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A uniform rod is set Up So that it can rotate about an axis at perpendicular to one of its ends. The length and mass of the rod are 0.719 m and 1.23 kg. respectivel...

Question

A uniform rod is set Up So that it can rotate about an axis at perpendicular to one of its ends. The length and mass of the rod are 0.719 m and 1.23 kg. respectively A force of constant magnitude F acts on the rod at the end opposite the rotation axis. The direction of the force is perpendicular to both the rod's length and the rotation axis_ Calculate the value of F that will accelerate the rod from rest t0 an angular speed of 6.95 radls in 9.13 s

A uniform rod is set Up So that it can rotate about an axis at perpendicular to one of its ends. The length and mass of the rod are 0.719 m and 1.23 kg. respectively A force of constant magnitude F acts on the rod at the end opposite the rotation axis. The direction of the force is perpendicular to both the rod's length and the rotation axis_ Calculate the value of F that will accelerate the rod from rest t0 an angular speed of 6.95 radls in 9.13 s



Answers

A thin, horizontal rod with length $l$ and mass $M$ pivots about a vertical axis at one end. A force with constant magnitude $F$ is applied to the other end, causing the rod to rotate in a horizontal plane. The force is maintained perpendicular to the rod and to the axis of rotation. Calculate the magnitude of the angular acceleration of the rod.

In a options in party only conservative 40 directing on the bar being a con, traditional energy off the board off the strong bitches. Okay, I place initial kinetic energy plus potential energy is opposed to Kiev. Plus, you have, for the evolution off reputational energy off the system would reject 40 can be murdered at all as a particle other center off mask. Take the zero configuration for potential energy for the bar to term with the bar for the genital. Under these conditions, final potential energy is Vero and initial potential energy is energy. L divided by using the conservation of energy question about which is the euro plus one by two mt off L, which is It was 21 by two. I may go find the lawmakers What and we can say will make a final music was two under Rudolf and D l devoted by I for about rotating about access through one and I is It was toe and most l squared divided by three. Therefore, Omega f is equals to Underwood off mgl divided by one by three off Amal's where statistic was toe under three g up on Capitol and know that we have so chosen clockwise rotation. As for 30 now in Barbie from National style music was Toe I ever for and I m into G into Alberto Logistical to one by three Emily, but it is close to $1. 30. Emma's quit into Alfa and Alfi. It was 23 g open to L in party. We have excellent goes to minus off a stink. Majid tickles Tonegative off our omega squared fugitive was toe negative off upon food. Multiply with three g upon. Al Majid was too negative with three g upon to Yes, that's a centripetal acceleration, which is a C. It is derived along the negative horizontal along. A by is opposed to minus off 30. It goes to minus off our Alfa, which is Elop onto into Alfa, which is minus three D of 14 From here we can find the values. A majestic was too big. There is a close to negative three by two g off I minus three very four g off Digga. No, In part B, the power are gathered our forces and that had asked on the road using Newton's second law. We find that effort succeeded was toe almost eggs. Is it closed toe minus three by toe, a. Most G and a long wire direction effort by minus energy. It goes to remove a by which was two minus three by four mg turned to act by. Majestic wants to MD minus three by four mg. We did the close to one by four off energy from here 1/4. It was to Mars in tow. Acceleration rejected was to minus three by two and most G I kept plus one by four AM off G J Cat. Once he did, this is the answer.

In this problem, we have toe find angler velocity off the road, which is a omega. Also, let's write relations for energies and worked in corresponding to the system. So initial candidate energy will be t I, which is equals to zero Joe as system is a West initially now for final candidate that you can write if which is equals toe one he wanted by two I know Tau Omega Square where, where this I noticed the moment off. Inertia on it can be written as I don't is equals to one divided by 12 m r l Square plus m r into evaluated by two whole square. Now, by inserting this value into described, and we can write TFs dear physicals to one divided by two scribe legs open here we can write to when he wanted by 12 and are elsewhere plus m r into LD Wanted by two whole square Andi. Then here we can write Omega Squared now, by inserting values into this, a quiet and we can write DF physicals to one divided by two square breaks up in one you wanted by 12 in tow. M R, which is a mess off the road, which is it goes to 10 kg into the let off. The road is a 3 m all square plus 10 kg into 3 m divided by two whole square into oh my God square. So from here we can white this TF stf physicals to 15 k g meter squared into Oh my God square. Now let's calculate the world by the force f and W s 04 the Force f the border will buf which is equals to f that's it. Now we know that this SF is equals to al Qaida so this can be written as a u F is equals two f help Tito. Now let's write the values into describing So you have physicals too. Averages equals to 115 utan into l, which is equals to 3 m into bye bye to radiance. So this is equals to so 106 point Add 5 ft. Yo, Now the wagon corresponding to the w Salutin as you w So it is equals to minus f m g. It well by two. You can call this as a potential energy to now by inserting values into the square. We can write minus into 10 kg into 9.1 m per second, squared into 3 m, divided by two. So this will give us a value for this, uh, you w which is the world in responding to the wedges minus 147 0.15 years. Now let's apply the work energy principle for which we can white submission off you initial to find And this is the total world been corresponding to the system which is equals to you have plus you w is equals two DF minus d i by setting the values into the square and we can write seven android and six point at 5 ft. Job minus 147 0.15 years old is equals to ar 15 KC meters square into, uh, Omega squared minus 00 So for me, a recon wide. The value for this omega omega is equals to 6.11 radian per second. So this is our answer and of the question. Thank you.

In this problem, we have to find England velocity off the road which is a Omega in order to calculate the singular closely, let's white the kinetic energies and they're working, Responding to the system eso the initial planet energy which is t i zero e o that the system is rest at initially. The final kinetic energy will be t f which is equals to one divided by two. I know. Oh, my God Square. But this I noticed that moment off financially and it is written as I notice equals two wonder why did by 12 Hamill Square Plus and Andi uh l divided by two whole square? No. By accepting this rather into the square and we can write dear friends dear physicals toe only won it by two square bags open one you wanted by 12 Amel Square Plus I am into LD wanted by two whole square described in schools Omega Square. Now let's slide the values into this equation. We can write it as a tear physicals to wonder wanted by to square rooms open one divided by 12 into the mass which is equals to 10 kg into 3 m hole square plus 10 kg into 3 m, divided by two whole square square books Clothes Omega Square So this can be written as a T F, which is equals to 15.0 k g meter square into Omega Square. Now let's kill fled the world in responding to the force f and the way w so for the world been responding to the full surface u f which is it goes to, if that's it, how we can ride the relation for sf sf physicals too. Uh, Elpida So this can Britain as a um, physicals too? If Healthy eater let's try the values into this. Quiet, um, physicals too. 150 new done into 3 m into 180 degree multiplied by. I'll bite it in you ordered by 1 80 Duty also, this will you are the value for this us as 141 3 1400 13 point so onto Joel Now the world and corresponding to the way it is written as you w which is equals to minus, uh, mgl. So let's try the values into describing so it will be minus into 10 kg head to 9.81 meter per second squared into tribute. So this will you are the value for this u W as a minus 294 0.3. Let's apply the work energy principle so we can write seven Sigma you initial toe. Find this is the network than corresponding to the system. In this case you have plus you w is equals to t f minus t I. Let's try the values for these individual world and and under the kind of technologies. So here we can write 1400 13.72 goals find this are 294.30 is equals to 15.0 k g meter square into Omega Square minus zero. So for me, we can ride the value for this omega Omega is equals to eight point 64 Radiance Parseghian. Radiant personally. So this is our answer. And after question

Exercise. We have a rod of mess M and length L that's initially oriented along the Y axis. And that starts, uh, rotating about a pivot. That's ah, represented here in figure by this green dot from rest And at the time when the road is horizontal, that is along the X axis. We want to know what is its angular speed Omega. In order to do that, I'm gonna use conservation of energy. So first of all, remember that the total energy is conserved. So this means that the initial kinetic energy ke I cause the initial potential energy you why it's people to the final kinetic energy was the final potential energy notice that the initial kinetic energy is zero since the rod is a breast, while the initial potential energy you why is equal to the potential energy at the center of mass? So it iss the mass times G times hello over to choosing the potential energy uh, the zero of the potential energy to lie at why equal zero to have mgl over to is able to KF that is the moment of inertia. Times w abs Air Omega squared over two plus the final potential energy, which is zero since the rod lies at why equals zero. So from here we obtained that omega is equal to the square root of M l g divided by I In the moment of inertia, I, uh, varada around one of its ends is equal to 1/3 of m l squared. So omega is equal to the square roots of mgl divided by AM L square over three and this is equal to the square root of three G. Overall, this ISS the angular speed. Then in question be we have to find the magnitude of the angular acceleration of the rod. And I'm going to use the torque in order to find it. So have the torque is able to the force times the distance from the axis of rotation where the force is applied. In our case, the force is just a gravitational force, which is mg and it supplied at the center of mass, which is located L over to and this is equal to I. That is a moment of inertia times the angular acceleration Alfa So Alfa is equal to mg yellow over to why, and as we saw Hi, is m l squared over three. So this is able to three G over too well. And this is the angular exploration we were looking for, then in question. See, we have to find the why an ex components of the exploration of the center of mass. And in order to do that, I'm gonna remind you that when the rod is horizontal, it's X component of the exploration is just the centripetal acceleration. So a C A X is equal to a C. That's this interpretive exploration which is equal to Omega, actually minus omega square over now over to because we're interested in the exploration of the center of mass in the minus here, because acceleration is pointing to the negative extraction and sins, Omega is the square root of three g or l, this years ago to minus three G over L l over two. So this is equal to three minus three g over to then for the Y axis acceleration is able to the general exploration Alfa Times. The radius that is the distance from the center of mass to the exceptional rotation which is a war to and s Alfa is equal to three after their the miners here again because the ah exploration is pointing downwards and says Alfa is able to three G over too well trying over to, um l yeah, that's right. Minus three g over too. Well, times l over to. So this is equal to minus three g over for this is the acceleration in the Why direction Finally, for question D. We have to find the force that the people people, um exerts on the rock both in the X and wide rations. So notice that the force off the pivot is responsible for the acceleration off the rod in the X direction because it's the only force that acts on the X direction. So effect is able to m times a X. And since a X is equal to minus three year or two, then this is equal to minus m r three m g over to if why, On the other hand, the force that the pivot exerts only y axis is not the only force that is exerted vertically. There's also the force of gravity. So have have Wyman is mg equals I am a why so if y is equal to em A Why plus G. So this is I am minus three G over four plus g. So this is going to mg over four and this concludes our exercise.


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