are going to have to do a little bit of integration.*1810

*We will integrate this so I’m going rewrite this first as the integral of V⁻² dv evaluated from some velocity V initial to final V = - μ /R *1815

*integral from T = 0 to T of dt which implies then the integral of V⁻² is -1 /V.*1832

*– V⁻¹ evaluated from V knot to V = - μ / R × T or I could write that as -1 / V -1 /V initial = -μ /RT *1844

*which implies then that, let us rearrange the order there, put that negative through.*1864

*1/ V knot -1 / V = - μ/ R T or getting a common denominator, multiplying everything by V knot, V knot /V knot - V knot / V = -V knot μ T / R.*1869

*Let us see if we can do little bit more rearrangement.*1893

*That is just 1 so we could say that – V knot / V = - V knot μ T / R -1 which implies then, let us multiply that to -1.*1896

*V knot / V is equal to, I’m going to write 1 as R/R.*1913

*R/R + V knot μ T /R or V knot / V is equal to R + V knot μ T / R which implies then, if V knot / V is that / that and V /V knot is the denominator/numerator.*1918

*V / V knot = R /r + V knot μ T.*1941

*V all by itself is V knot × r /r + V knot μ T.*1949

*All the work for that, alright hopefully that gets you a pretty good start on rotational kinematics.*1962

*Thank you so much for spending time with us here at www.educator.com and make it a great day everyone. *1968

1 answer

Last reply by: Professor Dan Fullerton

Sun Dec 20, 2015 8:43 AM

Post by Jim Tang on December 19, 2015

These lectures click so much more with the AP frq at the end. Your physics 1 and 2 would be so good if you had them, maybe some B frq? Anyways, great lectures!

1 answer

Last reply by: Professor Dan Fullerton

Mon Mar 9, 2015 6:07 AM

Post by Philip Schultz on March 9, 2015

Where are the practice problems like advertised?

0 answers

Post by Professor Dan Fullerton on December 31, 2014

Since r is a constant, it gets pulled out of the integration, for the rest of the derivation, I am using the chain rule. (Derivative of rcos (wt) = -rsin(wt)*d/dt(wt) = -wrsin(wt)

1 answer

Last reply by: Thadeus McNamara

Wed Dec 31, 2014 3:26 PM

Post by Thadeus McNamara on December 31, 2014

at around 9:20, can you rexplain how you found the derivative of r? why arent you using chain rule