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 torque.
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Our objectives include calculating the torque on a rigid object and applying conditions of equilibrium to analyze a rigid object under the influence of a variety of forces.
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Let us start off by defining torque.
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Torque which gets the symbol of a big letter co is a force that causes an object to turn and it is a vector.
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Torque must be perpendicular to the displacement in order to cause a rotation.
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And the further away the force is applied from the point of rotation, the more leverage it obtains.
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This distance is known as the lever arm r.
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If you look here, for an example on a wrench using that to turn this piece over here, replying the force at some angle θ with B we call the line of action.
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The distance from the center of our rotation to where we are applying the force is our lever arm r and
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the only force that is really going to matter here is this piece of the force, the one that is perpendicular to the line of action.
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That is going to be F sin θ if we draw that over here maybe it will be a little easier to see F sin θ at some distance r when we are trying to find the torque.
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You probably know by experience if you try and apply a lot of force here, that is not going to do a whole lot.
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Apply that same force further away, you get more rotation that is because you have more torque.
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Torque is we the r vector, the distance from that point where you are applying the force, that vector, crossed with your force vector.
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Let us do a number of cross products if we want the magnitude of the torque that is going to be r F sin θ.
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Now the direction of the torque vector again is a little bit counterintuitive.
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It is perpendicular to both the position vector r and the force vector f.
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You find the direction using the right hand rule, point the fingers of your right hand in the direction of the line of action you are
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and then bend your fingers in the direction of the force.
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Your thumb points in the direction of the positive torque, that is the direction of your torque vector.
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Positive torques causes counterclockwise rotation and negative torques cause clockwise rotation according to the standard sign convention.
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Where this starts to become really interesting, we have been doing all these parallels between translational rotational motion,
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from velocity to angular velocity, from mass inertial mass to rotational inertia.
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We have another one of those now for torque, net force in the translational world corresponds to net torque in the rotational world.
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Newton’s second law of the translational version, net force = mass × acceleration.
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In the rotational world, net torque= moment of inertia × angular acceleration.
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Just point out the parallels, linear acceleration to angular acceleration, inertial mass to rotational inertia or moment of inertia, and force to torque.
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That is going to allow us to solve and analyze a whole new set of problems and situations.
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Let us go over equilibrium again because we are going to start with some equilibrium problems.
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Static equilibrium implies that the net force and the net torque of an object is 0 and the system is at rest.
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Dynamic equilibrium implies that the net force and net torque is 0, the system is moving at constant translational and rotational velocity.
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It is moving but no net force or net torque.
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Let us start off with a see saw problem, the 10 kg tortoise, that is a big tortoise, sits on a see-saw 1m from the fulcrum,
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where must the 2 kg hare sit in order to maintain static equilibrium?
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What is the force on the fulcrum?
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Let us draw a diagram here of our seesaw first and we will put some fulcrum there.
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We know that our tortoise, its 1m from the fulcrum, so that distance there will be 1m.
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Over here, at this end, we are going to have our tortoise.
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Let us see, that means that the force from the tortoises is going to be its force due to gravity MG 10 kg × G we could just write this as 10 G for the force.
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We have got a 2kg hare where does it have to sit to maintain equilibrium?
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We will say that is going to be somewhere over here, we do not know exactly where that is going to be.
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Its force is going to be 2 G and we will call this distance x.
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For the purposes of this problem, we will ignore the mass of the fulcrum itself, its mass is the perfect fulcrum, the magic fulcrum.
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In order to solve this, one of the things I'm going to look at first is, understanding that it is in equilibrium, it is not rotating if they are balanced.
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Therefore, we can write that the net torque which is equal to moment of inertia × α must equal 0.
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We can replace our torques with the net torque with the some of our torques.
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We have over here a 10 G force at a distance 1 m that is in the counterclockwise direction so that would be a positive torque
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that is going to be from our tortoise, the force 10 G × its distance at which it acts 1 m and its perpendicular.
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We do not have to worry about that angle component.
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We have, due to our hare, we have a clockwise torque and that will be negative, so minus the force 2 G × the distance from our center of rotation x all of that has to equal 0.
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I have 10 G-2 Gx = 0 or 10 G = 2 Gx, x must equal 5m.
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We have a follow-up question, what is the force on the fulcrum?
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For that we can look at Newton’s second law in the translational world, net force = mass × acceleration equal 0.
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We look at our forces, we have over here, if we call up positive, we have -10 G from our tortoise.
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We have -2 G from our hare and we have some force up from our fulcrum so + the force of our fulcrum and all that must equal 0.
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Therefore, the force of our fulcrum must equal 12 G which is going to be 12 × G 10 m / s² is going to be 120 N.
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A fairly straightforward example but we will do some more here.
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Let us take a look at a beam, we have a beam of total mass M and length L, with the moment of inertia about its center of ML² / 12.
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The beam is attached to a frictionless hinge and angle of 45° and allowed to swing freely.
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Find the beams angular acceleration.
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The first thing I notice is it is giving us the moment of inertia about the center point not about the hinge.
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If you remember the moment of inertia of a uniform rod about the end, you could use that but let us just get some practice with a parallel axis theorem.
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Let us say that the moment of inertia about the N is the moment of inertia about the center of mass + mass × the shift² where this will be our distance D.
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That is going to be, we have ML² / 12 + M D is L /2².
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Put that together and I end up with 1/3 ML² so that is the moment of inertia in the current configuration we have for our beam here.
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Let us define a couple of things as we look at the problem.
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We said this is distance D, we have here the force of gravity on the center of our beam where it acts which is MG.
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But if we are looking at torques, only the piece that is perpendicular to our line of action counts.
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We are really after that component, that is going to be MG cos θ because that is our angle θ at 45° which matches our angle θ over here.
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We have got that figure out, MG cos θ, we know our distance D, we can go write our Newton’s second law equation for rotational motion.
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Net torque = moment of inertia × angular acceleration and I look at our net torques we have, let us see at the clockwise direction,
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some negative we have -MG cos θ, that force × the distance of which it acts over 2 must be equal to Iα.
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Which implies then that α must be equal to -MG cos θ L /2 × the moment of inertia is going to be -MG cos θ L/2.
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We found our moment of inertia over here was 1/3 ML² so that is going to be 1/2 ML².
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With just a little bit of simplification here, we have got an L, we got an L², we have M and M we can cancel out, that gives us -3 G cos θ/2 L.
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There is our angular acceleration using the parallel axis theorem to find the moment of inertia and Newton’s second law for rotation.
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We have done some Atwood problems with the ideal pulleys, now let us talk about real pulleys.
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We have a light string attached to a mass M wrapped around a pulley that has some mass Mt and radius R, find the acceleration of the mass.
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Alright, to do this what I'm going to start by drawing my pulley.
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There it is, it has some radius R and the forces acting on it, in the places where they are acting,
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if that is our tangent T, we have T acting that direction, we have the weight of the pulley.
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Mass of the pulley × the acceleration due to gravity and we must have some force of the pivot here, a normal force that is acting up.
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There is our pulley diagram.
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Starting there, let us take a look at Newton’s second law, net torque = Iα.
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As I look at our torque, our torque is going to be, we have T at a distance R and that is perpendicular.
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Our torque is going to be RT, I'm going to worry about magnitudes for now.
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Our moment of inertia for a disk is ½ MR² so our moment of inertia is going to be ½ MT R².
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Our torque, our T = ½ MT R² × α of course, which implies then that our tension T divide R from both sides is going to be equal to ½ MT Rα.
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But we also are looking for linear acceleration, we get angular acceleration, remember that α is equal to A /R or A= Rα.
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Which we can replace Rα with A to find that our tension is ½ mass of our pulley × A.
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Alright, now let us draw a free by the diagram for our mass here.
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We have our object, we have our tension up and we have force of gravity down.
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In writing Newton’s second law, we called down the positive y direction MG - T must equal MA or MG - we know our tension now is ½ MPA must equal MA or MG = ½ MPA + MA.
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I can pull out an A there to find that acceleration is going to be equal to MG /M + MP /2.
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We found the acceleration of the mass now that we have a real pulley that has some mass and rotational inertia.
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Looking a little bit more detail at torque, we have a system of 3 wheels fixed to each other that is free to rotate about an axis through its center here.
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Forces are exerted on the wheels as shown, what is the magnitude of the net torque on the wheels?
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Our net torque is just the sum of all our individual torques.
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Let us add those up, starting with this one up here.
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We have a force of 2 F acting at a distance of 2R and it is perpendicular so we have cos IR my sin of 90° which is 1.
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Since it is causing a clockwise torque, let us make sure we call that negative.
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We also have a force over here of 2 F and a distance 1.5 R still 90° but this one is in the counterclockwise direction so that is positive.
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We have a 2 F force that looks like it is at 1R.
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Let me draw that a little bit more carefully × 1R and we have our 3 F force which is operating at a radius of 1.5 R,
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also causing a counterclockwise rotation so that is positive.
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Our net torque M is -4 FR + 2 FR + 3 × ½ is 4.5 FR or net torque= 6 ½ - 4 2.5 FR.
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Let us do a ranking test, a constant force F is applied for 5s at various points of the uniform density object below.
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Rank the magnitude of the torque exerted by the force on the object about an axis located at the center of mass from smallest to largest.
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We are going to have the greatest torque when we are the furthest away and most perpendicular.
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As I look at these different spots, it looks like we are going to have our maximum torque when we start with the minimum,
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when we are applying net force right where the center B, then we will go to C, then we will go to A, because that is an angle.
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Finally D, because we got that one that is perpendicular or most close to perpendicular compared to the axis of rotation here.
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So B, C, A, D, would give us the ranking of the torque from smallest to largest, assuming we are rotating about that point in the center.
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We can also look at ranking angular acceleration.
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A variety of masses are attached to different points to a uniform beam attached to a pivot, write the angular acceleration of the beam from largest to smallest.
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If we want the largest angular acceleration, we want the most force, the furthest away from the axis of rotation.
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That is going to be at D, where we have 2 M at the very end and then M right beside it.
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D will be the most then it looks like C is the next most, we have got M at the very end and 2 M just inside that.
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Then, it looks like we are probably looking at A where we have 2 M at the very end and finally we have B 3 M half of the distance.
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D, C, A, B would give us the greatest angular acceleration from largest to smallest because we are looking at the ranking of the torques.
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Let us do a cafe sign example, a 3 kg cafe sign is hone from a 1 kg horizontal pole as shown and a wire is attached to prevent the sign from rotating.
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We are trying to find the tension in the wire.
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Let me just redraw that a little more simply over here and use a ruler just to make things nice and neat.
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If there is our pull that goes with their sign, it looks like it is a 4 m long pull so 1, 2, 3, 4 m.
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As I look at the different forces acting on it, it looks like we have a force that is 1 kg.
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The force from the center of mass, its gravitational force is going to be 1 kg × G or 1 G at the center.
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We have a 3 kg mass that is right over here, so that will be 3 G and we have a tension from the wire over here.
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We will draw that at the very end where it is acting.
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There is our tension and that angle right there is 30°.
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Since, it is an equilibrium we know the net torque must be 0 so we will start there.
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Net torque equal 0 which implies then, as we end add up the torques that is going to be we will have T sin 30 × the distance over which it acts,
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over that 4 m, that is counterclockwise and we will call that positive.
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We also have -3 G acting at 3m, negative because it is causing a clockwise torque.
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We have a -1 G at 2m negative because it is also causing a clockwise torque.
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Putting this together, sin 30 is 2, so that will be 2 T.
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We have 2 T – 9 G - 11 G so T is going to be equal to 11 G /4 sin 30 or 2 which is going to be 54 N.
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Let us finish up by looking at an old AP problem.
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Here we have the 2008 free response number 2 problem, you can find it here at the link on top.
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Go ahead and download that there and let us take a look at that.
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This looks mighty familiar, we got a horizontal rod with some length and mass.
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The left of the rod is attached to the hinge and we got a spring scale attached to our wire in order to determine the tension in the wire.
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First thing we are asked to do is to diagram, draw and label vectors to show all the forces acting on the rod.
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Let us start by drawing a rod here, something like that.
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And as I look, we are going to have the weight of the rod itself MG.
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We are going to have the weight of our block on the end, we will call that mg.
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We have a tension here at some angle 30° and we also have a force from the hinge which in order to balance all this out must be going somewhere up into the right.
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That is the force of our hinge.
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M we will call 2kg, m is 0.5 kg and the whole thing has a length of 0.6 m.
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There is A, looking at part B, calculate the reading on the spring scale.
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The net torque has to be equal to 0 which implies that let us add up our torque, we have got a TL sin 30.
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We are going to assume that pivot is around here so that length is L – we will have mgl negative
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because it is causing a clockwise torque - MGL /2 with a mass of our bar causing its torque, its force.
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All that has to equal 0.
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Therefore, our tension must equal, we have G L / M × M + M /2 divided by L sin 30° which implies that our tension must be L sin 30 is ½
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that will be 2 G, L/L cancel out × M + M /2 which is going to be 2 × 10 m/s² × our little mass 0.5 + M/2 2kg/ 2 is 1kg so 20 × ½ is going to be 30 N.
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There is part B, moving on to part C, the rotational inertia of a rod about its center is 1/12 ML² where M is the mass of the rod and L is its length.
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Find rotational inertia of the rod block system about the hinge.
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For C, the moment of inertia of our system is going to be the moment of inertia of the rod + the moment of inertia of the block.
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We can find the moment of inertia of the rod about its center of mass is 1/12 ML² if we want that about the hinge, we can use the parallel axis theorem.
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Just in case you did not remember what the moment of inertia of a rod is about its end.
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The moment of inertia of the rod about the hinge is going to be the moment of inertia of the rod about its center of mass + Md²
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because we got a parallel axis and our initial was about the center of mass.
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That is going to be 1/12 ML² + M × L/2² which is ML² /12 + ML² /4 or the moment of inertia of the rod about the hinge is just ML² /3.
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We know the moment of inertia of the block is ml².
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When I put that all together, the moment of inertia of the system is 1/3 ML² + ml² which is going to be L² × M /3 + m
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which is going to be 0.6² × 2kg /3 + ½ kg = 0.42 kg m².
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One more part to the problem, part D, if the cord that supports the rod is cut near the end of the rod, calculate the initial angular acceleration of the rod block system.
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Net torque = moment of inertia × angular acceleration.
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Therefore, our angular acceleration is the net torque/ the moment of inertia which is MGL + MG L /2 all divided by the moment of inertia.
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We no longer are worried about that tension, the wire, because we cut it.
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Which is going to be equal to, we can factor out a GL /I m + M /2 which is going to be 10 m/s² × length 0.6/moment of inertia 0.42 kg m² × 0.5 kg +
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2/2 is going to be 1, for a total angular acceleration of 21.4 radiance/s².
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Hopefully, that gets you a good feel for torque and Newton’s second law for things that are rotating.
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We will get more into that in our next lesson on rotational dynamics.
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Thank you for watching www.educator.com.
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We will see you again soon and make it a great day everyone.