AP Physics C: Mechanics 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 hand0154

or wrap the fingers of your right hand in the direction the object is moving around that circular path0160

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 uniformly0313

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 be0671

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 velocity0901

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 then1060

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 system1145

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 projectile1158

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 / 121322

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 energy1561

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 indicate1620

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 force1641

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 and1725

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 /R1815

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 /RT1844

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