AP Physics C: Mechanics Rotational Dynamics

Hello, everyone, I am Dan Fullerton and welcome back to www.educator.com.0000

In this lesson we are going to talk about rotational dynamics.0005

Our objectives include determining the angular acceleration of an object when extra net torque or force is applied.0009

Determining the radial and tangential acceleration of a point on a rigid object.0015

Analyzing problems involving strings and mass or real pulleys.0020

Analyzing problems involving objects that roll with in without slipping0024

and applying conservation of energy to objects undergoing both translational and rotational motion.0028

As we go through this lesson, realize there are not real fundamental concepts in this one.0034

What we are doing is taking what we have done in the last few lessons, pulling them together,0039

and doing a bunch of sample problems, different applications of the things you have already learned.0042

With that, let us review a couple, first of all kinetic translational energy is ½ MV².0048

In the rotational analog is ½ Iω², where we have the moment of inertia × the square of the angular velocity.0055

The total kinetic energy, we have to add the translational and the rotational components together.0063

Let us do an example here, we have got this disc of radius R which starts at rest and rolls down an incline of height H.0071

Let us see if we can find its velocity when it gets to the bottom, its speed.0081

As I look at it, a couple of things I’m going to notice right away, we have a disk and the moment of inertia of a disk rotating about its center of mass is ½ MR².0087

As we start our analysis here, we will use a conservation of energy approach, our initial kinetic energy + our initial potential energy0101

must = final kinetic energy + our final potential energy.0109

If it starts at rest, our initial kinetic energy must = 0.0114

If we call this our 0 point at the bottom of the ramp, we can say that our final gravitational potential energy is also 0.0118

Our initial potential energy must = our final kinetic energy.0126

Our initial gravitational potential energy is MGH and our final kinetic energy is going to have both translational and rotational components.0132

We are going to have ½ mass × the velocity of the center of mass², the translational component.0145

Add it to ½ the rotational inertia × the angular velocity², the rotational component of kinetic energy.0152

Since, we know that the moment of inertia of a disk rotating about its center is ½ MR², we can write this as MGH = ½ M velocity center of mass².0163

I’m going to pull that V² from that one, + ½ × ½ MR² ω².0178

It looks like we can factor the mass out of all of these which implies then that GH = V² /2 + we will have ¼ R² ω².0192

If you recall V = ω R so R² ω² must be = V², this implies then that GH must be equal to,0212

we have V² /2 + we will have V² /4 or GH = ¾ D² which implies then that V² = 4 GH /30224

or solving for just the speed at the bottom of the incline that will be the √4/3 GH.0246

Using our rotational and translational skills all to find the answer to this problem.0258

We also talked a little bit about rotational dynamics, Newton’s second law, net force = mass × acceleration and net torque = moment of inertia × angular acceleration.0265

We can put that into play as we analyze a string with massive pulleys.0277

We have 2 blocks connected by light string over a pulley of mass MT and radius R.0282

Find the acceleration of mass M2, if M1 sits on a frictionless surface.0288

A couple things to note here, our moment of inertia, our pulley is disk ½ FMP R² and also recall the angular acceleration is A/r.0293

With that, let us start by drawing a free body diagram for M1 here.0309

I will draw it as a box, there it is, we have the normal force acting up on it.0314

We have its weight M1 G down and we have T1 to the right.0320

As we do this, let us define that direction as our positive direction so what we have a consistent axis here.0326

Newton’s second law applied in the direction of motion is going to state that the net force which is T1 has to equal M1 a.0333

Let us do the same thing for our M2 here.0344

M2, we have T2 pulling it up, we have M2 G down, and down is the positive direction.0349

For this one M2 G - T2 = M2 a or T2 is going to be = M2 G – M2 a.0359

Finally, let us take a look here at the pulley.0372

If we draw our pulley in here, we have a tension to the left T1 by Newton’s third law.0375

We have a tension T2 there, we have the weight of the pulley acting at its center of mass MPG.0385

We have the force from this pivot and should be pretty easy to see, if it is going to be some direction roughly up into the right, force of our pivot.0395

There is our free body diagram for our pulley.0403

Let us wrote Newton’s second law for rotation as we look at that pulley and we can start with net torque = we will have T2 R - T1 R for our torques.0408

Those all must equal Iα but we know T2 is M2 G - M2 A so let us write that in there for T2 M2 G - M2 A × R - T10425

we know is M1 a × R = our moment of inertia ½ MP R² × our angular acceleration which is a/R.0446

When I put all that together, it looks like I can do some simplifications.0468

First off, I got some R’s that I can pull out of this equation.0471

I get, what do we have here, M2 G - M2 a - M1 a = our R is going to go and we are just going to have mpa / 2.0475

Or rearranging this to get all our a together M2 G = M1 a + M2 a + MP a /2.0497

We can factor the a out of there so that the M2 G = a × (M1 + M2 + mp / 2) which implies that our total acceleration is going to be = M2 G ÷ M1 + M2 + MP/2.0510

There is the acceleration of mass M2, noting that we had to take a look at Newton’s second law translationally and0540

the rotational version of that in order to put it all together and solve our problem.0546

Alright, let us take a look at an example where we have some rolling without slipping.0554

We have a disk of radius R rolling down an incline of angle θ without slipping, find the force of friction on the disk.0558

Again, it is helpful to know the moment of inertia of the disk about its center of mass is ½ MR².0566

I'm going to draw our free body diagrams here, do my best of the circle on this, so something like that there is our disk.0578

As I look at the forces acting on our disk, we have the normal force perpendicular to the ramp.0593

The force of friction opposing motion acting up the ramp and we have the weight of the disk straight down.0600

I could break that up into components with my pseudo free body diagram like we have been doing.0610

Let us do that, we will draw our disk again, normal force perpendicular to the surface, force of friction is parallel with the x axis.0617

We do not have to change that, MG is not however.0637

We are going to break it up into component parallel with the axis MG sin θ.0641

The component perpendicular to the ramp which would be MG cos θ.0646

As I look at this, we can start to analyze our free body diagrams and start by looking at the net force in the y direction and0654

realizing that it is not accelerating off of the ramp in any way.0663

We can right away recognize that the normal force in MG cos θ must be = in magnitude so N = MG cos θ.0667

In the x direction, we have MG sin θ - the force of friction must = MA, so MG sin θ - friction = mass × acceleration.0683

Let us also go and start looking at the net torque on our system.0700

As we look at the net torque here, net torque is going to be, we have F operating at some distance r and its perpendicular so just FR and that must = Iα.0704

We know that moment of inertia we said was ½ MR², we can write that FR = MR² α /2.0722

We also know still that α = a/R or a = R α.0741

Let us see here, we can write that FR = if A = R α that would be MRA /2 divide an R on both sides.0748

The force of friction must = MA/2 but we know what M is, we figure that out up here MA is MG sin θ – F, that is going to be MG sin θ - F ÷ 2.0768

This implies then that the force of friction, let us multiply the 2 over so we have 2 F = 2 MG sin θ – F.0790

If I add F to both sides, 3 F = MG sin θ divide both sides by 3, friction = MG sin θ ÷ 3, the force of friction on our disk.0802

Let us do another one, this one is pretty involved.0825

As we look at a bowling ball that is rolling with slipping has a mass M and radius R.0828

It is going to skate horizontally down the alley with an initial velocity V0, find the distance the ball skids before rolling given the coefficient of kinetic friction UK.0832

And realize as we get into this problem, this should be a pretty difficult APC level Mechanics problem.0843

Not impossible but certainly we are pushing things a little bit here.0848

First thing, solid sphere uniformly distributed, we know our moment of inertia about the center of mass0854

for that is going to be 2/5 MR² and we are just going to start with that as a given.0861

Hopefully, you have memorized it here.0865

We are not going to take the time to derive it in this problem, we will have enough to do.0867

If I go draw a free body diagram for our bowling ball and there we go.0872

The forces we have acting on it is on the alley, the normal force, its weight MG, and if with slipping this is a force of kinetic friction FK.0891

Let us start with Newton's laws here.0906

The net force in the x direction is going to be just - FK = mass × acceleration.0908

In the y direction, pretty easy to see that the normal force is going to be equal 2MG because it is not accelerating up off of the alley or down through it.0918

Net force in the y direction must be 0.0926

We can put those 2 together in order to write that - FK must equal – μ KM must = - μ K,0929

and I'm going to place normal force with MG and all of that must equal mass × acceleration.0945

Therefore, we can state that our acceleration must = - μ KG.0953

We can go to our kinematic equation for velocity, velocity = initial velocity + acceleration × time.0961

We know that our acceleration is – μ KG so for our velocity, we find velocity as a function of time is V initial - μ KGT.0972

That will come in handy in a little bit.0989

Let us take a look at the net torque here, our net torque is going to be, we got force of friction FK at some distance R, perpendicular again.0991

Our sin θ is going to be 1 = I α.1004

We can make a couple of substitutions here as well.1011

We know that our force of friction is μ K MG and we also know that our moment of inertia is 2/5 MR².1014

The left hand side becomes μ K MG × R, all of that has to be = our moment of inertia 2/5 MR² × R α .1030

We can simplify that up a little bit, we can divide the M out of there.1051

We can pull an R out of there to say that μ K × G = 2/5 R α or solving for α, α = 5 μ KG ÷ 2 R.1054

Let us go to our angular velocity equation, angular velocity = initial angular velocity + α T, which implies then since α we just said was 5 μ K G /2R.1077

Ω initial is 0, we could write that our angular velocity function is = 5mu KG /2 R × T.1097

Alright with those, let us give ourselves some more room here.1117

We have to recognize that it stops slipping when V = Rω.1121

When those are equal, we have got our condition when it stops slipping.1133

We can state then that our V initial – μ KGT = time × 5 μ KG /2 R × T.1136

And what we can do is solve this for T then by writing that V0 - μ KGT = we got an R, we got an R, so that is going to be 5mu KG/2T.1157

Therefore, V initial = μ KGT + 5/2 μ KGT which implies then that V initial is going to be = 7 ½ μ KGT or solving for T, T is going to be = 2 V0 /7 μ KG.1178

We can go back to our kinematics and we want to know how far it is gone.1209

Δ X is V initial T + ½ AT² which is going to be V initial × our T 2V knot /7 μ KG + ½ × R acceleration which was - μ KG × T 2V knot /7 μ KG².1215

It becomes an exercise in algebra, Δ x = we will have 2 V knot² /7 μ KG from this term.1254

-1/2 μ KG × 4 V knot² /49 μ KG, μ K² G² which is going to be = we still have 2 V knot² /7 μ KG.1269

If we take ½ of this on the right hand side, that looks like we can have 2 V knot² /7 μ KG² -, 2 and 2 that will become 2 V knot² /49 μ KG.1296

Let us put this in terms of a common denominator.1321

We have 7 μ KG and 49 μ KG, we can multiply the top and bottom here by 7 that will give us 14 V initial²/49 μ KG – 2 V initial² /49 μ KG.1323

Which implies then that Δ x is going to be 14 -2 that will be 12 V initial² /49 μ K × G.1340

Quiet involved problem there to figure out how far that ball goes before it starts connecting and stop slipping on the bowling alley.1356

Alright, let us take a look at the amusement park swing problem, I love this problem.1368

An amusement park ride of radius x allows children to sit in the spinning swing held by cable of length capital L.1373

It smacks from angular speed the cable makes an angle of θ with a vertical as shown, determine the maximum angular speed of the rider in terms of G θ X and L.1380

We will start with a free body diagram, for our rider we are going to have MG for our rider down and a tension that being our angle θ point them up to the right,1390

or breaking it up in the components in our pseudo free body diagram we will have MG down,1407

we will have T sin θ to the right, and the vertical component T cos θ.1415

From there, we can start to write our Newton’s second law equations net force in the x direction is going to be = T sin θ which is MA1422

and since it is going in a circle MB² /R.1433

Hand in the y direction, it is pretty easy to see that T cos θ - MG = 0 force is accelerating up or down.1439

Therefore, we could write that T cos θ = MG.1450

Let us give ourselves a little bit more room to continue with the math here.1457

Let us divide that first equation by the second, we had T sin θ = MV² /R and we are going to divide that by our second equation T cos θ = MG, sin θ /cos θ1462

is going to give us the tan θ = I’m going to get V² /GR.1482

Let us convert translational speed to angular speed and solve for ω, our angular velocity.1494

If tan θ = V² /GR and we know that V = ωr then we can write that tan θ = ω² R² /GR1501

which is going to be ω² R /G or solving for ω, ω is going to be G tan θ ÷ √R.1519

Now what we have to do is replace R which is one of our defined variables with the entire distance here.1534

Replacing R, knowing that R is = to that x + the opposite side here which is the sin, L sin θ I have ω = we had G tan θ /√R, that is going to be G tan θ/x + √L sin θ.1542

There is our angular velocity as a function of our parameters.1572

Alright, let us close out with a bunch of AP practice problems to see how you can use this in so many different situations.1580

We will start by looking at the 2002 exam free response question number 2.1588

I will give you a minute to download it, you can find it up here at this link or google search it.1593

It should be easy to find on the web and a bunch of different places, print it out, try it for a minute or 2 then come back here and play to see how you did.1597

Looking at this problem for part A, find a rotational inertia of all four tires.1609

The moment of inertia of the tires is going to be = 4 × the moment of inertia for a single tire which is 4 × the moment of inertia for single tire is ½ ML²1615

which is going to be 4 × ½ our mass of the tires is going to be m/4.1629

That is m/4 × R² or total of ½ m R² for the moment of inertia for all 4 tires.1638

For part B, find the speed of the car when it reaches the bottom of the incline.1653

I would look at this from a conservation of energy perspective.1657

The initial gravitational potential energy must = the kinetic rotational energy + the kinetic translational energy at the bottom,1661

which implies then that its total mass of the top is 2m and so we have 2m GH =rotational kinetic energy ½ Iω² + ½ × our mass 2mv²,1670

which implies then that 2 mgh = ½ × moment of inertia ½ m R² ω² + ½ × 2 is just going to give us MV².1691

I will leave it as it is for now, ½ 2m V² which implies then, since we know that V = ω R and ω = V /R,1713

we are going to have the 2 mgh = ¼ mr² and ω² will be V² /R² + ½ × 2= rmv².1728

This implies then that 2 mg H = ¼ mv² as our R² /R² make a ratio of 1 + MV² or 2 mgh = 5/4 MV² divide our M out,1746

2 GH × 4/5 = V² or V² = 8 GH /5 which implies then that our velocity must be √8 GH /5.1771

Alright, let us take a look at part C, after rolling down the incline across the surface,1794

the cart collides with a bumper of negligible mass attached to an ideal spring which has some spring constant K.1801

Find the distance the spring is compressed before it all comes to rest.1807

That is another conservation of energy problem.1811

Its initial potential energy must equal the final compressed energy in the spring so we can say that 2 mgh = ½ K × xm².1814

A little bit of math, 4 mgh /K = xm².1830

Therefore, xm = √mgh /K.1838

For part D, assume the bumper has a non negligible mass after the collision,1853

the spring is compressed to a maximum distance of about 90% of the value of xm in part C, give an explanation for that.1860

As I think about that, the inelastic collision between the bumper and spring results in some sort of energy transfer to the bumper,1865

whether it is in the form of deformations, sound, and heat, you have got some energy that is given to the bumper and it is not accounted for.1875

And that is where you would have lost that about 10% of that energy.1882

I would explain in words something to that effect and I think you would be pretty well covered there.1887

Alright, there is a 2002 problem, let us move on to 2006 exam.1893

We will take a look at the 2006 exam, you can find it the same place, free response number 3.1902

Here we have got a thin hoop of mass M and rotational inertia MR² at the top of a ramp that is on the table.1909

We are asked to first derive an expression for the acceleration of the center of mass of the hoop as it rolls down there.1915

But I'm going to start with diagrams again because those always help me.1921

As we are looking on the table, it is on this ramp, there it is.1925

If I am going to draw a free body diagram, let us take a second and do that.1933

As we look at our forces here, we have the normal force out of the plane of the ramp, we have MG, its weight, and we will have the force of friction up the ramp.1943

Or if I want to do our pseudo free body diagram, normal force still out of the plane on the ramp MG sin θ,1958

down the ramp MG cos θ into the plane on the ramp, and we still have our force of friction.1978

As I look at this, derive an expression for the acceleration of the center of mass.1988

We will start by taking a look at what different equations we have.1993

We know that the net torque is going to be Iα, net torque = I α or - FR = - Iα.1998

Therefore, FR = Iα.2016

Let us take a look now over here on the left at the Newton’s second law for translational motion.2023

Let us say that the net force in the x direction is going to be MG sin θ - our force of friction F which is M8 for the center of mass.2030

And we can also see pretty easily that the normal force must = MG cos θ.2046

We have got a bunch of equations here, we will have to see where we can go with them.2051

As I look over here, we have MG sin θ - F = mass × acceleration.2057

Let us see, FR = I α, we can write then that the acceleration of the center of mass is MG sin θ – F.2066

Let us rearrange this a little bit, the acceleration for the center of mass is going to end up being α R because a = α R2078

that implies then that α = the acceleration of the center of mass ÷ R.2089

we also know the moment of inertia here is mr², it is given in the problem.2095

As we put these together, this implies then along with that piece that FR = mr² × R α which is the acceleration of the center of mass ÷ R.2101

Therefore, we have FR = mr a × the center of mass or F = mass × acceleration, our frictional force = Ma.2118

Let us pull that in over here into our blue equation and we can then write MG sin θ - Ma = Ma.2136

We can divide an M out of all of these G sin θ = 2a or a = G sin θ /2.2154

Derive an expression for the speed of the center of mass when it reaches the bottom.2170

We are getting the acceleration first, I think that covers us for part A, G sin θ /2.2175

Let us take a look at B, giving ourselves some more room on the next page.2181

Here as I look at part B, we are finding the speed of the center of mass.2186

With that, we have got the acceleration we can go to our kinematic equations VF² = V initial² + 2a Δ x.2192

V initial is going to be 0 it starts to rest so that is just VF² = 2 × the acceleration of our center of mass Δ x or VF = √2a Δ x.2203

Since we know a, VF = √GL sin θ.2221

It was a lot easier than part A.2231

Alright C, derive an expression for the horizontal distance from the edge of the table to where the hoop lands on the floor.2234

That is a projectile problem.2240

For part C, let us start by figuring out how long it is in the air.2242

Vertically, we know V initial = 0, Δ Y is going to be H, the height of the table, the acceleration in the y is going to be G.2249

You might be able to do this just off the top your head by now.2259

But Δ y is going to be V initial T + ½ a YT² where our V initial is 0.2262

This implies then that Δ y = ½ a YT² or T = √2 Δ y/a, which is going to be √2 H/G.2271

We can look at the horizontal motion.2288

If we want to know how far it goes, Δ x = the vertical velocity Vx or horizontal velocity Vx × the time.2292

We just found our horizontal velocity S √GL sin θ multiplied by our time √2 H/G, to give us √2 HL sin θ as our G's make a ratio of 1 or cancel out.2301

Part D, suppose the hoop is replaced by a disk with the same mass and radius, how will the distance from the edge of the table2330

toward the disk lands on the floor compare with the distance from part C?2338

It is going to be greater, the moment of inertia of the disk is ½ MR², less kinetic energy is taken up by the rotation leaving more translational kinetic energy2343

so it has a greater horizontal velocity as it goes off the edge of the table.2356

Therefore, it is going to travel further.2360

It has the same amount of time in the air, more horizontal velocity, it will cover more distance.2362

It is greater than and make sure you explain in detail how you would come to that sort of conclusion.2368

That covers the 2006 question number 3.2375

Let us do another one, let us go to the 2010 exam free response number 3.2379

It looks a little bit familiar, a bowling ball of mass 6kg is released from rest, from the top of the slant roof that is 4m long, an angle that 30° as shown.2387

The ball rolls without slipping and it gives us the moment of inertia there.2398

On the figure that it gives us, it asks us to draw the forces not the components acting on the ball at there points of application.2403

A free body diagram, on the diagram to give us.2415

Let us draw that something like that, we have got our ball right on there and I would label things like we have the normal force perpendicular to the table,2421

we have the force of friction up the ramp, and MG the weight of the ball.2433

That to me would be A.2441

For B, calculate the force due to friction as it rolls along the roof.2445

If you need to draw anything other than, use the space, do not do anything to this figure.2449

We are leaving our free body diagram together if we want to make a pseudo free body diagram and telling us to draw it again.2453

Let us draw our pseudo free body diagram over here for part B, a separate diagram.2461

We will draw our ball again, we have our normal force, we have our frictional force acting up the ramp,2469

and by now you are probably pretty good at seeing right away that we will have MG sin θ down the ramp and MG cos θ into the plane of the ramp.2478

Looking at Newton’s second law here, let us start with Newton’s second law in the x direction.2493

Net force in the x direction is going to be, we will have MG sin θ - the force of friction = MA.2499

Let us take a look at the Newton’s second law for rotation.2511

The net torque is going to be Iα and our net torque is going to be FR.2516

FR = Iα but we know I is 2/5 MR², it is given in the problem so that means that FR = 2/5 MR² α2525

and the third equation to relate our angular acceleration and linear acceleration A = R α .2542

Let us start by combining 2 and 3 here to see what we get.2552

If we put 2 + 3 together, I have that FR = 2/5 MR² instead of α, I'm going to write α as A/R, that implies then that FR = 2/5 MR × A,2555

which implies then that MA is just going to be =5 F ÷ 2.2583

Let us call that equation 4.2592

We can combine 1 and 4 to state and that we have MG sin θ - F = Ma, where MA is 5 F /2 which implies then that MG sin θ = 7 F /2,2595

which implies then that F is going to be = 2 MG sin θ /7.2620

Or plugging in my values, friction that is going to be 2 × mass 6kg × G 10 m/s² × sin of our angle 30° ÷ 7 comes out to be something right around2630

about 8.5 or 8.6 N, depending on how you round, whether you use 10 or 9.8.2648

There is that one for part B, for part C, let us give ourselves more room here.2660

Part C, calculate the linear speed of the center of mass of the ball when it reaches the bottom edge of the roof.2672

Here, I think I would be looking at a conservation of energy approach.2677

We will say that the potential energy at the top must = the kinetic energy, the translational kinetic energy at the bottom2682

+ the rotational kinetic energy at the bottom or MGH = ½ MV² + ½ I ω².2689

Which implies that MG H is going to be D sin θ, MG D sin θ is going to be = ½ MV² + ½ RI is 2/5 MR² × ω².2705

That implies then if V = ωr, then V² = ω² R², we could write MG D sin θ = ½ MV² + 1/5 R² ω² is just V².2724

We are going to have 2/10 or 1/5 MV² which is 7/10 MV² or V² is going to be = 10 MG D sin θ /7 M.2745

We can cancel our M's out to say that V² = 10 GD sin θ /7 or to get V all by itself, that is going to be the square root of all of that which is 10 × G 9.8 m /s².2768

Our distances for meters, sin 30° that is going to be ½ ÷ 7 or about 5.29 m/s.2786

Part D, a wagon containing a box is at rest on the ground below the roof, the ball falls 3m and sticks in the center of the box .2806

The mass of the wagon and box is 12 kg, find the speed after the balls land in.2813

A conservation momentum problem there.2818

Momentum before equals the momentum after, we are only worried about the x components here.2822

The initial mass × the initial velocity = the final mass × the final velocity.2828

Therefore, final velocity is going to be = MI /MF × our initial velocity.2835

Where our initial velocity is 5.29 m/s but we want the x component of that.2843

So that times the cos sin 30° which is about 4.58 m/s.2848

Our initial mass is 6kg, our final mass with the ball in there, you have to add this together to come up with 18 kg.2857

This implies then that our final velocity is going to be MI 6kg /our final mass 18 kg × our velocity 4.58 m /s, for a total of 1.53 m/s.2866

And that finishes the 2010 question 2, let us do one more.2891

Looking at the 2013 free response question 3, we have a disk of mass M and radius R, supported by rope of negligible mass is shown.2900

The rope is attached to the ceiling and passes under the disk, the other end of the rope is pulled upward with some force FA.2917

It tells us the rotational inertia of the disk about its center is MR² /2.2923

Part A, find the magnitude of the force necessary to hold the disk at rest.2928

Alright, as we take a look at this, we have got our disk, we have tension 1, we will call that over to the left, and on the right we have some FA.2934

We must also have the force of gravity here MG calling down.2947

Newton’s second law in the y direction T1 + FA = MG,2951

Therefore, FA = MG - T1.2958

We also know that the net torque has to = 0.2965

Therefore, T1 R - FAR must = 0.2972

Therefore, we can state that T1 must = FA, common sense there.2981

Putting those together, we can state that FA - MG - T1 then substituting for T1 FA to say that FA =MG – T12987

which implies FA = MG - FA or FA = 2 FA = MG, FA = MG /2 which is going to be 2 × 10 ÷ 2 or just 10 N.3003

Which probably makes sense, if it is a 20 N disk, each of the cords is going to have half of the tension, 10 N on each one while it is sitting there in equilibrium.3023

For part B, we are going to liven things up a little bit.3035

At time T = 0, the force FA is increased to 12 N causing the disk to accelerate upward.3038

The rope does not slip on the disk as it rotates, find the acceleration of the disk.3044

For part B, now we have T1 + FA - MG = MA.3051

Their net torque equation is FAR - T1 R= I α .3061

We know that our moment of inertia given in the problem is MR² /2, moment of inertia of a disk and = α =A /R.3073

We can rewrite this as FA × R - T1 R = I is going to be MR² /2 × α A/R, which implies then that FA - T must be equal to MA /2.3083

We can substitute that in over here so this equation together with this equation, since right that T1 + FA - MG = MA.3110

We also have down here, FA - T1 = MA /2.3125

We will combine those 2 equations, adding up the left hand side, I’m going to get 2 FA = we will have 3 MA /2.3134

I’m going to add MG over on the right side + MG which implies then that 3 MA /2 must = FA - MG which implies then that our acceleration A is 2/3 M × 2 FA – MG,3149

which implies then that acceleration is going to be 4 FA 3 M as we distribute that through -2 G/3, which is going to be 4 × 12 N/ 3 × 2kg - 2 × 10 m s² ÷ 3,3161

which is 48/6 -20/3, which is going to be 8/6 or 4/3 m/s².3180

There is part B.3205

Let us take a look at part C, calculate the angular speed of the disk at time T = 3s.3209

Our angular speed is just ω initial + α T and α = A /R so that implies then that ω is going to be = if ω is 0, ω initial is 0,3219

that means we just have AT /R which is going to be 4/3 m /s², the acceleration we just found × our time 3s ÷ our radius 0.1 m or 40 radiance/s.3235

And D, calculate the increase in total mechanical energy of the disk from T = 0 to T = 3s.3256

Alright, to do that, we are going to look at a couple different quantities.3264

We have got the change in height first, Δ H that is going to be ½ AT² which is ½ × are acceleration 4/3 × 3s² which is 46 × 9 36/6 is going to be 6m.3268

That implies that the change in gravitational potential energy is going to be MG Δ H which is 2 × 10 × 6 m or 120 joules.3288

Now we have also got shifts in kinetic energy.3302

We will have a change in rotational kinetic energy which is ½ I ω² which is going to be ½ × I, MR² / 2 × ω² which is going to be 40²3305

so that is ½ × mass we have got 2 × 0.1² × 0.1 ÷ 2 × 40²= 1600 is going to be 8 joules.3322

We have a change in translational kinetic energy which is ½ MV² or ½ × our mass × V is ω R² which is going to be ½ × our mass 2 × 40² × 0.1² or 16 joules.3340

To get the total change in energy that is going to be 120 + 8 + 16 or 144 joules.3367

Part E, the disk is replaced by a hoop of the same mass and radius,3382

indicate whether the linear acceleration of the hoop is greater than or less than or the same as the disk.3387

The linear acceleration does not depend on moment of inertia which is what increases when you switch from the distance of the hoop.3392

Because there is no dependence on moment of inertia there, it is going to be the same.3400

Make sure you justify your answer.3405

Alright, hopefully that gets you a good start on rotational dynamics with lots of in-depth examples of sample problems.3408

Thank you so much for watching www.educator.com.3413

I look forward to seeing you soon and make it a great day everybody.3416