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(II) A pendulum 2.00 $\mathrm{m}$ long is released (from rest) at anangle $\theta_{0}=30.0^{\circ}$ (Fig. 14). Determine the speed of the$70.0-\mathrm{g}$ bob: $(a)$ at the lowest point $(\theta=0) ; \quad$ (b) at$\theta=15.0^{\circ},(c)$ at $\theta=-15.0^{\circ}$ (i.e., on the opposite side).d) Determine the tension in the cord at each of thesethree points. $(e)$ If the bob is given an initial speed $v_{0}=1.20 \mathrm{m} / \mathrm{s}$ when released at $\theta=30.0^{\circ},$ recalculate the Speeds for parts $(a),(b),$ and $(c)$

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A) $2.29 \mathrm{m} / \mathrm{s}$B) $1.98 \mathrm{m} / \mathrm{s}$C) $1.98 \mathrm{m} / \mathrm{s}$D) A. $0.687 \mathrm{N}$B. $0.594 \mathrm{N}$C. $0.594 \mathrm{N}$E) A. $2.59 \mathrm{m} / \mathrm{s}$B. $=2.31 \mathrm{m} / \mathrm{s}$C. $2.31 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 8

Conservation of Energy

Work

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Yusuff H.

September 27, 2020

Thr answer in the video does not seem right

University of Washington

Simon Fraser University

Hope College

University of Winnipeg

Lectures

04:05

In physics, a conservative force is a force that is path-independent, meaning that the total work done along any path in the field is the same. In other words, the work is independent of the path taken. The only force considered in classical physics to be conservative is gravitation.

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

05:43

Figure $8-23$ shows a pend…

09:26

The 2-kg pendulum bob move…

06:02

The pendulum bob $B$ has a…

10:01

02:55

A pendulum is released fro…

06:52

05:52

Figure $8-34$ shows a pend…

02:59

The length of a simple pen…

Okay, so today we're going to talk about energy conservation. So energy conservation is a special case that happens when the Onley forces acting on a system. The system that we are studying our conservative forces. Okay. And conservative forces are the ones that can be expressed in terms of a potential an example. This is gravity. Okay. Okay. Eso conservation of energy is going to say that the energy of your system in an initial state e I is going to be equal to the energy off a final state E f. Okay, this is the This is the expression for energy conservation. Now, in this exercise, we're going to send application off this principle in the following. So suppose we have a pendulum. Sorry. So we have a pendulum of length on which you have Ah, a wire of length l. That is not to be equal to 2 m at in the edge of the wire. We have a small bowl. Okay. His mobile off mass equal to 70 Graham. So 70 times. 10 to the minus three kilograms UK. So this is a panda long. Now suppose that we know that we release our ball in the angle theta zero released from rest such that that the zero is study degrees. Okay, so given this what is going to be the velocity off the ball when Satya is equal to zero? So what is going to be developed of the ball in this point? Okay, okay. So we won't find the velocity. So remember, first, that the velocity for rotations can be expressed in terms off the angler velocity. This is the linear velocity. And it's going to be the angler velocity that I want to call fed adult times the distance between the particle that is rotating and the center of rotation. So here l Okay. Now, um, in this case, what are the forces acting in our system? So the only force that is acting in our system, that is the the small ball is the gravitational force. And I told you that gravity is a conservative force. So since the only forces a conservative force, we can use conservation off energy, we have conservation off energy. Okay. Now, a new expression we can use for conservation of energy is that the total energy off the system at this point when the ball is released from rest has to be equal to the total energy of the system when the ball is making a zero angle with the horizontal axis. Okay, so, e I has to be equal to e f. Okay, eso Let's look at the initial state. So when the ball is making a 30 degree angle with the horizontal Sorry with the vertical axis. Um Okay, So since there is no velocity, the kinetic energy, the initial kinetic energy is going to be equal to zero. And we're going to have on Lee the gravitational energy they're going to call you. Okay, so what is going to be the gravitational energy in this case? First to to find the gravitational energy, we have to choose a place, a place on our system to be the zero for gravitational energy. Because since gravitational is a potential, the only thing that interests to us is the difference. Okay, So if we put this point making the zero angle to be our point off zero gravitational energy, then we're going to have that gravitational energy in this point. Here is going to be m g times H on reach. Age is going to be the distance between the vertical projection. So let me put this in in the green and the total length off the off the wire. So it's going to be this thing here. Okay? Okay. Now we can see that. H is l minus l the projection into the vertical access or l Times CO sign off theta zero. Okay, so let me just put l, um, in front of everything, I'll sign a fight. A zero. Okay, so we have the initial energy of the system is going to be on lee mg times L a times one minus. Good sign after the degrees are theater zero. And this is going to be equal to the energy when the ball is making zero degrees. Okay, when the ball is making zero degrees, the now the potential energy is going to be zero. Because we set this specific point to be our zero gravitational potential energy. Uh um, point and the Onley energy Acting on the system will be the kind of energy okay, and the kinetic energy for rotation. If you remember ISS, the moment of inertia times the angler velocity squared, divided by two. But the moment of inertia for a particle that is rotating around the point that has a the that isn't at a distance. L between to the particle is going to be the mass of the particle. Times the distance between the particle and the center of rotation squared, They they squared. Okay, we can isolate the angular velocity. So first you can notice that we can cancel out the mass off the ball. We're going to have the The angular velocity squared is going to be to G L one minus co sign off. 30 degrees, um, divided by l squared. Okay, we can cancel out l with l squared and we can rewrite the angular velocity in terms off the linear velocity, which is the quantity during that we want to fight. So it's going to be V square divided by l squared. OK, now taking the square root, We're going to have that V. It's going to be sorry. Yeah, this it's going to be l times the square root off this term. So to G, which is 9.81. Okay, times one minus co. Sign off. 30 degrees. The divided by the length off the wire, which is 2 m. So doing this calculation, they're going to find that the velocity at the bottom for the ball in the pendulum is going to be 2.29 m per second. Okay, now, question being we want to find again the velocity. But now, for not for the point that makes zero a zero angle with the vertical axis. But for, um, the point that makes sorry a 15 degree angle with the horizontal. So we want to find Yeah, it's this point here that makes an angle. Ferre that's equal to 15 degrees. Okay, we once again use conservation of energy. So the initial energy off the system we calculated in question a is M G. Um, sorry, it's m G one minus. Sorry is mgl Times one minus co. Sign off theta zero, on which I'm calling theta zero 30 degrees. Okay, so this is the initial energy Now, the final energy is going to be the kinetic term, uh, moment of inertia time. State of dots square, divided by two plus the potential energy for from gravity. Now you can see that this time the height is not going to be zero because we set zero to be here. Okay, so this is are you equal to zero? Making a zero degree with the vertical axis. So H he's going to be this height here, which is just one miles. Co sign off 15 degrees. Now we can, um we can equal these two expressions. Okay, we have that M g l one minus. Go sign off. Theater zero has to be equal to AM l squared. Data not squared. Divided by two plus mg, sir. There's an l hear M G l one minus. Go sign off data. Then we can isolate the angular velocity. So theta dot squared Notice that we can also cancel out the mass and cancel out this l here. Okay, um, you're going to have the data that squared is going to be equal to two over. L times one minus. Go. Sign off. Data zero, minus one. Blood's co Sign off. Data. Okay. We can cancel out one. No, we can again rewrite the angular velocity in terms off the linear velocity. Right? And s o b stood the values that we already know. So we is going to be, um l times the square root. Oh, sorry. There is a G in here. Square root of two times G divided by l. Sorry. I told you that I will. I'm going to assume is to develop the two times nine 0.81 divided by two Okay, times go sign off. 15 degrees minus. Because sign off 30 degrees. Kate, do this calculation. We're going to find the velocity is when you've seen one point, uh, 98 meters per second. Okay, now questions see, Want us once again finds the velocity. But this time we want to find a velocity when the ball is making a 15 degree angle to the left. Okay? Here. Previously, we calculated to the right Now we want to calculate to the left. So data, right? Yes. Okay. So you can notice that since the angle and in respect to the vertical access is the same. No matter if you're looking to the left or to the right, then the answer is also going to be the same as the one that we calculated in question. Be because data here is 15 degrees. So no matter if it is left or right, if it's the same angle in respect to the vertical the The answer is going to be the same. Sophie here. 1.98 meters per second. Again? Yeah. Now question Dean question. The wants us to find the tension on the wire for each point. For the points off question a question be and questions see. Okay, so d A what is attention when the ball is making a zero degree angle with the horizontal sorry and with the vertical axis. So let's draw it here again. Our system We have our ball like this. The only first acting is the gravity, and the gravitational force is pointing to the same direction as attention. Since the ball is not moving upwards and Lord downwards, that is, since the ball is not floating or thinking is standing still like this, we're going to have that, uh, the net force in the Y direction is equal to zero and is equal to the tension minus the gravitational force. So the tension is equal to M G. On which I am is seven d times sent to the ministry kilograms and this times the gravitational acceleration. So you're going to find that attention for the point of question. A is going to be coaches. 0.68 6. You teams okay, Pretty simple. Now we want to find the tension in the point that we calculated on question be so attention When Now the ball is making a 15 degree angle. So the bodies like this. Okay, so we have the gravitational force putting downwards, and you're going to have attention pushing in the direction off the wire. Okay, Now we can see the that there's going to be a component of the gravitational force that is pointing in the direction off the tension. I'm going to call w f Y. Sorry. Okay. And from Newton's second law, we're going to find that. Okay, since the ball is not going in this direction. Lord, going like this, we're going to have that from the same argument. We didn't hear that. The projection off the gravitational along the direction off the wire. He's going to be equal to the attention. Okay. And and is going to be mg times the single here status. So good. Sign off 50 degrees. Now the tension is going to be equal to 0.59 for mutants. This and for the point in question, see, So the tension when the wire is making a 15 degree angle to the left. Okay, since theater is the same angle so you can see that gravity downwards tension along the wire and we're going to have the projection off the gravitational force here. The angle here is also theta, so it's going to be the same thing as we did for D. B. So tea is what it should be. Also equals 0.594 Newtons. Okay, now our last question Question E yes. Three questions E, um asks us to re calculate the velocity, the linear velocity off the ball in the same points off questions A, B and C. But now we must consider that the bowl have ah initial velocity of 1.2 m per second. Okay, so let's see how this will affect our calculations. Okay, so in question A, we had that from conservation off energy, we have that The final energy was I data dot square divided by chiu and this was equal Chew mgl one minus cool sign off theta zero, which is study degrees. Now we're going to have to also add a kinetic term for the initial configuration. So I over to data zero that squared. Okay, The initial angler velocity. Okay. Um okay, Now let's try to isolate data dot So thought about, um is going to be just, um sorry. So theta dot squared is going to be two divided by l times G. Well minus. Go sign off. Data plus V zero. Sorry. That's right. In terms off. Off data plus data dot Squared, um, divided by two. Okay, so this is what happens when we isolate data recalling that I is m l squared here. Such that we can cancel out. I am. And l Okay, Um okay. I'm sorry. I was forgetting term state a zero adult squared times l. Okay, Um okay, now we can to be stupid to the angular velocity for the linear velocities and take the square root off everything we're going to have that we the new velocity is going to be I'm going to write just the expression I'm not going to be stood The values it is going to be l times The square root of two divided by L. Times G times one minus school Sign off data nos v zero squared, divided by two Well, square root off everything. So you can notice that the only difference between this expression and the expression from questions A and B is basically that we are adding disturbing here. Yeah, Okay. We're adding zero squared, divided by two l inside our square root. Okay, Now, if we should be suit everything for the values that we have that is help for 2 m G for 9.81. Uh, data zero. Sorry. Theta zero for 30 degrees V zero for 1.2 m per second. We're going to find out the velocity is able to, um, 2.58 m per second. Okay, So noticed that this is greater than the value from question A. So when we put a initial velocity in our ball than the velocity that we're going to measure at the lowest point is going to be greater than if the ball was falling from the rest. Okay, um, now find the same thing for the point off question being, but with the initial velocity of 1.2. Okay, so we want to find fourth data equal to 15 degrees. True. Okay, Now you can you could see from what? What I did in Question e A. That I told you that the only thing that was going to change in our calculation is that inside the square root, we're going to add disturbing here. So basically, what we're going to do is we're going to just rewrite the expression that we found in question being okay, this expression here but here. Sorry, but adding a term off plus V zero square divided by two l Okay, Select Do this can see that we're going to save lots of time by doing by noticing this this trick. So this case is going to be so we found to be too divided By L Times Gene Times Co sign off 15 degrees minus CO. Sign off 30 degrees. Okay. And now we're going to add plus. Sorry. Yeah, plus the extra term coming from the initial velocity V zero square divided by two l. Okay, you can just be stood the values immigrants, you'll find that develops is going to be 3.31 m per second. Which again is great. Greater than the value that we found For a question, be when the bubbles falling from rest. Okay, lastly, we're going to populate our velocity. But when the ball is making a 15 degree to the left. Okay. Now, as I told you in the beginning question be in question. See? Sorry. Is that no matter if the ball is to the left or to the right from the horizontal from the vertical access, what matters is if the angle it makes with the vertical axis is the same. And this is what actually happened. So this angle here is the same as the angle off question be. So if the angle are the same, if the angle is the same, then the question the answer is also going to be the same. So here the same thing. Two point study, 1 m per second. And this is the

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