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Two Masses on a BalanceIwo Dlocis eonmass kg are suspended from the ends of & rigid weightless rod length Initlal acceleration the mass attached t0 Assume posit...

Question

Two Masses on a BalanceIwo Dlocis eonmass kg are suspended from the ends of & rigid weightless rod length Initlal acceleration the mass attached t0 Assume positive and down negative122.0 cm and I2 10.9 cm. The rod held the horizonta position shown and then released Calculate theJOMccoddcooddcooddccoddcooccoooccco_

Two Masses on a Balance Iwo Dlocis eon mass kg are suspended from the ends of & rigid weightless rod length Initlal acceleration the mass attached t0 Assume positive and down negative 122.0 cm and I2 10.9 cm. The rod held the horizonta position shown and then released Calculate the JOMccoddcooddcooddccoddcooccoooccco_



Answers

Two Blocks on a Rod Figure $11-37$ shows two blocks, each of mass $m$, suspended from the ends of a rigid massless rod of length $L_{1}+$ $L_{2}$, with $L_{1}=20 \mathrm{~cm}$ and $L_{2}=80$ $\mathrm{cm}$. The rod is held horizontally on the fulcrum and then released. What are the magnitudes of the initial accelerations of (a) the block closer to the fulcrum and (b) the other block?

So we're gonna consider counterclockwise to be positive. And if this is the case, we can then say the net torque is gonna be equal to M. G else of one minus mg. Else up to this would be equal to the moment of inertia times the angular acceleration, the moment of inertia would be m else of one squared plus m Elsa two squared again, multiplied by the angular acceleration. So we can then say the angular acceleration would be equal to G times else of one minus else of two divided by else of one squared plus else of two squared. Given that the masses all cancel out and essentially, this is equaling 9.80 meters per second squared multiplied by 0.20 meters minus 0.80 meters, divided by 0.20 meters quantity squared plus 0.80 meters quantity squared. The angular acceleration is equaling negative 8.65 radiance per second squared and it being negative. We chose counterclockwise to be positive, so the angular acceleration would be in the clockwise direction. Ah, say t equals zero seconds. We have an essentially an instantaneous velocity equaling 20 meters per second. And at that instant, the radio acceleration is equaling zero meters per second as well. So here, um when we're trying to find be 10 essentially for part a the magnitude of the net acceleration, we would only be taken into account the tangential acceleration. So here this would be equal to the Magna Leah. Rather the absolute value of the angular acceleration times else of one. This would be 8.65 radiance per second squared, multiplied by point 20 meters. This is giving us one 0.7 meters per second squared and then for part B, the magnitude of the net acceleration sub two would be equal to again. The absolute value of the angular acceleration times else up to this would be equal to 8.65 radiance per second squared multiplied by 0.80 meters. Ah, this is giving us 6.9 meters per second squared. These would be your two answers Answer for party. An answer for party. That is the end of the solution. Thank you for watching

So we have a diagram here with a fulcrum and whoops, two masses. L one is here, L two is here. Okay, So we're told that L one is 20 centimeters and L two is 80 centimeters. Oops, I do. Okay with the magnitude of the accelerations On particles one and 2. Okay, so um the inertia of this system would be the some of the masses times the distance is squared. Um We I don't think we're told numbers for the mass, but they're the same. So the inertia would be and I'm making this the positive direction anyway, it would be, well the total inertia, it doesn't matter on the direction, it's just going to be some of the M. R squares, which would be mass times L one squared plus mass times all to square. But I'm just going to change that to a factor at the M. Okay, now, um the torque net, this is where I need the direction is going to be um mm times G at L one, So MGL one minus M G L two. And that's going to be I alpha. We're trying to figure out Alpha. So Alpha is going to be factoring out the MG who MG? It's not a not a subscript. L 1 -6 2 over I. But I is M L one squared plus L two. Where'd whoops squared? Uh race? Uh draw. Okay. So Then the acceleration at one is going to be Alpha times the distance to one, which would be L one. The acceleration at two is going to be Alpha Times The Distance to L two. So putting this all together in a calculator. Um I'm going to write, first of all, I'm going to declare L one and L two. And I finally figured out if you just type L one, it puts in a sub script automatically equals 0.2 Okay. L two equals 0.08. Ah While I'm at it, G Is 9.81. Okay. Did I cancel out the ems yet? I did not the M's cancel out. Yeah. So the man says don't really matter. So this is going to be alpha equals G times L one -L. 2. I don't know where my two went. It must have gotten erase. Yeah. L one minus L. Two over hell one squared plus L. Two. Where? Okay, so alpha, that makes sense. It's going in the negative direction. Um So that would be clockwise. Um A one A 1 is going to be alpha alpha L. One And then a two is going to be alpha L. Two. Okay, but something's not right here because definitely a one is accelerating upward. A two is accelerating downward. So a one has to be positive 1.73, And this would be a negative 6.92 m/s squared meters per second squared. We know since it's rotating in this direction because the acceleration was negative, this is accelerating downward and this is accelerating upward. Um All right, so this was the answer to 56. Thank you for watching.

Hello and welcome to this video solution of enumerate. This question is based on the principle self simple harmonic motion. We are given a small compact mass him that is attached to a end of a uniform railroad of mass M. And lengthen that is promoted at the top here we can see it. Uh huh. So we have this mask capital M. And we have this road with the lintel that is devoted at this top. Right? Yes. The people Now party, we have to find out the tension in the rod at the people. So this is and when it's stationary obviously. So the force is pretty simple for this, which is equal to the total mass of the system. Right? So the mass of this bowl is MG. And also the mass of this road is given us him. So Both of the weights will add up to give the total force at these periods, which is equal to two MG. So first question we have solved next, let's say. Now we have to calculate the weight at this pivot point B. That means you have to consider this complete weight of the below mass and the mass of this. Uh the weight of this section wide total weight of the section way. So let us write that down. Now we have got F. B. The force at what point which is equal to MG. This is for the pillow mass plus. Now here we have to calculate the mass per length of the adult. That means him by L. Is A. You mass per length of this road at the unit mask for this rod times the amount of length were required to calculate that is why times. Thank you. Will give us the total weight at P. That we can say we have M. G. Plus mm G. Why by L. Queen. So the force at the pivot point we obtain an expression of the next. We have to calculate the period of oscillation for small displacement From equilibrium and determined the period for l. equal to two m length of that out. No. We have to calculate the time period of the motion. So part B. First of all to calculate the time period of motion. We need to calculate the moment of inertia of the system about this period. That is about the point of rotation, which is I equal to the moment of energy. Is the eye of this Villa mass Let's say uh See you guys. A ball plus I. R. Is a dog. Okay, so we have got the amount of energy of the ball equal to M. L. Square right? Sometimes the total length and plus I. R. Which is a moment of inertia of this road. About one point about the end to be specific. So it is equal to one third of ml square This give us for 3rd of M. L. Squared. Right? So the moment of inertia we calculate us for 30 mile square about the pivot point. Now we have to we now let us write down the expression of time period of uh compound body or a physical pendulum we can say. So it is two pi you know what else? Bye bye. And giddy. And this setup actually constitute physical physical went along we can say so M. G. D. Now I we have actually calculate now we have to calculate the D. B. Is actually the distance of this people out from the center of mass of the system. Okay, so let's say the center of mass will be somewhere around definitely draw green point like this. Okay, so the distances uh this is the D From here to the center of mass of the system. Now do we have to calculate? So B is calculated as equal to him? The mass of the road by Times L x two because we assumed that the center of muscle this route is at the center of it plus M. Times ill for that ball by to tell herself to Oh sorry, sorry, sorry to put the hospital. Yeah, I am but I am right divided by the total mass system. This gives us Ds Equal to three L. Before now we will simply plug in these values here and we will see how much we're getting to pay Right over i. e. s. for 3rd. Four by three. A. Male square Bye. Uh here we have the mass M. So here I am will be quoted to a total mass of the system. Fine. So here we have to um G and they will be three L by four, the top I'm waiting for. And that will give us the total valuers. four x 3, five. We'll take it off. We're waiting. So the expression of time period for this complex set up is four by three By route over 12 Kg. Now we have to calculate this for L equal to two m, right? So it's pretty simple For private three through the wall of two times 2 x 9.8 2nd. This gives us 2.68 seconds. I hope this is clear to you and have a very good taste of that.

Silver party. We're going to apply Equation 10 34 and we can say that the kinetic energy is equaling 1/2 times the moment of inertia times the angular acceleration or the sorry, the unclear velocity squared. Ah, this is equaling 1/2 times the moment of inertia of a rod rotating about one end going 1/6 um, I'll square Omega squared and now we can solve. So the kinetic energy would be 1/6 times 0.42 kilograms multiplied by 0.75 meters. Quantity squared, multiplied by 4.0 radiance per second. Quantity squared. Ah, this is giving us 0.63 jewels for part B. We apply the conservation of energy and we can say that here, the rotational, uh, simply equal the gravitational potential energy MGH. So we can say that H would simply be equal to be a magnitude of the kinetic energy divided by the weight so this would be equal to M. L squared Omega squared, divided by mg rather six mg. The mass is cancelled out, and we find that this is gonna be l squared Omega squared, divided by six g we can solve, So h I would be equal 2.75 meters Quantity squared times 4.0 radiance per second. Quantity squared, divided by six times 9.80 meters per second squared. Ah, we have an H A maximum height, rather equaling two point 153 meters. This would be a final answer for part B. That is the end of the solution. Thank you for watching.


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