WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, I am Dan Fullerton and welcome back to www.educator.com.
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In this lesson we are going to talk about rotational dynamics.
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Our objectives include determining the angular acceleration of an object when extra net torque or force is applied.
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Determining the radial and tangential acceleration of a point on a rigid object.
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Analyzing problems involving strings and mass or real pulleys.
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Analyzing problems involving objects that roll with in without slipping
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and applying conservation of energy to objects undergoing both translational and rotational motion.
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As we go through this lesson, realize there are not real fundamental concepts in this one.
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What we are doing is taking what we have done in the last few lessons, pulling them together,
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and doing a bunch of sample problems, different applications of the things you have already learned.
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With that, let us review a couple, first of all kinetic translational energy is ½ MV².
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In the rotational analog is ½ Iω², where we have the moment of inertia × the square of the angular velocity.
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The total kinetic energy, we have to add the translational and the rotational components together.
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Let us do an example here, we have got this disc of radius R which starts at rest and rolls down an incline of height H.
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Let us see if we can find its velocity when it gets to the bottom, its speed.
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As I look at it, a couple of things I’m going to notice right away, we have a disk and the moment of inertia of a disk rotating about its center of mass is ½ MR².
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As we start our analysis here, we will use a conservation of energy approach, our initial kinetic energy + our initial potential energy
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must = final kinetic energy + our final potential energy.
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If it starts at rest, our initial kinetic energy must = 0.
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If we call this our 0 point at the bottom of the ramp, we can say that our final gravitational potential energy is also 0.
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Our initial potential energy must = our final kinetic energy.
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Our initial gravitational potential energy is MGH and our final kinetic energy is going to have both translational and rotational components.
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We are going to have ½ mass × the velocity of the center of mass², the translational component.
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Add it to ½ the rotational inertia × the angular velocity², the rotational component of kinetic energy.
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Since, we know that the moment of inertia of a disk rotating about its center is ½ MR², we can write this as MGH = ½ M velocity center of mass².
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I’m going to pull that V² from that one, + ½ × ½ MR² ω².
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It looks like we can factor the mass out of all of these which implies then that GH = V² /2 + we will have ¼ R² ω².
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If you recall V = ω R so R² ω² must be = V², this implies then that GH must be equal to,
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we have V² /2 + we will have V² /4 or GH = ¾ D² which implies then that V² = 4 GH /3
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or solving for just the speed at the bottom of the incline that will be the √4/3 GH.
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Using our rotational and translational skills all to find the answer to this problem.
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We also talked a little bit about rotational dynamics, Newton’s second law, net force = mass × acceleration and net torque = moment of inertia × angular acceleration.
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We can put that into play as we analyze a string with massive pulleys.
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We have 2 blocks connected by light string over a pulley of mass MT and radius R.
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Find the acceleration of mass M2, if M1 sits on a frictionless surface.
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A couple things to note here, our moment of inertia, our pulley is disk ½ FMP R² and also recall the angular acceleration is A/r.
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With that, let us start by drawing a free body diagram for M1 here.
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I will draw it as a box, there it is, we have the normal force acting up on it.
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We have its weight M1 G down and we have T1 to the right.
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As we do this, let us define that direction as our positive direction so what we have a consistent axis here.
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Newton’s second law applied in the direction of motion is going to state that the net force which is T1 has to equal M1 a.
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Let us do the same thing for our M2 here.
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M2, we have T2 pulling it up, we have M2 G down, and down is the positive direction.
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For this one M2 G - T2 = M2 a or T2 is going to be = M2 G – M2 a.
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Finally, let us take a look here at the pulley.
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If we draw our pulley in here, we have a tension to the left T1 by Newton’s third law.
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We have a tension T2 there, we have the weight of the pulley acting at its center of mass MPG.
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We have the force from this pivot and should be pretty easy to see, if it is going to be some direction roughly up into the right, force of our pivot.
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There is our free body diagram for our pulley.
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Let us wrote Newton’s second law for rotation as we look at that pulley and we can start with net torque = we will have T2 R - T1 R for our torques.
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Those all must equal Iα but we know T2 is M2 G - M2 A so let us write that in there for T2 M2 G - M2 A × R - T1
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we know is M1 a × R = our moment of inertia ½ MP R² × our angular acceleration which is a/R.
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When I put all that together, it looks like I can do some simplifications.
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First off, I got some R’s that I can pull out of this equation.
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I get, what do we have here, M2 G - M2 a - M1 a = our R is going to go and we are just going to have mpa / 2.
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Or rearranging this to get all our a together M2 G = M1 a + M2 a + MP a /2.
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We can factor the a out of there so that the M2 G = a × (M1 + M2 + mp / 2) which implies that our total acceleration is going to be = M2 G ÷ M1 + M2 + MP/2.
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There is the acceleration of mass M2, noting that we had to take a look at Newton’s second law translationally and
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the rotational version of that in order to put it all together and solve our problem.
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Alright, let us take a look at an example where we have some rolling without slipping.
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We have a disk of radius R rolling down an incline of angle θ without slipping, find the force of friction on the disk.
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Again, it is helpful to know the moment of inertia of the disk about its center of mass is ½ MR².
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I'm going to draw our free body diagrams here, do my best of the circle on this, so something like that there is our disk.
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As I look at the forces acting on our disk, we have the normal force perpendicular to the ramp.
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The force of friction opposing motion acting up the ramp and we have the weight of the disk straight down.
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I could break that up into components with my pseudo free body diagram like we have been doing.
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Let us do that, we will draw our disk again, normal force perpendicular to the surface, force of friction is parallel with the x axis.
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We do not have to change that, MG is not however.
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We are going to break it up into component parallel with the axis MG sin θ.
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The component perpendicular to the ramp which would be MG cos θ.
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As I look at this, we can start to analyze our free body diagrams and start by looking at the net force in the y direction and
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realizing that it is not accelerating off of the ramp in any way.
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We can right away recognize that the normal force in MG cos θ must be = in magnitude so N = MG cos θ.
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In the x direction, we have MG sin θ - the force of friction must = MA, so MG sin θ - friction = mass × acceleration.
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Let us also go and start looking at the net torque on our system.
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As we look at the net torque here, net torque is going to be, we have F operating at some distance r and its perpendicular so just FR and that must = Iα.
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We know that moment of inertia we said was ½ MR², we can write that FR = MR² α /2.
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We also know still that α = a/R or a = R α.
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Let us see here, we can write that FR = if A = R α that would be MRA /2 divide an R on both sides.
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The force of friction must = MA/2 but we know what M is, we figure that out up here MA is MG sin θ – F, that is going to be MG sin θ - F ÷ 2.
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This implies then that the force of friction, let us multiply the 2 over so we have 2 F = 2 MG sin θ – F.
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If I add F to both sides, 3 F = MG sin θ divide both sides by 3, friction = MG sin θ ÷ 3, the force of friction on our disk.
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Let us do another one, this one is pretty involved.
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As we look at a bowling ball that is rolling with slipping has a mass M and radius R.
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It is going to skate horizontally down the alley with an initial velocity V0, find the distance the ball skids before rolling given the coefficient of kinetic friction UK.
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And realize as we get into this problem, this should be a pretty difficult APC level Mechanics problem.
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Not impossible but certainly we are pushing things a little bit here.
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First thing, solid sphere uniformly distributed, we know our moment of inertia about the center of mass
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for that is going to be 2/5 MR² and we are just going to start with that as a given.
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Hopefully, you have memorized it here.
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We are not going to take the time to derive it in this problem, we will have enough to do.
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If I go draw a free body diagram for our bowling ball and there we go.
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The forces we have acting on it is on the alley, the normal force, its weight MG, and if with slipping this is a force of kinetic friction FK.
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Let us start with Newton's laws here.
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The net force in the x direction is going to be just - FK = mass × acceleration.
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In the y direction, pretty easy to see that the normal force is going to be equal 2MG because it is not accelerating up off of the alley or down through it.
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Net force in the y direction must be 0.
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We can put those 2 together in order to write that - FK must equal – μ KM must = - μ K,
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and I'm going to place normal force with MG and all of that must equal mass × acceleration.
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Therefore, we can state that our acceleration must = - μ KG.
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We can go to our kinematic equation for velocity, velocity = initial velocity + acceleration × time.
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We know that our acceleration is – μ KG so for our velocity, we find velocity as a function of time is V initial - μ KGT.
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That will come in handy in a little bit.
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Let us take a look at the net torque here, our net torque is going to be, we got force of friction FK at some distance R, perpendicular again.
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Our sin θ is going to be 1 = I α.
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We can make a couple of substitutions here as well.
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We know that our force of friction is μ K MG and we also know that our moment of inertia is 2/5 MR².
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The left hand side becomes μ K MG × R, all of that has to be = our moment of inertia 2/5 MR² × R α .
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We can simplify that up a little bit, we can divide the M out of there.
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We can pull an R out of there to say that μ K × G = 2/5 R α or solving for α, α = 5 μ KG ÷ 2 R.
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Let us go to our angular velocity equation, angular velocity = initial angular velocity + α T, which implies then since α we just said was 5 μ K G /2R.
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Ω initial is 0, we could write that our angular velocity function is = 5mu KG /2 R × T.
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Alright with those, let us give ourselves some more room here.
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We have to recognize that it stops slipping when V = Rω.
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When those are equal, we have got our condition when it stops slipping.
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We can state then that our V initial – μ KGT = time × 5 μ KG /2 R × T.
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And what we can do is solve this for T then by writing that V0 - μ KGT = we got an R, we got an R, so that is going to be 5mu KG/2T.
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Therefore, V initial = μ KGT + 5/2 μ KGT which implies then that V initial is going to be = 7 ½ μ KGT or solving for T, T is going to be = 2 V0 /7 μ KG.
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We can go back to our kinematics and we want to know how far it is gone.
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Δ X is V initial T + ½ AT² which is going to be V initial × our T 2V knot /7 μ KG + ½ × R acceleration which was - μ KG × T 2V knot /7 μ KG².
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It becomes an exercise in algebra, Δ x = we will have 2 V knot² /7 μ KG from this term.
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-1/2 μ KG × 4 V knot² /49 μ KG, μ K² G² which is going to be = we still have 2 V knot² /7 μ KG.
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If we take ½ of this on the right hand side, that looks like we can have 2 V knot² /7 μ KG² -, 2 and 2 that will become 2 V knot² /49 μ KG.
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Let us put this in terms of a common denominator.
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We have 7 μ KG and 49 μ KG, we can multiply the top and bottom here by 7 that will give us 14 V initial²/49 μ KG – 2 V initial² /49 μ KG.
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Which implies then that Δ x is going to be 14 -2 that will be 12 V initial² /49 μ K × G.
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Quiet involved problem there to figure out how far that ball goes before it starts connecting and stop slipping on the bowling alley.
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Alright, let us take a look at the amusement park swing problem, I love this problem.
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An amusement park ride of radius x allows children to sit in the spinning swing held by cable of length capital L.
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It smacks from angular speed the cable makes an angle of θ with a vertical as shown, determine the maximum angular speed of the rider in terms of G θ X and L.
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We will start with a free body diagram, for our rider we are going to have MG for our rider down and a tension that being our angle θ point them up to the right,
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or breaking it up in the components in our pseudo free body diagram we will have MG down,
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we will have T sin θ to the right, and the vertical component T cos θ.
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From there, we can start to write our Newton’s second law equations net force in the x direction is going to be = T sin θ which is MA
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and since it is going in a circle MB² /R.
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Hand in the y direction, it is pretty easy to see that T cos θ - MG = 0 force is accelerating up or down.
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Therefore, we could write that T cos θ = MG.
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Let us give ourselves a little bit more room to continue with the math here.
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Let us divide that first equation by the second, we had T sin θ = MV² /R and we are going to divide that by our second equation T cos θ = MG, sin θ /cos θ
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is going to give us the tan θ = I’m going to get V² /GR.
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Let us convert translational speed to angular speed and solve for ω, our angular velocity.
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If tan θ = V² /GR and we know that V = ωr then we can write that tan θ = ω² R² /GR
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which is going to be ω² R /G or solving for ω, ω is going to be G tan θ ÷ √R.
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Now what we have to do is replace R which is one of our defined variables with the entire distance here.
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Replacing R, knowing that R is = to that x + the opposite side here which is the sin, L sin θ I have ω = we had G tan θ /√R, that is going to be G tan θ/x + √L sin θ.
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There is our angular velocity as a function of our parameters.
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Alright, let us close out with a bunch of AP practice problems to see how you can use this in so many different situations.
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We will start by looking at the 2002 exam free response question number 2.
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I will give you a minute to download it, you can find it up here at this link or google search it.
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It should be easy to find on the web and a bunch of different places, print it out, try it for a minute or 2 then come back here and play to see how you did.
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Looking at this problem for part A, find a rotational inertia of all four tires.
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The moment of inertia of the tires is going to be = 4 × the moment of inertia for a single tire which is 4 × the moment of inertia for single tire is ½ ML²
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which is going to be 4 × ½ our mass of the tires is going to be m/4.
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That is m/4 × R² or total of ½ m R² for the moment of inertia for all 4 tires.
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For part B, find the speed of the car when it reaches the bottom of the incline.
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I would look at this from a conservation of energy perspective.
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The initial gravitational potential energy must = the kinetic rotational energy + the kinetic translational energy at the bottom,
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which implies then that its total mass of the top is 2m and so we have 2m GH =rotational kinetic energy ½ Iω² + ½ × our mass 2mv²,
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which implies then that 2 mgh = ½ × moment of inertia ½ m R² ω² + ½ × 2 is just going to give us MV².
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I will leave it as it is for now, ½ 2m V² which implies then, since we know that V = ω R and ω = V /R,
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we are going to have the 2 mgh = ¼ mr² and ω² will be V² /R² + ½ × 2= rmv².
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This implies then that 2 mg H = ¼ mv² as our R² /R² make a ratio of 1 + MV² or 2 mgh = 5/4 MV² divide our M out,
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2 GH × 4/5 = V² or V² = 8 GH /5 which implies then that our velocity must be √8 GH /5.
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Alright, let us take a look at part C, after rolling down the incline across the surface,
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the cart collides with a bumper of negligible mass attached to an ideal spring which has some spring constant K.
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Find the distance the spring is compressed before it all comes to rest.
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That is another conservation of energy problem.
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Its initial potential energy must equal the final compressed energy in the spring so we can say that 2 mgh = ½ K × xm².
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A little bit of math, 4 mgh /K = xm².
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Therefore, xm = √mgh /K.
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For part D, assume the bumper has a non negligible mass after the collision,
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the spring is compressed to a maximum distance of about 90% of the value of xm in part C, give an explanation for that.
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As I think about that, the inelastic collision between the bumper and spring results in some sort of energy transfer to the bumper,
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whether it is in the form of deformations, sound, and heat, you have got some energy that is given to the bumper and it is not accounted for.
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And that is where you would have lost that about 10% of that energy.
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I would explain in words something to that effect and I think you would be pretty well covered there.
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Alright, there is a 2002 problem, let us move on to 2006 exam.
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We will take a look at the 2006 exam, you can find it the same place, free response number 3.
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Here we have got a thin hoop of mass M and rotational inertia MR² at the top of a ramp that is on the table.
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We are asked to first derive an expression for the acceleration of the center of mass of the hoop as it rolls down there.
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But I'm going to start with diagrams again because those always help me.
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As we are looking on the table, it is on this ramp, there it is.
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If I am going to draw a free body diagram, let us take a second and do that.
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As we look at our forces here, we have the normal force out of the plane of the ramp, we have MG, its weight, and we will have the force of friction up the ramp.
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Or if I want to do our pseudo free body diagram, normal force still out of the plane on the ramp MG sin θ,
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down the ramp MG cos θ into the plane on the ramp, and we still have our force of friction.
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As I look at this, derive an expression for the acceleration of the center of mass.
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We will start by taking a look at what different equations we have.
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We know that the net torque is going to be Iα, net torque = I α or - FR = - Iα.
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Therefore, FR = Iα.
00:33:43.600 --> 00:33:50.300
Let us take a look now over here on the left at the Newton’s second law for translational motion.
00:33:50.300 --> 00:34:05.700
Let us say that the net force in the x direction is going to be MG sin θ - our force of friction F which is M8 for the center of mass.
00:34:05.700 --> 00:34:11.500
And we can also see pretty easily that the normal force must = MG cos θ.
00:34:11.500 --> 00:34:17.200
We have got a bunch of equations here, we will have to see where we can go with them.
00:34:17.200 --> 00:34:26.400
As I look over here, we have MG sin θ - F = mass × acceleration.
00:34:26.400 --> 00:34:37.900
Let us see, FR = I α, we can write then that the acceleration of the center of mass is MG sin θ – F.
00:34:37.900 --> 00:34:48.700
Let us rearrange this a little bit, the acceleration for the center of mass is going to end up being α R because a = α R
00:34:48.700 --> 00:34:55.400
that implies then that α = the acceleration of the center of mass ÷ R.
00:34:55.400 --> 00:35:01.200
we also know the moment of inertia here is mr², it is given in the problem.
00:35:01.200 --> 00:35:18.200
As we put these together, this implies then along with that piece that FR = mr² × R α which is the acceleration of the center of mass ÷ R.
00:35:18.200 --> 00:35:36.100
Therefore, we have FR = mr a × the center of mass or F = mass × acceleration, our frictional force = Ma.
00:35:36.100 --> 00:35:53.800
Let us pull that in over here into our blue equation and we can then write MG sin θ - Ma = Ma.
00:35:53.800 --> 00:36:10.400
We can divide an M out of all of these G sin θ = 2a or a = G sin θ /2.
00:36:10.400 --> 00:36:14.800
Derive an expression for the speed of the center of mass when it reaches the bottom.
00:36:14.800 --> 00:36:21.300
We are getting the acceleration first, I think that covers us for part A, G sin θ /2.
00:36:21.300 --> 00:36:25.800
Let us take a look at B, giving ourselves some more room on the next page.
00:36:25.800 --> 00:36:31.800
Here as I look at part B, we are finding the speed of the center of mass.
00:36:31.800 --> 00:36:43.300
With that, we have got the acceleration we can go to our kinematic equations VF² = V initial² + 2a Δ x.
00:36:43.300 --> 00:37:01.300
V initial is going to be 0 it starts to rest so that is just VF² = 2 × the acceleration of our center of mass Δ x or VF = √2a Δ x.
00:37:01.300 --> 00:37:10.900
Since we know a, VF = √GL sin θ.
00:37:10.900 --> 00:37:13.800
It was a lot easier than part A.
00:37:13.800 --> 00:37:20.000
Alright C, derive an expression for the horizontal distance from the edge of the table to where the hoop lands on the floor.
00:37:20.000 --> 00:37:22.400
That is a projectile problem.
00:37:22.400 --> 00:37:28.800
For part C, let us start by figuring out how long it is in the air.
00:37:28.800 --> 00:37:39.100
Vertically, we know V initial = 0, Δ Y is going to be H, the height of the table, the acceleration in the y is going to be G.
00:37:39.100 --> 00:37:42.300
You might be able to do this just off the top your head by now.
00:37:42.300 --> 00:37:51.400
But Δ y is going to be V initial T + ½ a YT² where our V initial is 0.
00:37:51.400 --> 00:38:08.200
This implies then that Δ y = ½ a YT² or T = √2 Δ y/a, which is going to be √2 H/G.
00:38:08.200 --> 00:38:12.000
We can look at the horizontal motion.
00:38:12.000 --> 00:38:20.900
If we want to know how far it goes, Δ x = the vertical velocity Vx or horizontal velocity Vx × the time.
00:38:20.900 --> 00:38:50.400
We just found our horizontal velocity S √GL sin θ multiplied by our time √2 H/G, to give us √2 HL sin θ as our G's make a ratio of 1 or cancel out.
00:38:50.400 --> 00:38:58.600
Part D, suppose the hoop is replaced by a disk with the same mass and radius, how will the distance from the edge of the table
00:38:58.600 --> 00:39:02.700
toward the disk lands on the floor compare with the distance from part C?
00:39:02.700 --> 00:39:15.900
It is going to be greater, the moment of inertia of the disk is ½ MR², less kinetic energy is taken up by the rotation leaving more translational kinetic energy
00:39:15.900 --> 00:39:20.200
so it has a greater horizontal velocity as it goes off the edge of the table.
00:39:20.200 --> 00:39:22.000
Therefore, it is going to travel further.
00:39:22.000 --> 00:39:27.800
It has the same amount of time in the air, more horizontal velocity, it will cover more distance.
00:39:27.800 --> 00:39:35.000
It is greater than and make sure you explain in detail how you would come to that sort of conclusion.
00:39:35.000 --> 00:39:39.500
That covers the 2006 question number 3.
00:39:39.500 --> 00:39:47.000
Let us do another one, let us go to the 2010 exam free response number 3.
00:39:47.000 --> 00:39:58.500
It looks a little bit familiar, a bowling ball of mass 6kg is released from rest, from the top of the slant roof that is 4m long, an angle that 30° as shown.
00:39:58.500 --> 00:40:03.100
The ball rolls without slipping and it gives us the moment of inertia there.
00:40:03.100 --> 00:40:14.800
On the figure that it gives us, it asks us to draw the forces not the components acting on the ball at there points of application.
00:40:14.800 --> 00:40:20.800
A free body diagram, on the diagram to give us.
00:40:20.800 --> 00:40:32.800
Let us draw that something like that, we have got our ball right on there and I would label things like we have the normal force perpendicular to the table,
00:40:32.800 --> 00:40:41.000
we have the force of friction up the ramp, and MG the weight of the ball.
00:40:41.000 --> 00:40:45.100
That to me would be A.
00:40:45.100 --> 00:40:49.400
For B, calculate the force due to friction as it rolls along the roof.
00:40:49.400 --> 00:40:53.600
If you need to draw anything other than, use the space, do not do anything to this figure.
00:40:53.600 --> 00:41:01.300
We are leaving our free body diagram together if we want to make a pseudo free body diagram and telling us to draw it again.
00:41:01.300 --> 00:41:09.200
Let us draw our pseudo free body diagram over here for part B, a separate diagram.
00:41:09.200 --> 00:41:18.600
We will draw our ball again, we have our normal force, we have our frictional force acting up the ramp,
00:41:18.600 --> 00:41:33.000
and by now you are probably pretty good at seeing right away that we will have MG sin θ down the ramp and MG cos θ into the plane of the ramp.
00:41:33.000 --> 00:41:39.000
Looking at Newton’s second law here, let us start with Newton’s second law in the x direction.
00:41:39.000 --> 00:41:51.000
Net force in the x direction is going to be, we will have MG sin θ - the force of friction = MA.
00:41:51.000 --> 00:41:56.400
Let us take a look at the Newton’s second law for rotation.
00:41:56.400 --> 00:42:05.300
The net torque is going to be Iα and our net torque is going to be FR.
00:42:05.300 --> 00:42:21.800
FR = Iα but we know I is 2/5 MR², it is given in the problem so that means that FR = 2/5 MR² α
00:42:21.800 --> 00:42:31.700
and the third equation to relate our angular acceleration and linear acceleration A = R α .
00:42:31.700 --> 00:42:35.500
Let us start by combining 2 and 3 here to see what we get.
00:42:35.500 --> 00:43:03.100
If we put 2 + 3 together, I have that FR = 2/5 MR² instead of α, I'm going to write α as A/R, that implies then that FR = 2/5 MR × A,
00:43:03.100 --> 00:43:11.700
which implies then that MA is just going to be =5 F ÷ 2.
00:43:11.700 --> 00:43:14.900
Let us call that equation 4.
00:43:14.900 --> 00:43:40.500
We can combine 1 and 4 to state and that we have MG sin θ - F = Ma, where MA is 5 F /2 which implies then that MG sin θ = 7 F /2,
00:43:40.500 --> 00:43:49.900
which implies then that F is going to be = 2 MG sin θ /7.
00:43:49.900 --> 00:44:08.300
Or plugging in my values, friction that is going to be 2 × mass 6kg × G 10 m/s² × sin of our angle 30° ÷ 7 comes out to be something right around
00:44:08.300 --> 00:44:20.100
about 8.5 or 8.6 N, depending on how you round, whether you use 10 or 9.8.
00:44:20.100 --> 00:44:31.900
There is that one for part B, for part C, let us give ourselves more room here.
00:44:31.900 --> 00:44:37.600
Part C, calculate the linear speed of the center of mass of the ball when it reaches the bottom edge of the roof.
00:44:37.600 --> 00:44:42.200
Here, I think I would be looking at a conservation of energy approach.
00:44:42.200 --> 00:44:49.600
We will say that the potential energy at the top must = the kinetic energy, the translational kinetic energy at the bottom
00:44:49.600 --> 00:45:05.100
+ the rotational kinetic energy at the bottom or MGH = ½ MV² + ½ I ω².
00:45:05.100 --> 00:45:24.600
Which implies that MG H is going to be D sin θ, MG D sin θ is going to be = ½ MV² + ½ RI is 2/5 MR² × ω².
00:45:24.600 --> 00:45:45.500
That implies then if V = ωr, then V² = ω² R², we could write MG D sin θ = ½ MV² + 1/5 R² ω² is just V².
00:45:45.500 --> 00:46:07.800
We are going to have 2/10 or 1/5 MV² which is 7/10 MV² or V² is going to be = 10 MG D sin θ /7 M.
00:46:07.800 --> 00:46:26.500
We can cancel our M's out to say that V² = 10 GD sin θ /7 or to get V all by itself, that is going to be the square root of all of that which is 10 × G 9.8 m /s².
00:46:26.500 --> 00:46:45.700
Our distances for meters, sin 30° that is going to be ½ ÷ 7 or about 5.29 m/s.
00:46:45.700 --> 00:46:53.200
Part D, a wagon containing a box is at rest on the ground below the roof, the ball falls 3m and sticks in the center of the box .
00:46:53.200 --> 00:46:58.600
The mass of the wagon and box is 12 kg, find the speed after the balls land in.
00:46:58.600 --> 00:47:01.800
A conservation momentum problem there.
00:47:01.800 --> 00:47:08.000
Momentum before equals the momentum after, we are only worried about the x components here.
00:47:08.000 --> 00:47:14.900
The initial mass × the initial velocity = the final mass × the final velocity.
00:47:14.900 --> 00:47:22.900
Therefore, final velocity is going to be = MI /MF × our initial velocity.
00:47:22.900 --> 00:47:28.600
Where our initial velocity is 5.29 m/s but we want the x component of that.
00:47:28.600 --> 00:47:36.700
So that times the cos sin 30° which is about 4.58 m/s.
00:47:36.700 --> 00:47:46.300
Our initial mass is 6kg, our final mass with the ball in there, you have to add this together to come up with 18 kg.
00:47:46.300 --> 00:48:10.800
This implies then that our final velocity is going to be MI 6kg /our final mass 18 kg × our velocity 4.58 m /s, for a total of 1.53 m/s.
00:48:10.800 --> 00:48:20.300
And that finishes the 2010 question 2, let us do one more.
00:48:20.300 --> 00:48:36.900
Looking at the 2013 free response question 3, we have a disk of mass M and radius R, supported by rope of negligible mass is shown.
00:48:36.900 --> 00:48:43.100
The rope is attached to the ceiling and passes under the disk, the other end of the rope is pulled upward with some force FA.
00:48:43.100 --> 00:48:48.400
It tells us the rotational inertia of the disk about its center is MR² /2.
00:48:48.400 --> 00:48:53.900
Part A, find the magnitude of the force necessary to hold the disk at rest.
00:48:53.900 --> 00:49:06.900
Alright, as we take a look at this, we have got our disk, we have tension 1, we will call that over to the left, and on the right we have some FA.
00:49:06.900 --> 00:49:11.200
We must also have the force of gravity here MG calling down.
00:49:11.200 --> 00:49:18.100
Newton’s second law in the y direction T1 + FA = MG,
00:49:18.100 --> 00:49:25.300
Therefore, FA = MG - T1.
00:49:25.300 --> 00:49:32.500
We also know that the net torque has to = 0.
00:49:32.500 --> 00:49:40.700
Therefore, T1 R - FAR must = 0.
00:49:40.700 --> 00:49:47.400
Therefore, we can state that T1 must = FA, common sense there.
00:49:47.400 --> 00:50:03.000
Putting those together, we can state that FA - MG - T1 then substituting for T1 FA to say that FA =MG – T1
00:50:03.000 --> 00:50:22.700
which implies FA = MG - FA or FA = 2 FA = MG, FA = MG /2 which is going to be 2 × 10 ÷ 2 or just 10 N.
00:50:22.700 --> 00:50:35.100
Which probably makes sense, if it is a 20 N disk, each of the cords is going to have half of the tension, 10 N on each one while it is sitting there in equilibrium.
00:50:35.100 --> 00:50:38.300
For part B, we are going to liven things up a little bit.
00:50:38.300 --> 00:50:44.100
At time T = 0, the force FA is increased to 12 N causing the disk to accelerate upward.
00:50:44.100 --> 00:50:50.800
The rope does not slip on the disk as it rotates, find the acceleration of the disk.
00:50:50.800 --> 00:51:01.100
For part B, now we have T1 + FA - MG = MA.
00:51:01.100 --> 00:51:13.600
Their net torque equation is FAR - T1 R= I α .
00:51:13.600 --> 00:51:23.400
We know that our moment of inertia given in the problem is MR² /2, moment of inertia of a disk and = α =A /R.
00:51:23.400 --> 00:51:50.300
We can rewrite this as FA × R - T1 R = I is going to be MR² /2 × α A/R, which implies then that FA - T must be equal to MA /2.
00:51:50.300 --> 00:52:04.700
We can substitute that in over here so this equation together with this equation, since right that T1 + FA - MG = MA.
00:52:04.700 --> 00:52:14.300
We also have down here, FA - T1 = MA /2.
00:52:14.300 --> 00:52:29.100
We will combine those 2 equations, adding up the left hand side, I’m going to get 2 FA = we will have 3 MA /2.
00:52:29.100 --> 00:52:41.100
I’m going to add MG over on the right side + MG which implies then that 3 MA /2 must = FA - MG which implies then that our acceleration A is 2/3 M × 2 FA – MG,
00:52:41.100 --> 00:53:00.400
which implies then that acceleration is going to be 4 FA 3 M as we distribute that through -2 G/3, which is going to be 4 × 12 N/ 3 × 2kg - 2 × 10 m s² ÷ 3,
00:53:00.400 --> 00:53:25.600
which is 48/6 -20/3, which is going to be 8/6 or 4/3 m/s².
00:53:25.600 --> 00:53:29.200
There is part B.
00:53:29.200 --> 00:53:38.800
Let us take a look at part C, calculate the angular speed of the disk at time T = 3s.
00:53:38.800 --> 00:53:55.100
Our angular speed is just ω initial + α T and α = A /R so that implies then that ω is going to be = if ω is 0, ω initial is 0,
00:53:55.100 --> 00:54:16.200
that means we just have AT /R which is going to be 4/3 m /s², the acceleration we just found × our time 3s ÷ our radius 0.1 m or 40 radiance/s.
00:54:16.200 --> 00:54:23.900
And D, calculate the increase in total mechanical energy of the disk from T = 0 to T = 3s.
00:54:23.900 --> 00:54:28.000
Alright, to do that, we are going to look at a couple different quantities.
00:54:28.000 --> 00:54:47.900
We have got the change in height first, Δ H that is going to be ½ AT² which is ½ × are acceleration 4/3 × 3s² which is 46 × 9 36/6 is going to be 6m.
00:54:47.900 --> 00:55:01.800
That implies that the change in gravitational potential energy is going to be MG Δ H which is 2 × 10 × 6 m or 120 joules.
00:55:01.800 --> 00:55:04.900
Now we have also got shifts in kinetic energy.
00:55:04.900 --> 00:55:21.900
We will have a change in rotational kinetic energy which is ½ I ω² which is going to be ½ × I, MR² / 2 × ω² which is going to be 40²
00:55:21.900 --> 00:55:39.900
so that is ½ × mass we have got 2 × 0.1² × 0.1 ÷ 2 × 40²= 1600 is going to be 8 joules.
00:55:39.900 --> 00:56:07.000
We have a change in translational kinetic energy which is ½ MV² or ½ × our mass × V is ω R² which is going to be ½ × our mass 2 × 40² × 0.1² or 16 joules.
00:56:07.000 --> 00:56:22.100
To get the total change in energy that is going to be 120 + 8 + 16 or 144 joules.
00:56:22.100 --> 00:56:26.900
Part E, the disk is replaced by a hoop of the same mass and radius,
00:56:26.900 --> 00:56:32.400
indicate whether the linear acceleration of the hoop is greater than or less than or the same as the disk.
00:56:32.400 --> 00:56:39.900
The linear acceleration does not depend on moment of inertia which is what increases when you switch from the distance of the hoop.
00:56:39.900 --> 00:56:45.300
Because there is no dependence on moment of inertia there, it is going to be the same.
00:56:45.300 --> 00:56:47.700
Make sure you justify your answer.
00:56:47.700 --> 00:56:53.600
Alright, hopefully that gets you a good start on rotational dynamics with lots of in-depth examples of sample problems.
00:56:53.600 --> 00:56:56.200
Thank you so much for watching www.educator.com.
00:56:56.200 --> 00:56:58.000
I look forward to seeing you soon and make it a great day everybody.