WEBVTT physics/ap-physics-1-2/fullerton
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Hi, folks. I am Dan Fullerton and I would like to welcome you back to Educator.com.
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Our next topic -- Rotational Dynamics.
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Our objectives are going to be to understand the moment of inertia or rotational inertia of an object or system -- depends upon the distribution of mass within the object or system, to determine the angular acceleration of an object when an external torque or force is applied.
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We will calculate the angular momentum for a point particle, utilize the Law of Conservation of angular momentum and analyzing the behavior of rotating rigid bodies, and finally calculate the kinetic energy of a rotating body.
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With that, let us talk about types of inertia.
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So far, we have talked about inertial mass or translational inertia, which is an object's ability to resist the linear acceleration.
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Well, in the rotational world, we have an analog of that as well. It's called moment of inertia or rotational inertia.
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That is an object resistance to a rotational acceleration or an angular acceleration.
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Now, objects of that most of their mass near their center of rotation tend to have smaller rotational inertias than objects with more mass farther from their axis of rotation.
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Think of a figure skater spinning on the ice. While their arms are out, they tend to go slower.
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To go faster they pull their arms in; they are shrinking their moment of inertia as they do that.
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Smaller moment of inertia means easier to accelerate.
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The formula for moment of inertia is the sum of mass times the square of the radius.
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Now, if you have an object that is more complex than a simple particle, you have to add up all of the little bitty pieces of mass times the square of their distance from that axis of rotation.
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Add them all up and you get the moment of inertia.
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Let us talk about the moment of inertia for a couple of common objects.
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For any object, if you take the sum of the all masses times the square of the distance from the axis of rotation that formula will work for any object.
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But that is not always easy to apply, so for some common objects -- things like a disc, the moment of inertia is 1/2 times the mass of the disc times the square of its radius, assuming it is a uniform mass density distribution.
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A hoop on the other hand is mr². A solid sphere is 2/5 mr².
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A hollow sphere on the other hand, where all of the mass is on the outside almost like a spherical shell, is 2/3 mr².
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A rod rotated about its center is going to be about 1/12 mL² where L is the length of the rod, but if you rotate it about its end, then it becomes 1/3 mL².
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The moment of inertia goes up because more of the mass is situated away from that axis further away from that axis of rotation.
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Let us take a look of how could we calculate moment of inertia.
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We have two 5 kg bowling balls joined by a meter long rod and we are going to say that rod is of negligible mass.
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If we rotate it about the center of the rod we can find its moment of inertia this way.
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The moment of inertia is going to be the sum for all the different particles of mr², which in this case -- let us call this m1 and we will call this m2.
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We will call this distance r1 and this distance r2.
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That is going to be m1(r1)² + m2(r2)².
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In this case, m1 here is going to be 5 and if this whole distance -- we call 1 m then r1 must be 1/2, so that is .5² + m2(5) × r(.5)².
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This gives me a moment of inertia equal to 2.5 kg × m².
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Now, let us take the same object and rotate it now by the end under one of the bowling balls -- so putting more of the mass further away.
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That to me, just theoretically, I would think you know that is going to be harder to spin.
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I am thinking we are going to have a larger moment of inertia. Let us find out.
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Once again, moment of inertia, capital I, is the sum of mr², which will be m1r1² + m2r2².
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Once again, m1, m2, but now r1 is this entire distance, 1 m and r2 is going to be 0.
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So I end up with mass1(5) × r1(1)² + mass2(5) × distance from the axis of rotation, 0²
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That is just going to be 5 kg-m².
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So the moment of inertia here doubled compared to when we spun it about its center of mass.
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We can take a look at this in terms of Newton's Second Law as well.
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Newton's Second Law said that the net force on an object was equal to its mass -- its linear inertia times the acceleration.
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The angular acceleration of an object, on the other hand, was the net torque applied divided by the object's moment of inertia.
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Again we have the same parallels -- force, torque. Linear inertia -- rotational inertia. Linear acceleration -- rotational acceleration
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It all works the same way. Let us take another example here. Let us talk about a rotating top.
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A top with moment of inertia .001 kg-m² is spun on a table by applying a torque of .01N-m for 2 seconds.
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If the top starts from rest find the final angular velocity of the top.
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Well, let us figure out what information we know to begin with.
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The initial angular velocity is 0. It starts at rest.
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We are trying to find the final angular velocity.
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We do not know the angular displacement; we do not know α, and we don't know time.
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Pardon me. We do know time, it is 2 s.
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Well, it is sure be helpful to know that angular acceleration.
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Let us take a look and say that the net torque is equal to I(α) -- Newton's Second Law for Rotation.
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That means then that α is going to be the net torque divided by the moment of inertia and our net torque was .01N-m and our moment of inertia .001 kg-m².
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That tells me then that my angular acceleration must be 10 rad/s².
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I can plug that in over here for my α as 10 rad/s².
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Now, I can use my kinematics to find what final angular velocity is.
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Final angular velocity is initial angular velocity plus α times time
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That is going to be -- well this is 0, so 10 rad/s² × 2s is going to give us a final angular velocity of 20 rad/s.
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That is Newton's Second Law in kinematics, all put together -- This time though for rotation.
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Let us take another example.
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What is the angular acceleration experience by a uniform solid disc of mass 2 kg and radius .1 m when the net torque of 10N-m is applied?
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Assume the disc spins about its center, which we can see from the diagram there as well.
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Well, net torque is moment of inertia or rotational inertia times angular acceleration.
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Now, because this is a disk we can look up its moment of inertia which is going to be 1/2mr², where there is our (r) and it has some total mass (m).
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The net torque equals -- well I, we have 1/2mr² × α.
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Therefore, α must be equal to 2 times our net torque divided by mr².
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Now, we can substitute in our values to find that α is equal to 2 times our net torque (10N-m) divided by the mass (2 kg) times the square of the radius .1 m².
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20/2 × .1² = .01, which should give us 1000 rad/s².
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The same basic sort of problem -- now, we are just solving for angular acceleration and we had to go look up the formula for the moment of inertia, which you saw a couple of slides ago for some common objects.
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Linear momentum -- the product of an object's inertial mass and its velocity -- is conserved in a closed system.
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That is the conservation of linear momentum. We have talked about that already.
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Linear momentum describes how difficult it is to stop a moving object.
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There is an analogy in the rotational world, too.
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Angular momentum -- a vector (capital L), which is the product of an object's moment of inertia or rotational inertia and its angular velocity about the center of mass -- is also conserved in a closed system when there are not any external torques.
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That describes how difficult it is to stop a rotating object.
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We have angular momentum equals moment of inertia times angular velocity.
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That fits right along with our analogy, linear momentum equals linear inertia (mass) times linear velocity.
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Here are the analogs -- angular momentum, linear momentum; rotational inertia, linear inertia; angular velocity, linear velocity -- same sort of parallels again.
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How do we calculate angular momentum?
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Well, what we are going to do is, we are going to talk about a mass moving along with some velocity (v) at some position(r) about point (Q).
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Angular momentum depends on their point of reference.
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We are going to start by setting a reference point (Q).
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In that case, the object has some angular momentum (L) about (Q) and we could find that by multiplying the vectors (r) and (p) with the vector cross product -- the vector product, which will give us another vector, which is a lot like we did the talking about torque.
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The angular momentum vector (r) cross (p) -- we determine its direction by the right-hand rule.
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Point the fingers of your right hand in the direction of (r) where (r) is the vector from your reference point to the object.
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Now, bend your fingers in the direction of the velocity. Your thumb then will point in the direction of the positive angular momentum.
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It is another right-hand rule, so that would be into or out of the plane of the page.
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In this case, if I point the fingers of my right hand in the direction of (r), bend them in the direction of (v), my thumb is going to point into the plane of the page or the screen here.
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The direction of the angular momentum vector would be into the plane of the page.
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Its magnitude is given by (mvr) sin(θ) -- mass times velocity times its distance (r) times the sine of the angle between this continued line and velocity -- very, very similar to torque.
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We have two ways to find out our angular momentum.
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Now, total angular momentum -- if we have a bunch of particles -- is just the sum of all the individual angular momenta.
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Let us take a look quickly at a special case here -- what about for an object traveling in a circle?
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Now, if we have some mass traveling in a circle with some velocity at a given point (V) and it is located some radius (r) from the center of the circle -- and let us call that point (C), our reference point...
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...Then the angular momentum about point (C) is going to be (mvr) sin(θ).
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But notice because (v) is always going to be tangent to the circle and (r) is always 90 degrees from that -- sin(θ) is always going to be 90 degrees -- sin 90 degrees is 1.
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So that is just going to be (mvr), but, also remember when we do our translation between linear and angular variables that (v) is equal to = ω(r).
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I can replace (v) with ω(r) so that is (m) ω(r) times another (r) or (r)².
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If I rewrite that, I could rewrite that as ω (mr)², but if you recall for a point particle mr² is the moment of inertia.
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(L) about point (C) is equal to ω times (I), or as we wrote it earlier that is I(ω).
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That is where that comes from.
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Angular momentum is equal to rotational inertia or moment of inertia times angular velocity.
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Let us take a look at how we could calculate angular momentum for a couple of particles.
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We are trying to find the angular momentum for a 5 kg point particle located at 2-2 with a velocity of 2 m/s to the East.
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We want to find it about three different points though, so first, let us find it about this point (O).
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The angular momentum about point (O) -- and let us just stick with this magnitude now to make life nice and simple.
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The magnitude of the angular momentum about point (O) is going to be equal to (mvr) sin(θ), where our mass is 5 and our velocity is 2 m/s.
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Our distance from our point -- well if this is 2 and this is 2, the Pythagorean Theorem says right here that our hypotenuse must be 2 square roots of 2.
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The sin of θ -- well that is going to be an angle here of 45 degrees and that is equal to square root of 2 over 2.
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When I do all of this -- 5 × 2 = 10 × 2 = 20 and square root of 2 × and the square root of 2/2 = 1, so I end up with 20 kg-m²/s.
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Now, let us find it about point (P).
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Angular momentum about point (P) -- same formula (mvr) sin(θ).
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Our mass is still same 5 and our velocity is still 2.
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Now, about point (P) though -- our (r) distance is just 2 units (2) and the sine here is going to be sin 90 degrees which is 1, so 5 × 2 = 10 × 2 = 20 × 1 = 20 -- 20 kg-m²/s.
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So for the moment of inertia about (O) and about (P), you get the same thing.
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Now, let us do it about point (Q) -- Moment of inertia about point (Q) is going to be (mvr) sin(θ), but in this case, about point (Q), notice our (r) vector and (v) vector are in the same direction -- the angle between them then is 0.
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Since the sine of 0 degrees equals 0, the angular momentum about point (Q) is going to be 0.
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Angular momentum depends on your point of reference.
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Let us take a look at an example with a rotating pedestal.
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Angelina spins on a rotating pedestal with an angular velocity of 8 rad/s.
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Bob throws her an exercise ball which increases her moment of inertia from 2kg-m² to 2 1/2 kg-m².
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What is Angelina's angular velocity after she catches exercise ball?
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We are going to neglect any external torque from the ball just to keep the problem simple.
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Well, what I do here is realize by conservation of angular momentum -- since she is spinning about her center, her axis of rotation -- we can say that the total angular momentum before she catches the ball must be equal to the total angular momentum after she catches the ball.
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So, (L)initial equals (L)final, but angular momentum is moment of inertia times angular velocity initial, so that must equal moment of inertia final times angular velocity final.
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Well, initial moment of inertia, we know is 2 and final is going to be 2 1/2, so ω must change.
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In this case, (I)initial is 2, (ω)initial is 8, so that must equal (I)final (2.5) times whatever her final angular velocity is.
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16 divided by 2 1/2 -- I am going to come up with an angular velocity of 6.4 rad/s.
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By increasing her rotational inertia, her angular velocity decreases.
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Let us take some example of some rotating discs.
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We have a disc with moment of inertia 1 kg-m² spinning about an axle through its center.
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It has an angular velocity of 10 rad/s.
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An identical disc which is not rotating is slid along the axle until it makes contact with the first disc.
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If the 2 discs then stick together, what is their combined angular velocity?
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Well, I will go back to conservation of angular momentum, which will work because they are rotating about their centers of mass.
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Initial angular momentum equals final angular momentum or initial moment of inertia and initial angular velocity must equal final moment of inertia, final angular velocity.
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I want to know what the final angular velocity is.
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That is going to be equal to I(0), ω(0) over I-final.
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I-initial was 1, ω-initial was 10, and I-final -- well if we double that, it is going to go from 1 kg-m² to 2 kgm², so 10/2 = 5 rad/s.
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This should make some amount of intuitive sense -- one objects spinning at 10 rad/s, the other is still, but identical object, and you put them together -- What happens?
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Again you get the twice the mass, twice the rotational inertia, and half the angular velocity.
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Let us talk about angular momentum with respect to heavenly bodies.
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Really what we are talking about here is orbits.
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We want to develop a relationship for the velocity and radius of a planet in an elliptical orbit about any point in that orbit.
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Now, right away when we look at this, we know angular momentum must be conserved because there is no external torque in the system.
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We will go put something like a planet over here, and call up the mass.
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It has some velocity right at that point.
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At this point, it has some (r) vector r(1), we will call that v(1) and there is our mass.
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Another point in time -- say it is down over here -- now it has velocity (2) and it has a different position vector r(2).
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Since the net torque is 0 though, the total angular momentum must be the same.
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The angular momentum about point (S) is going to be -- well when it is at point (1) that it is going to be (m1v1r1) sin(θ)(1) where that angle there is θ(1).
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But that also must be equal to the angular momentum over here at point (2) -- (m2v2r2) sin(θ)(2).
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But the mass is the same. That has not changed.
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We can divide out the mass and then state that (v1r1) sin(θ)(1) must equal (v2r2) sin(θ)(2).
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Our relationship between the velocity is the distance from the Sun and the angle at any point in that orbit.
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Now for the special case, when the planet is at this point, which is known as the apogee point -- so let us call that point (A) or when it is over here at perigee, you can call that point (P).
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Well, at those points we have a special situation, because if you look here -- the velocity and the (r) vectors -- we are going to have an angle of 90 degrees and the same thing over here.
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We have (r) versus velocity and our angle here again is (θ) 90 degrees, so at apogee and perigee, we can simplify this even further.
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The velocity at apogee times the radius of the apogee times the sine of θ at apogee must equal the velocity at perigee times the radius with a position vector at perigee times the sine of θ(P).
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But since these are both 90 degrees and the sine of 90 degrees is 1, we can simplify this to say that the velocity at (A) times the length of the position vector at (A) must equal the velocity at perigee times the position vector at perigee.
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That works when we are at these special points where we have got that 90 degree angle.
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It is a nice relationship between velocity and the position vector and the angle.
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All right, let us talk now about types of kinetic energy.
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We briefly talked about the kinetic energy of an object as the energy an object has due to its state of motion.
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The translational kinetic energy we talked about was 1/2 mv² -- mass times the square of speed.
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Objects traveling with a translational energy must have a translational kinetic energy.
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Similarly again another parallel to rotational motion, objects that are spinning must have a rotational kinetic energy.
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Again as we look here, rotational kinetic energy is 1/2 instead of mass or linear inertia -- we have rotational inertia or moment of inertia.
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Instead of linear velocity squared, we have angular velocity squared.
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Same parallels again just swapping the linear variables for the rotational variables.
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If we wanted to put this all together into a nice table -- displacement in the translational world, we called Δ(s) or Δ(x) depending on what we were talking about.
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In the angular world, Δ(θ) - angular displacement; velocity (v) - angular velocity ω; linear acceleration A - angular acceleration α; and time, the same in both worlds...
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...Force (F) linear, the angular equivalent torque and mass or moment of inertia -- (m) in the translational world is (I) -- rotational inertia in the angular world.
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In our equations, we can expand too -- (S) = r(θ), θ equals (s) over (r).
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We have done this translations between linear and angular quantities before.
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Time is the same, but now Newton's Second Law -- F = ma and torque = I(α).
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For momentum -- linear momentum (P) equals mass times velocity and angular momentum equals moment of inertia times angular velocity.
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And kinetic energy -- kinetic energy is 1/2 mv² and rotational kinetic energy is 1/2I(ω)².
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Let us put this together to talk about the kinetic energy of a basketball.
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A .62 kg basketball flies through the air with a velocity of 8 m/s.
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Find its translational kinetic energy.
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Well, kinetic energy to the translation linear kinetic energy is 1/2 mv², which is going to be 1/2 times our mass (.62 kg) times our velocity (8 m/s²) or 19.84 and the units of energy are joules (J).
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The same basketball -- knowing that its radius is .38 m -- also spins about its axis as it is traveling with an angular velocity of 5 rad/s.
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Let us determine its moment of inertia and its rotational kinetic energy.
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Well, we can model it as a hollow sphere and going back to our table of formulas for moments of inertia, the moment of inertia of a hollow sphere is 2/3 mr².
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That is going to be 2/3 times its mass (.62) times its radius (.38²) or .0597 kg-m².
00:27:09.000 --> 00:27:10.000
Determine its rotational kinetic energy.
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Well, kinetic energy for rotational motion is 1/2 I(ω²).
00:27:18.000 --> 00:27:37.000
For our moment of inertia, we just determined as .0597 and its angular velocity is 5 rad/s, so 5² -- multiply that out and I come up with 0.75 J.
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What is the total kinetic energy of the basketball?
00:27:41.000 --> 00:27:48.000
Well, to get its total kinetic energy, all we are going to do is we are going to combine its translational and its rotational.
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Total kinetic energy is the translational kinetic energy plus the rotational kinetic energy, so that is going to be 19.84 J + 0.75 J or about 20.6 J in total.
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That is kinetic energy of a rotating object that is also moving translationally.
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Let us take a look at a playground roundabout again.
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A roundabout on the playground with a moment of inertia of 100 kg-m² -- (I) = 100 kg-m² -- starts at rest and is accelerated by a force of 150N at a radius of 1 m from its center.
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If the force is applied at an angle of 90 degrees from the line of action for a time of.5 s that equals 90 degrees times half of a second, what is the final rotational velocity of a roundabout?
00:29:00.000 --> 00:29:09.000
Well, as I look here, any time I start seeing final angular velocity in initial, I am starting to think about 'You know probably looking at a kinematics equation.'
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But it would sure be nice to have the angular acceleration.
00:29:12.000 --> 00:29:17.000
Well to do that I probably need to go to Newton's Second Law for Rotation.
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Net torque equals I(α), therefore, α is going to be equal to our net torque over our moment of inertia.
00:29:28.000 --> 00:29:32.000
I do not have net torque, but I do have force, radius and the angle.
00:29:32.000 --> 00:29:39.000
Our net torque is going to be F(r) sin(θ) over our moment of inertia.
00:29:39.000 --> 00:29:56.000
Now, I can substitute in to find angular acceleration equal to our force -- 150 times our radius (1) times the sine of 90 degrees and that is going to be 1 all over our moment of inertia -- 100.
00:29:56.000 --> 00:30:08.000
So I get 150/100 or 1.5 rad/s².
00:30:08.000 --> 00:30:11.000
We want the final rotational velocity of the roundabout though.
00:30:11.000 --> 00:30:22.000
I am going to go back now to my kinematics for rotation and say that final angular velocity is initial angular velocity + α angular acceleration times time.
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That is going to be 0 + α -- we just determined is 1.5 rad/s² times our time of 0.5 s.
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Therefore, our final angular velocity is going to be 1.5 × 1/2 or 0.75 rad/s.
00:30:52.000 --> 00:30:53.000
Let us take a look at another one.
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The ice skater is a famous problem in physics around rotational dynamics and moment of inertia.
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Here without getting into the numbers, we have an ice skater that 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 to half of its original value.
00:31:11.000 --> 00:31:13.000
What happens to her angular velocity?
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Well, as the skater pulls her arms and legs in, moment of inertia is going to decrease to the point that it is half of its original value, but angular momentum remains constant.
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As she spins around, her center of mass remains constant.
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Why? There is no external torque in this problem.
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Therefore, angular momentum about the center of mass -- the axis of rotation to the center of mass -- is conserved.
00:31:45.000 --> 00:31:57.000
If (L) equals I(ω) and we are going to cut (I) in half, (L) must remain the same -- it is conserved.
00:31:57.000 --> 00:32:05.000
In that case, ω must double, so if we cut that in half ω doubles.
00:32:05.000 --> 00:32:11.000
All right, that explains what happens to our angular velocity, but what about a rotational kinetic energy?
00:32:11.000 --> 00:32:21.000
Well, for that, let us go to our formula for rotational kinetic energy. It is 1/2 I (ω)².
00:32:21.000 --> 00:32:31.000
In this case again, we have 1/2 -- (I) became a lot smaller. It got cut in half, but ω doubled.
00:32:31.000 --> 00:32:46.000
Do not forget ω is squared, so if that cut in half and that got doubled -- 1/2 × 2 × 2 -- we are going to double the rotational kinetic energy of the entire system.
00:32:46.000 --> 00:32:55.000
Kinetic energy rotational doubles while angular velocity gets cut in half.
00:32:55.000 --> 00:32:58.000
Wait. Where did that energy come from?
00:32:58.000 --> 00:33:01.000
This rotational energy doubled as she pulled her arms in.
00:33:01.000 --> 00:33:04.000
Well, the skater must have done work to pull her arms in.
00:33:04.000 --> 00:33:08.000
That must have required a force applied for some distance in order to do that.
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That is where we got this extra rotational kinetic energy.
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Let us take a look at the example of a bowler.
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Gina rolls a bowling ball of mass 7 kg -- m = 7 kg and radius (10.9 cm), which is .109 m, down a lane with a velocity of 6 m/s.
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Find the rotational kinetic energy of the bowling ball assuming it does not slip.
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What is its total kinetic energy?
00:33:40.000 --> 00:33:45.000
Well, the first thing I am going to do -- it is a solid bowling bowl, and we will assume the mass is uniformly distributed.
00:33:45.000 --> 00:33:51.000
I am going to find the moment of inertia of the bowling ball by modeling it as a solid sphere.
00:33:51.000 --> 00:33:57.000
Moment of inertia first for a solid sphere is 2/5 mr².
00:33:57.000 --> 00:34:14.000
That will be 2/5 times its mass (7) times the square of its radius (.109²) or about 0.033 kg-m².
00:34:14.000 --> 00:34:26.000
Now, it would be helpful to find its angular velocity and we can do that by recognizing angular velocity as its linear velocity divided by the radius, assuming it is not slipping and we can make that assumption. It does not slip.
00:34:26.000 --> 00:34:40.000
That is going to be our 6 m/s divided by its radius (.109 m) or about 55 rad/s.
00:34:40.000 --> 00:34:46.000
Well, from here let us find its rotational kinetic energy.
00:34:46.000 --> 00:35:11.000
Rotational kinetic energy is 1/2 I(ω²) or 1/2 × our (I) .033 kg-m²× our angular velocity (55 rad/ms²) or about 50 J.
00:35:11.000 --> 00:35:14.000
What about its total kinetic energy?
00:35:14.000 --> 00:35:35.000
Well, kinetic energy -- total is going to be 1/2 mv² + the rotational, 1/2 I (ω²), which is going to be 1/2 × our mass (7) × the velocity (6²) +...
00:35:35.000 --> 00:35:54.000
Well, 1/2 I(ω²) -- we all ready said was 50 J, so 1/2 × 7 × 6², 36 + 50, I come out with about 176 J for its total kinetic energy -- rotational + translational.
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Thanks for watching Educator.com.
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Hopefully this gets you started with rotational motion and conservation of angular momentum and putting that all together with rotational dynamics as well.
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Make it a great day. We will see you again.