WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I’m Dan Fullerton and in this lesson we are going to talk about angular momentum and conservation of angular momentum.
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Our objectives include calculating the angular momentum vector for moving particle.
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Calculating the angular momentum vector for rotating rigid object or angular momentum is parallel to the angular velocity.
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Recognizing conditions under which angular momentum is conserved.
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Stating the relationship between net torque and angular momentum.
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Analyzing problems in which the moment of inertia changes as an object rotates.
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Finally, analyzing collisions between moving particles and rigid rotating objects.
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Let us take a look by starting off with a quick review of linear momentum.
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Linear momentum with the symbol P is a vector describing how difficult it is to stop a moving object.
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The total momentum is the sum of individual momentum when you are dealing with the system comprised of smaller objects.
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A mass with velocity V has momentum P equal the MV, our formula for linear momentum.
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Looking in the rotational world, we have a rotational analog linear momentum and its angular momentum.
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Angular momentum given the symbol capital L is a vector describing how difficult it is to stop a rotating object.
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Just like with the linear momentum, the total angular momentum is the sum of the individual angular momentum when you have a more complex object.
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A mass with velocity V, moving at some position R about 0.2 has angular momentum L with respect to q, the angular momentum about point q.
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Our calculation of angular momentum is going to get a little bit more complicated.
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Calculating angular momentum, let us assume that we have some mass M situated at distance from point q, we will define an R vector,
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position vector from q our reference point to our mass M.
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Our mass M is moving with some velocity V in this direction which means its momentum, its linear momentum must also be in that direction.
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The angle between Rand V we will define as θ.
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By definition, the angular momentum about point q is equal to our position vector from our reference to our object crossed with its momentum vector,
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its linear momentum, which is going to be R crossed with MV.
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Our definition of linear momentum.
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Mass is a constant so we can pull it out and get our cross V × the mass.
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The direction of angular momentum is given by the right hand rule.
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Let us make sure we specify that.
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Direction by right hand rule because it is a cross product.
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Take the right hand, point the fingers of your right hand in the direction of the position vector, bend them in the direction of the velocity of the particle
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and you will have your thumb pointing in the direction of the angular momentum.
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In the case of this diagram here, our angular momentum vector is going to be into the plane of the screen.
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If we wanted to look at the magnitude of the angular momentum about point q, we can say that that is going to be MVR sin θ.
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We also know, if we start to look at something like an object that is rotating in a circle, let us make that point q,
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we will define our position vector up that way with an object here moving with the velocity there.
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Our angle is 90°, if that is the case, and we also know that V = ω R for something traveling in a circular path,
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we can say the magnitude of the angular momentum is going to be M ω R².
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Or if we rearrange that a little bit, the magnitude of the angular momentum vector about point q is MR² ω.
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Note that MR², how that is starting to look like a moment of inertia, we are not exactly there yet but
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you are starting to see a relationship that is coming out as we define angular momentum.
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For an object rotating about its center of mass, the angular momentum is equal to the moment of inertia × the angular velocity.
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That is known as an object's spin in angular momentum.
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Spin angular momentum is constant regardless of your reference point, regardless of what point you pick about which you are measuring the angular momentum.
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Regular angular momentum depends on your reference points.
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Spin angular momentum, when it is rotating about its center of mass does not.
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Let us take a look at a couple examples here to get started.
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Find the angular momentum of a planet orbiting the sun assuming that perfectly circular orbit.
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Let us define our planet and make a couple of different positions.
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If we started over here with some tangential velocity V, we will call that V1 that position 1 and
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maybe define a vector R1 from our reference point our center point we will call the q.
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There is our R1 vector, our angle here must be 90°, it is moving with some angular velocity ω.
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In a later point in time it is over here.
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We will define our vector R2 to our mass moving with velocity V2.
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Our angle there is 90°, our angular momentum about point q is R crossed P which is going to be R crossed with MV or R crossed with V × M.
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The magnitude of that is just going to be MVR sin θ but since θ = 90° we can write that as MVR and
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the direction again point the fingers in the direction of the position vector and bend in the direction of velocity,
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I have an angular momentum vector pointing into the screen again.
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It is a pretty straightforward, when the angular momentum of a planet orbiting the sun MVR into the plane.
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How about some particles?
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Find the angular momentum for 5 kg point particle located at 2, 2.
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There is a particle with the velocity of 2 m/s east and first we are going to find it about the origin, about 0, 0.
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Let us start there, the angular momentum and we will define the magnitude of it.
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The magnitude of the angular momentum about the origin is going to be MVR sin θ.
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Our mass is 5 kg, our velocity 2 m/s east, our distance, our position vector from 0 to there, if that is 2 and that is 2,
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by the Pythagorean Theorem that is going to be 2√2 and sin θ is going to be sin 45° which is √2/2.
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That is going to be 20 kg m²/s.
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Let us find it about point P over here at 2, 0.
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The angular momentum to about point P same formula MV R sin θ, same mass 5, same velocity 2.
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The position vector, the magnitude of R is just 2 this time, sin θ all that is going to be sin 90°, the angle between our position vector and the velocity vector.
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sin 90° is just 1, 5 × 2 × 2 is going to be 20 kg m²/w.
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Let us find the magnitude of the angular momentum about 0.2 here at 0, 2.
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As I look at that, the magnitude of the angular momentum about point q MV R sin θ which is going to be 5 × our velocity 2 × R again, 2 × the sin of,
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Now, our position vector and our velocity vector, angle between them is 0.
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That is going to be sin 0, sin of 0 is just 0 so our angular momentum about point q is 0.
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Notice, how we have a changing angular momentum depending upon our reference point.
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Let us take a look at angular momentum and net torque.
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If we start off with angular momentum, our definition is the position vector crossed with its momentum and we take the derivative of both sides.
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We will have to remember how we do the derivative of a cross product.
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The derivative with respect to T of A cross B is going to be the derivative of A with respect to T crossed with B + A crossed with the derivative of B with respect to T.
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So that implies if we take the derivative of both sides with respect to time, on the left hand side I have the derivative of my angular momentum
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with respect to q must be equal to, we will take DR DT cross with momentum + R crossed with DP DT.
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If you recalled DR DT, this is what we call velocity.
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The derivative of momentum with respect to time, that, we know as force.
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We can write this as the derivative of the angular momentum with respect about point q with respect to T is equal to
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we will have our velocity V crossed B + our position vector cross with F.
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Another simplification here, V cross P, the velocity and the momentum vector are in the same direction.
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Their cross product is going to be 0.
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Since, we know V crossed P= 0, determine that the derivative, the angular momentum about point q with respect to T = R cross F.
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R cross F should look familiar though.
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R cross F was our definition of torque.
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We could write then that the derivative of the angular momentum about point q with respect to T is going to be equal to the torque about point q.
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What does that mean?
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A torque on an object is going to change the objects angular momentum or if you have a change in angular momentum that must be caused by some torque.
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Torque change angular momentum.
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Onto the conservation of angular momentum and other conservation law.
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Spin angular momentum, the product of an object's moment of inertia and
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angular velocity about the center of mass is conserved in the close system with no external net torque applied.
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Spin angular momentum Iω remains constant unless you have an external torque.
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We can use that in many situations.
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We are going to look at one of them right now.
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An ice skater spins with a specific angular velocity.
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She brings her arms and legs closer to her body reducing her moment of inertia that half its original value.
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You have probably seen that if you watch a figure skater spinning.
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They have their arms out, they bring them in, and they start spinning faster.
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What happens to her angular velocity and what happens to her rotational kinetic energy?
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We know angular momentum is Iω and that has to remain constant if we do not have external torque and we do not.
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If we bring moment of inertia down, if we make that half, and angular momentum must stay the same, we have to double the angular velocity.
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Angular velocity doubles.
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What happens to rotational kinetic energy?
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Rotational kinetic energy is ½ I ω².
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If moment of inertia is cut in half but we double angular velocity and it is squared, we are going to end up with double the kinetic energy.
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Angular velocity is doubled, rotational kinetic energy is doubled.
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Where the extra energy come from?
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She had some kinetic energy K knot, she got twice that amount.
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The skater must do work to bring arms and legs and reduce that moment of inertia.
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That work becomes the kinetic energy of the skater.
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Taking a look at another example, we have a disk with moment of inertia 1 kg m² spinning about
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an axle through its center of mass with an angular velocity of 10 radians/s.
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An identical disk here on the right which is not rotating is sitting along the axle until it makes contact with the first disk.
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If the 2 disks stick together what is their combined angular velocity?
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We could take a look at this from conservation angular momentum that the initial angular momentum must equal the final angular momentum.
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Or initial moment of inertia × initial rotational velocity must equal final moment of inertia × final angular velocity.
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If I solve for final angular velocity that is going to be I initial ω initial /I final.
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Our initial moment of inertial was 1 kg m², our initial angular velocity was 10 radians/s, and our final moment of inertia,
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if the moment of inertia of one of these is 1, the moment of inertia of two of these must be 2, so 2 kg m².
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Therefore, ω final must equal 5 radians/s.
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Let us take another example, Angelina spins on a rotating pedestal with an angular velocity of 8 radiance/s,
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Bob throws an exercise ball which increases her moment of inertia from 2 kg m² to 2.5 kg m².
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What is our regular velocity after catching the exercise ball and we are going to neglect any extra net torque from the ball’s forces
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caused by the ball and assuming magically she catches the ball and her moment of inertia goes up without having any other outside effects.
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Since, there is no net external torque we know the initial spin angular momentum must equal the final spin angular momentum.
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We can solve for Angelina’s final angular velocity, L initial = L final which implies that I initial ω initial, let us write the final to make it easier, equal I final ω final.
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Ω final again = I initial ω initial/ the final moment of inertia.
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Initial moment of inertia was 2 kg m², initial angular velocity was 8 radiance/s and final moment of inertia is 2.5 kg m² after she catches the ball.
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That is going to be 4/5 of 8 or 6.4 radians/s.
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A constant force F is applied for a constant time of various points of the object below.
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Write the magnitude of the change in the object’s angular momentum due to the force from smallest to largest.
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Remember, its torque that changes angular momentum.
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We need to really look at this in terms of the torque and once we know that in terms of torque this becomes a pretty easy ranking test.
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Our smallest torque, of course, is going to be from B, next we will have C and then we will have A and then we will have D,
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the most force applied the furthest distance and at the closest angle to that center of mass line.
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B, C, A, D would be the ranking of the objects angular momentum due to the force from smallest to largest.
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Alright, let us finish up with a couple of old AP free response problems.
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We will start off with a 2005 exam Mechanics question number 3.
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You can find it here, take a minute to download it, look it over, give it a try, and come back here and see if we can do it together.
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Alright, in this problem we have a ball that is at rest to the uniform rod that is swinging in the pivot point initially rotating at an angular speed ω.
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And it looks like we are looking initially for the angular momentum of the rod about point P before they collide.
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The angular momentum of the rod about point P is just its moment of inertia × its angular velocity.
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It tells us the moment of inertia of the rod is M1 D² /3 ω.
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There is our answer.
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Moving on to part B, derive an expression for the speed of the ball after the collision.
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Here we can use conservation of angular momentum.
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A moment of inertia of the ball about point P must equal the moment of inertia of the rod about point B.
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This is while the ball is moving, this while the rod is moving, which implies then that the moment of inertia or the angular momentum of our rod,
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we just figured out was M1 D² /3 ω and that must be equal to the angular momentum of the ball about point P, is going to be its mass × its velocity × its distance.
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We do not have a sin θ there because that is going to be 90° according to our diagram.
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Therefore, all we have to do is solve for V.
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I find that V is going to be equal to, we have M1 D² /3 ω.
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We have got a D down here and then M2, which implies then that our V is going to be M1 D ω /3 M2.
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Alright there is B, let us give ourselves a little more room here for part C.
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C, says, assuming that this collision is elastic, find the numerical value of the ratio of M1/M2.
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If it is elastic collision that means the total kinetic energy before the collision must equal the total kinetic energy after the collision.
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Kinetic energy initial = kinetic energy final which implies that the kinetic energy of the rod before the collision must equal the kinetic energy of the ball after the collision.
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Or kinetic energy of the rod ½ I ω² must equal ½ M2 V², which implies then looking at our rods kinetic energy that is going to be M1 D² ω² /2 × 3 = ½ M2 V²,
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which is going to be M2 M1² D² ω² / 2 × 9 M2².
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Let me square our V from our previous answer.
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As I look to see what I can simplify here, we have got M1 here, we have got an M1² there, we have got an M2², we can get M2.
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We got 2 × 3 vs. 18 over here.
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This can be 6, that will be 18 D² ω².
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I get that 1/6 = M1/18 M2 or M1/M2 must equal 3.
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Alright, let us take a look here at D, a new ball with the same mass M1 as the rod is now placed at distant x from the pivot,
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assuming the collision is elastic for what value of x will the rod stop moving after hitting the ball?
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This looks like a conservation of angular momentum problem again.
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Angular momentum of the ball at point P = angular momentum of the rod at P so we had M1 D² /3 ω = M1 Vx.
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Therefore, the velocity of the ball is going to be D² /3x ω.
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When we go to look at our kinetic energy, we have kinetic energy of the rod before the collision must equal
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the kinetic energy of the ball after the collision because it tells us it is elastic,
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which implies that ½ I ω² = ½ M1 V², which implies that, let us see, we have ½ M1 D² /3 ω² for our left hand side.
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In the right hand side, we have ½ M1 and now V² is going to be D⁴/9 x² ω²,
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which implies then that M1 D² ω² /3, factor out the 1/2 must equal, we will have M1 D⁴ ω²/9x²,
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which implies then as we do a little bit of simplification M1 M1, that will be D² ω² ω², 3 in the 9, 13.
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I get that we have 1= D² /3x² which implies the net x² is going to be equal to D² /3 or x = D/ √3.
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That finishes up that problem.
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Let us take a look at one more.
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Let us go to the 2014 exam free response number 3.
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I will give you a minute to get that printed out and look it over.
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In this problem, we have got a large circular disk of mass M, radius R, a mass of M/2, standing on the edge of the disk.
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It got a large stone mass M/20, throws that stone horizontally, how long will it take the stone to strike the ice?
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This is a kinematics problem, we have been doing this problem for quite a while now.
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We called down the positive y direction from vertically the initial is 0, Δ with y is that height H, acceleration in the y is GD acceleration due to gravity.
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Δ y = V initial T + ½ AYT², V initial is 0.
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This implies then that our Δ y is H is going to be ½ GT² or T is going to be equal to √2 H/G.
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Part B, assuming the disk is free to slide on the ice, find an expression for the speed of the disk and person after the stone is thrown.
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That looks like a conservation of momentum problem, an explosion in one dimension.
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For part B, our initial momentum must equal our final momentum.
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Our initial momentum, our mass is M + M/2, the man + the disk × V + M/20, our stone × V initial, all of that is going to equal 0, that is our final momentum,
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which implies then that 3 M/2 V + M/20 V 0 = 0.
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In just a little bit of algebra here, 30 MV + MV initial =0.
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Therefore, 30 V + V0 = 0 which implies then that V0 is just going to be equal to V=-V0 /30.
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We have our expression for the speed of the disk and person after the stone is thrown.
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Let us take a look now at part C, let us give ourselves some more room here.
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For part C, derive an expression for the time it will take the disk to stop sliding.
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I’m going to start with a free body diagram for the disk and then we have got normal force,
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we have got its weight which is going to be 3 M /2 G and the force of kinetic friction.
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Net force in the x direction is going to be -FK which is -friction is fun, μ × the normal force which is –μ.
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Our normal force is going to be 3M /2 G and all of that has to equal our total mass 3 M/ 2 × the acceleration in the x direction,
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which implies then, 3 M /2 we can divide out and find out that ax = -μ G.
00:27:49.500 --> 00:28:02.500
If we want to know how long it is going to take to stop sliding, we can go to our kinematics V final = V initial + acceleration × time
00:28:02.500 --> 00:28:17.600
or time is going to be our V final – V initial/ acceleration which is 0 - V initial V knot /30 ÷ μ G.
00:28:17.600 --> 00:28:29.600
Complies then that our time is just going to be V0 /30 μ G.
00:28:29.600 --> 00:28:40.800
Moving on to part B, it looks like we have got a fixed poll through the center of our disk so it can rotate on the ice.
00:28:40.800 --> 00:28:50.600
The person throws the same stone horizontally, tangential direction, initial speed, and a rotational inertia of the disk is MR²/2.
00:28:50.600 --> 00:28:55.900
Find the angular speed of the disk after the stone is thrown.
00:28:55.900 --> 00:29:07.100
For part D, the moment of inertia of the disk is ½ MR²,R².
00:29:07.100 --> 00:29:19.900
We have got the moment of inertia of the man on the edge of the disk and that is going to be M/2 R², his mass × the distance from that center point.
00:29:19.900 --> 00:29:26.500
That the total moment of inertia is going to be, we have got ½ MR² + ½ MR².
00:29:26.500 --> 00:29:30.500
The man + the disk is going to be MR².
00:29:30.500 --> 00:29:39.400
We can use conservation of angular momentum, the initial angular momentum of our system must equal the final angular momentum of our system,
00:29:39.400 --> 00:29:49.100
which implies that the moment of inertia × the initial angular velocity must equal the moment of inertia × the final angular velocity.
00:29:49.100 --> 00:30:10.400
Initially, our angular momentum is 0 so that has to equal our total after, our stone M /20 V knot R + we have our moment of inertia of our man disk system MR² × ω.
00:30:10.400 --> 00:30:22.100
Solving this for ω, ω = MV knot R/ 20 MR².
00:30:22.100 --> 00:30:27.100
It looks like I’m not going to worry about magnitudes here.
00:30:27.100 --> 00:30:43.200
M will make a ratio of 1, we will have 1 /R, ω = V knot /20 R .
00:30:43.200 --> 00:30:45.800
Alright, that looks good for D.
00:30:45.800 --> 00:30:55.700
Now for part E, the person now stands on the disk at rest R/ 2 from the center of the disk, halfway to the outside.
00:30:55.700 --> 00:31:05.000
Throws a stone horizontally with speed V knot again, in the same direction, what happens to the angular speed of the disk after throwing the stone?
00:31:05.000 --> 00:31:10.100
Let us see, for part E, our moment of inertia of the man disk system is changing again.
00:31:10.100 --> 00:31:23.600
It is going to be ½ MR², the moment of inertia of the disk but the moment of inertia of the man now is his mass M/2 × the square of his distance from the center which is R/2².
00:31:23.600 --> 00:31:30.800
That is going to be ½ MR² + MR² /8.
00:31:30.800 --> 00:31:37.000
4/8 + 1/8 that is going to be 5 MR² /8 for the new moment of inertia.
00:31:37.000 --> 00:31:40.800
Our moment of inertia is going down.
00:31:40.800 --> 00:31:45.100
Let us apply conservation and angular momentum again and see what happens.
00:31:45.100 --> 00:32:06.800
L initial must equal L final, which implies that I ω initial = I ω final or again 0 = M /20 V knot, now our distance here is R/2 +
00:32:06.800 --> 00:32:14.900
our moment of inertia 5 MR² / 8 × our new angular velocity ω.
00:32:14.900 --> 00:32:36.000
A little bit of math here, MV knot /40 = 5 MR/8 ω dividing by one of those R, complies then that ω is going to be equal 8 V knot /200 R.
00:32:36.000 --> 00:32:43.200
That is going to be 4/102 or 50/125 V knot /25 R.
00:32:43.200 --> 00:32:51.100
That went down so it has to be less than.
00:32:51.100 --> 00:32:57.000
Hopefully, that gets you a pretty good start in understanding of angular momentum and conservation of angular momentum.
00:32:57.000 --> 00:32:59.000
Thank you for joining us here on www.educator.com.
00:32:59.000 --> 00:33:02.000
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