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 am Dan Fullerton and in this lesson we are going to talk about rotational kinematics.
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To begin with our objective, understand and apply relationships between translational and rotational kinematics.
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Write and apply relations among the angular acceleration, angular velocity,
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and angular displacement of an object rotating about a fixed axis with constant angular acceleration.
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Use the right hand rule to determine the direction of the angular velocity vector.
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Let us start by talking about radians in degrees.
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I know a couple of these are going to be a bit of review but as we get back in the rotation in more depth,
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it will probably take a minute or 2 to make sure we have got these fundamentals down.
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In degrees, once around the circle is 360° and once around the circle is 2π in radians.
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The radians measures the distance around an arc equivalent to the length of the arc’s radius.
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That distance around δ S is circumference or 2π r or if you are measuring diameter it would just be π × d.
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Let us do a couple conversions again very quickly, 90° radians, 90° if we want that in radians, 2π radians is equal to 360°.
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We would get π /2 radians or 6 radians to degrees.
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If we start off with 6 radians, 2π radians is 360° and those cancel out 360 ° × 6/2π is 344°.
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We have got our conversions, angular vs. linear displacement.
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Linear position displacement we have talked about as δ r and δ s.
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Angular position or displacement we give by δ θ.
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As you go around the circle, you have increasing amounts of θ, where S the linear distances r × θ or δ S is our δ θ.
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If we talked about velocity in the same way, linear speed or velocity is given by the V vector.
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Angular speed or velocity is given by the squiggly W, the ω vector where velocity is the derivative of position with respect to time.
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Angular velocity is a derivative of angular position or displacement with respect to time.
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We talk about these angular vectors, the direction is given by the right hand rule, something that is very non intuitive.
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If we think about an object going around a path like this, the radius to the side, take the right hand
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or wrap the fingers of your right hand in the direction the object is moving around that circular path
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and your thumb will give you the direction of the positive angular velocity vector.
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The angular velocity vector does not point in the direction the object is actually moving.
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Converting linear to angular velocity, we have velocity as the rate of change of position or displacement with respect to time.
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But we know that S is our r × θ, the radius × θ.
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Therefore, we can write that this is equal to D / DT of the derivative of r θ.
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But r is a constant, our radius is not changing so we can write this then as V = r D θ dt which is r ω.
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D θ dt is our ω, so V = r ω or if we want ω, ω = V/ r.
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Doing an example where we look at the angular velocity of the Earth.
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Find the magnitude of Earth’s angular velocity in radians per second.
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Ω is δ θ /δ T which is 2 π radians / 24 hours which is going to be π radians/ 12 hours but we know that 1 hour his 3600s to get this in more standard units.
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Ω would be equal to 7.27 × 10⁻⁵ radians /s.
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All the things that we have done before but useful to get just as firm foundation before we get a little bit more in depth here.
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If we want to talk about linear vs. angular acceleration, if linear acceleration is given by A vector, angular acceleration is given by the Α vector,
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where if A is the derivative of velocity, the angular acceleration is a derivative of angular velocity.
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In this case, it is how quickly you are changing your angular velocity is what we call angular acceleration.
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Just like we did with the angular velocity, as far as finding the direction, the direction of the angular acceleration vector is also given by a similar right hand rule.
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Let us do an angular acceleration problem.
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Our friend rides a unicycle, if the unicycle wheel begins at rest and accelerates uniformly
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in a counterclockwise direction to an angular velocity of 15 rpm to the time of 6s,
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Find the angular acceleration of the unicycle wheel.
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First, let us convert rpm to radians/ s, 15 rpm is 15 revolutions / 60s but there are 2 π radians in each revolution.
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That is going to be 1.57 radians /s.
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Our angular acceleration is our change in angular velocity with respect to time which will be our final - our initial angular velocity with respect to time,
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Or 1.57 radians/ s ÷ 6s which will be 0.26 radians /s² and because it is accelerating counterclockwise that is what we are going to call a positive angular acceleration.
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As we go through and look at rotational kinematics, it is helpful to talk about some of these variables.
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We are talking about translational motion, we have had δ S.
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We are talking about angular, we have δ θ, a linear velocity translational velocity V, angular velocity ω.
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Translational or linear acceleration A and angular acceleration Α and time is the same across both of these paradigms.
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Where it starts to get useful is when we look at the variable translations, you will start to see a pattern.
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If S = r θ, V = r ω, A = r α.
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All we are doing is just multiplying the angular version by the radius to get the linear or the translational version.
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Similarly, θ = s /r, ω = V /r, α = A /r.
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You take the linear version divided by the radius to get the angular version.
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Of course, time is time regardless of which paradigm you are doing.
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When we get to kinematic problems, this makes our formulas much simpler.
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Our kinetic equations that we have derived earlier V = V initial + AT, if we want to look at the rotational equivalent,
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all we do is we replace any velocities with angular velocity.
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We replace any displacements with angular displacements.
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We replace any acceleration with angular accelerations.
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This becomes ω = initial ω + Α × T or δ x = V knot t + ½ at² becomes δ θ = ω knot t + ½ α t² .
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Or finally, V² = V initial² + 2a δ x, ω² = ω initial² + 2α δ θ.
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You are just replacing the variables but the form of those kinetic equations and they are used that is exactly the same.
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We have also talked about how we derive that centripetal acceleration.
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Probably, we are taking a minute and doing it again just to make sure we have it down.
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So if we look at some specific point that you get after some angular displacement θ, that sometime from T0 to T,
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we could call its x position will be r cos θ and its y position would be r sin θ.
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Our r vector is going to be r cos where θ = ω t, that would be ω t I hat + r sin ω t j hat.
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Velocity is just going to be the derivative of that, it is the derivative of r with respect to t,
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that is going to be the derivative with respect to t of r cos ω t i hat + r sin ω t j hat.
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Which implies then that V is equal to, if we take the derivative of this we will get the derivative of the first + the derivative of the second.
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That is going to be V equal to, we will have r I hat, derivative of cos is going to be opposite of the sin.
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This will be in r I hat × ω – sin ω t.
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We have our term over here, we will have + r j hat ω cos ωt.
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If we rearrange them a little bit and make it look a little bit more formal, V = -ω r sin ω t I hat + ω r cos ω t j hat.
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There is our velocity but we can take that a step further and let us do that.
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If V (t), let us give a little room for that, if V (t)= - ω r sin ω t I hat + ω t cos ω t j hat.
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And our acceleration is going to be the derivative velocity with respect to time which is going to be
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the derivative of all of this is going to be - ω² r cos ω t I hat - ω² r sin ωt j hat.
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Which implies then that acceleration = - ω² and then we are left with r cos ω t I hat + r sin ω t j hat.
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If we recalled this is our initial r vector.
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A is - ω² r or we also know ω = V /r so that means A = - V /r² × r or V² /r - V² /r.
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Now the negative sign, why are we worried about that?
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We are talking about the centripetal acceleration, we are defining toward the center of the circle as positive so that would V² /r.
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When we are talking about the vectors, we define it this way where r is from the center out to that position point while the centripetal acceleration is opposite of that.
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That is where the negative comes.
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R goes from the center to the circle where as the acceleration is from the object to the center.
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It is easier as we do this and stop worrying about our vector signs and directions, A is V² /r.
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Let us do an example here.
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An object of mass M moves in a circular path of radius r according to θ = 2 t2 + t + 4, where θ is measured in radians and t is in seconds.
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Find the angular velocity of the object that equals to seconds.
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As I look at this, ω = d θ dt which is going to be the derivative with respect to time of 2t³ + t + 4, which is going to be 6t² + 1,
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which implies then since t = 2s, that ω T = 2s is going to be 6 × 2² + 1 or 25 radians/s.
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Find the object’s speed at this time.
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The velocity, the speed at T = 2s is going to be r ω at t = 2s which is just going to be 25r m/s.
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Let us take a look at an example with the wheel.
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The wheel of radius r and mass capital M undergoes a constant angular acceleration of magnitude α,
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what is the speed of the wheel after it is completed one complete turn assuming it started from rest?
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This is a kinematics problem so let us figure out what we know.
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Our initial angular velocity is 0 because it starts from rest, we are trying to find its final angular velocity
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or trying to find its final linear velocity but angular velocity we will get to it later.
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We know our displacement is 2π once around the circle and we have some angular acceleration α.
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With what I know, I would go to my kinematic equation by rotational version ω final² = ω initial² + 2 α δ θ or ω final² = ω initial² is 0 so this becomes 2 × α δ θ is 2π.
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This is 4π α, ω final is just going to be √4π α.
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If we want that, our speed, that is going to r ω that would be r√4π α.
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Let us finish up by doing a couple of AP problems from old past AP exams.
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We will take a look first at the 2003 exam Mechanics question 3.
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Take a minute, pull that out, you can find it here at the link above or google it, download it and give it a shot and then come back here and see what we have got.
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It looks like we first are plotting some data points.
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I plotted the data points first and we are supposed to draw the best fit curve.
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It is kind of a goofy, easy, initial question there.
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I would draw at something like this and to draw a curve, the shape, I have this kind of like that.
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It says using that best fit curve determine the distance traveled by the projectile if a 250 kg is placed in the counterweight bucket.
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To do that, all I do is go up here to wherever it happens to be 250 kg and come here and read off on the graph and I get an x of about 33 m.
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All right, going to part B.
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For part B, says students are assuming that the mass of the arm, the cup, and the counterweight bucket can be neglected and then
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they develop a model for x as a function of mass using x = Vxt, where Vx is the horizontal velocity of the projectile as it flies off the top of the cup and T is the time.
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First off, how many seconds after leaving the cup the projectile strike the ground?
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That sounds like a kinematics question to me, so B1 we have an initial, if we look vertically, initial vertical velocity is 0.
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We do not know our final vertical velocity, δ y is 15m, our acceleration is 10 m /s².
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We are calling down the positive Y direction and T that is what we are trying to find.
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I would use δ y = V initial t + ½ Ay T², where V initial is 0 so T is going to be 2 δ y / √A which is 2 × 15m / 10m/s², 30/10² that is about 1.73s.
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Let us go onto part B2, derive the equation that describes the gravitational potential energy of the system
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relative to the ground assuming the mass in the counterweight bucket is M.
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For B2, as I look at that, our initial potential energy is going to be equal to the potential energy in the bucket + the potential energy of our projectile
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which is going to be mass B, G × the height of B + mass of the projectile G × the height of the projectile, which is, let us see what we have, 10 m/s² for G × 3m × M + 10 kg.
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And all of that is going to be equal to 300 K + 30 M.
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Now let us take a look at part 3, derive the equation for the velocity as it leaves the cup.
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We are getting a little bit more involved here.
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Part 3, our final potential energy + our final kinetic energy must equal V initial and we are looking at when it leaves the cup there.
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We can say that our final potential is going to have to equal, we have got 1 × 10 × 110 M, 1 × 10 × M + 15 × 10 m/s² × 10 =1500 + 10 M.
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Our kinetic final, we know is ½, which implies there we are just coming up with it again, we are just coming up with the different pieces.
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Our kinetic finals is going to be ½ × mass × square root of our velocity V x² + ½ mass of our bucket × the velocity of our bucket².
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Our initial potential, we said was 300 + 30 M from part B2 up above.
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Putting all of this together, we have 300 + 30 M must be equal to that 10 M + 1500 from up here + we have got 5 Vx² + ½ M VB².
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The key to solving this problem at this point is realizing that both ends of that catapult, they are swinging with the same angular velocity.
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If that is the case, ω B must equal ω A and since V = ω r and ω = V /r, we can write that VB / 2 must equal Vx / 12
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and then we get our relationship between the velocity of bucket and our Vx.
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VB must equal Vx/6 so now we can go and we can put that back in our blue equation there to solve for the velocity as it leaves that Vx.
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300 + 30 M = 10 M + 1500 + 5 Vx² + ½ M and VB² now is just going to be Vx² / 36.
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It is an algebra exercise, 20 M = we take that 300 out, 20 M = 1200 + 5 Vx² + M Vx² / 72 which implies that 20 M - 1200 is going to be equal to Vx² × 5 + M / 72.
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Or getting vxx all by itself, Vx is going to be equal to 20 M - 1200 ÷ 5 + M / √72.
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And there is probably a way to simplify that further but that looks like plenty to me.
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Onto part C, complete the theoretical model by writing a relationship for x as a function of the counterweight mass using the results from B1 and B3.
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That is just x = velocity × time from our horizontal kinematics which is just going to be, time was 1.73.
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This is 1.73 × √20 M - 1200/5 +√M / 72.
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Part C2, compare the experimental and theoretical values of x for a counterweight bucket mass of 300 kg.
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Theoretically, when we plug in for M = 300kg in our formula, I come up with about 39.6m and we write it off the graph,
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if we read it off the graph actual M at 300kg was about 37m.
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Where did that difference come from?
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The difference could be from a lot of things, you could have had friction at the axis, you could have air resistance.
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Remember, how we neglected the masses of the arm, the bucket, and the cup, any of those can all contribute to this difference there.
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That covers that free response, let us take a look at one more here.
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Let us go to the 2014 Mechanics exam free response number 2.
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Again there is the link there, we can google it, take a minute and print it out, check it out, and try it, and come back here and see how you do.
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It is kind of an interesting set up on this one.
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The first thing I have to do is find an expression for the height of the ramp in terms of the V knot M and fundamental constants.
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I would use conservation of energy where the final kinetic energy must be equal to the initial gravitational potential energy
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or ½ M × what they call V initial² = MGH.
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Therefore, just solving for H that is going to be V knot² / 2G, pretty straightforward for part A.
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Move on to B, short time after passing point T, the block is in contact with the wall and moves with the speed of V.
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Is the vertical component of the net force on the block upward, downward, or 0?
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There is no vertical acceleration so Ay = 0, which means you can say that the net force in the y direction must equal 0 so 0.
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In part 2, it says on that figure, draw an arrow starting on the block to indicate
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the direction of the horizontal component of the net force on the moving block when it is that position shown.
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We have got it moving in a circular path and it is right here at that position.
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When it is at that position, we have the normal force acting directly to the left so that is the direction of the normal force
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but we also have some amount of frictional force that is opposing the motion.
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When I put those 2 together, we will add them up tip to tail normal friction, I get something that must be down into the left for my net force.
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I would draw something like that for my net force.
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That leads us to part C, determine an expression for the magnitude of the normal force exerted on the block by the wall as a function of velocity.
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The net centripetal force is MV² /r and the only centripetal component is provided by the normal force that is M.
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Therefore, I would say that N = MV² /r.
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The frictional force is perpendicular to that is not going to come into play when we are talking about the centripetal force.
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We have here part D, derive an expression for the magnitude of the tangential acceleration of the block at the instantaneous speed V.
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Tangential acceleration has to do with our frictional force.
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Our frictional force is μ × the normal force and we just found the normal force so that means that our frictional force is going to be me MV² /r and
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that must be equal to mass × the tangential acceleration or the magnitude of the tangential acceleration is just going to be μ V² /R.
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It looks like we got one more part of the question.
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Let us give ourselves a little more room here.
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For part E, derive an expression for the velocity as a function of time after passing point T.
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To start with, we know that the magnitude of our tangential acceleration is – μ V² /r, that negative because the speed is in the opposite direction of the acceleration.
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We also know that acceleration is the derivative of velocity with respect to time, dv dt must equal – μ V² / R or dv / V² is going to be equal –μ/R dt.
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If we want to get our velocity then it looks like we are going to have to do a little bit of integration.
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We will integrate this so I’m going rewrite this first as the integral of V⁻² dv evaluated from some velocity V initial to final V = - μ /R
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integral from T = 0 to T of dt which implies then the integral of V⁻² is -1 /V.
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– V⁻¹ evaluated from V knot to V = - μ / R × T or I could write that as -1 / V -1 /V initial = -μ /RT
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which implies then that, let us rearrange the order there, put that negative through.
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1/ V knot -1 / V = - μ/ R T or getting a common denominator, multiplying everything by V knot, V knot /V knot - V knot / V = -V knot μ T / R.
00:31:33.100 --> 00:31:36.200
Let us see if we can do little bit more rearrangement.
00:31:36.200 --> 00:31:53.300
That is just 1 so we could say that – V knot / V = - V knot μ T / R -1 which implies then, let us multiply that to -1.
00:31:53.300 --> 00:31:58.100
V knot / V is equal to, I’m going to write 1 as R/R.
00:31:58.100 --> 00:32:21.300
R/R + V knot μ T /R or V knot / V is equal to R + V knot μ T / R which implies then, if V knot / V is that / that and V /V knot is the denominator/numerator.
00:32:21.300 --> 00:32:29.500
V / V knot = R /r + V knot μ T.
00:32:29.500 --> 00:32:42.000
V all by itself is V knot × r /r + V knot μ T.
00:32:42.000 --> 00:32:48.400
All the work for that, alright hopefully that gets you a pretty good start on rotational kinematics.
00:32:48.400 --> 00:32:52.000
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