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Ommodc mhanging 0.25 mpulley 0.0rpulley mass (between 0.01 kg and kg) is hung by string from the edge of massive (between kg and kg) disk-shaped pulley (with radius...

Question

Ommodc mhanging 0.25 mpulley 0.0rpulley mass (between 0.01 kg and kg) is hung by string from the edge of massive (between kg and kg) disk-shaped pulley (with radius between 0.1 and meters) as shown (position is given in meters_ time is given in seconds_ and angular velocity is given in radians _ second) Restart_Set the hanging mass to 0.25 kg_ the radius of the pulley to m, and vary the mass of the pulley. How does the magnitude of the angular acceleration of the pulley depend on the mass (and t

Ommodc mhanging 0.25 mpulley 0.0 rpulley mass (between 0.01 kg and kg) is hung by string from the edge of massive (between kg and kg) disk-shaped pulley (with radius between 0.1 and meters) as shown (position is given in meters_ time is given in seconds_ and angular velocity is given in radians _ second) Restart_ Set the hanging mass to 0.25 kg_ the radius of the pulley to m, and vary the mass of the pulley. How does the magnitude of the angular acceleration of the pulley depend on the mass (and therefore moment of inertia} of the pulley? How does the magnitude of the acceleration of the hanging mass depend on the mass (and therefore moment of inertia) of the pulley? How are your answers to (a) and (b} related? Set the mass of the pulley to 0.5 kg_ the radius of the pulley to 2 m and vary the hanging mass How does the magnitude of the angular acceleration of the pulley depend on the hanging mass? How does the magnitude of the acceleration of the hanging mass depend on the hanging mass? How are your answers to (d) and (e) related? Set the hanging mass to 0.25 kg_ the mass of the pulley to 0.5 kg, and vary the radius of the pulley. How does the magnitude of the angular acceleration of the pulley depend on the radius of the pulley? How does the magnitude of acceleration of the hanging mass depend on the radius of the pulley? How are your answers to (g) and (h) related? Set the mass of the pulley to 0.5 kg_ the hanging mass to 0.25 kg, and the radius of the pulley to Determine the acceleration of the hanging mass and the angular acceleration of the pulley. k From Newton = second aw, determine the tension in the string How much torque does this tension provide to the pulley?



Answers

As shown in Fig. $10-3$, a mass $m=400$ g hangs from the rim of a frictionless pulley of radius $r=15 \mathrm{~cm}$. When released from rest, the mass falls $2.0 \mathrm{~m}$ in $6.5 \mathrm{~s}$. Find the moment of inertia of the wheel. The hanging mass linearly accelerates downward due to its weight, and the pulley angularly accelerates clockwise due to the torque produced by the rope. The two motions are linked by the fact that $a_{T}=r \alpha$. Consequently we will need to determine $a_{T}$, and then $\alpha$, and then $F_{T}$, and then $\tau$, and then $I$. Remember that Newton's Second Law is central here (i.e., $\tau=I \alpha$ for the wheel and $F=m a$ for the mass). First we find $a$ using $y=v_{i} t+\frac{1}{2} a t^{2}$, since the mass accelerates down uniformly: $$ 2.0 \mathrm{~m}=0+\frac{1}{2} a(6.5 \mathrm{~s})^{2} $$ which yields $a=0.095 \mathrm{~m} / \mathrm{s}^{2}$, and that equals the tangential acceleration $\left(a_{T}\right)$ of a point on the rim of the pulley, which equals the acceleration $a$ of the rope. Then, from $a_{T}=\alpha r$, $$\alpha=\frac{a_{T}}{r}=\frac{0.095 \mathrm{~m} / \mathrm{s}^{2}}{0.15 \mathrm{~m}}=0.63 \mathrm{rad} / \mathrm{s}^{2}$$ The net force on the mass $m$ is $m g-F_{T}$ and so $F=m a$ becomes $$\begin{aligned} m g &-F_{T}=m a_{T} \\ (0.40 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right) &-F_{T}=(0.40 \mathrm{~kg})\left(0.095 \mathrm{~m} / \mathrm{s}^{2}\right) \end{aligned} $$ from which it follows that $F_{T}=3.88 \mathrm{~N}$. Now $\tau=I \alpha$ for the wheel: $$\left(F_{T}\right)(r)=I \alpha \quad \text { or } \quad(3.88 \mathrm{~N})(0.15 \mathrm{~m})=I\left(0.63 \mathrm{rad} / \mathrm{s}^{2}\right)$$ from which we get $I=0.92 \mathrm{~kg} \cdot \mathrm{m}^{2}$.

In this question, you are given this set up. Two blocks are connected by a string off negligible mass, um, passing over fully. Then the acceleration off the blocks are equal to 2 m per second square. And then there are five parts this question and we want to find moments off inertia after fully. Okay, so in part, A. Yeah, they're six months. Sorry. Yeah. Eso impact a What analysis model is appropriate for the blocks. We'll be using the particle under the net force model. Okay, man in Park City, What analysis model is appropriate for the pulley. You will be rigid object under Annette talk model. Okay, then. Proxy, um, using the model in parts A find attention, T one. So using Newton's second law. Okay, so that's net, because two Emmy. So I'm going to apply this, uh, equation along the inclined Mhm. So, um, t one minus m one g science data 02 and one A So t one equals two and one A class G. Santa. Okay. Putting in the numbers and one is 15 kg. Is too. She is not going to eat. Data is 37 degrees, and then you get uh, 118 Newton's. Okay, So this is the answer for part C, then Pastie. Um, similarly Look fine, Tito. So 42. Okay, so just look at the free body diagram for T two. So this is m two g and then t two is pointing upward. Acceleration is outward. Okay, So using Newton's second law, he hath not Sequels to me. I'm to a in this case, so mtg my next tee to it goes to and to a the T two it goes to and to G minus A is equal to 20 nine point times 9.8 minus two. This is equal to 156 Newton's. Okay, so this is the answer parts. Mhm. Okay. And then in part e um, using the nexus model in part B. Alright. Right in the equation. Um, for I the moment off initial, they're pulling in terms off tensions on Taiwan. He to regis our acceleration. Okay, so, um, he's looking at the pulley, so you have t to and then you have t one. And then the angular acceleration is from this, you know, clockwise direction. So, um, that talk because thio I alpha. Okay, then, uh, network is he to minus? He won. Hence our you go to I times a over our Okay, So I is he go to p two minus t one, right? Are square if I buy a mhm. Okay, So this is the expression we need. Uh, this is the expression of your team for party, and then we're going to use this expression for parts else. Hey, find a numerical value off the moment of inertia off the pulley. Okay, so I goes to t two t one times are square develop, but a so key to is 156. He wanted 118. R is 0.25 square and divide by two. Okay. And this is 1.17 low brands that square. Okay, so this is answer for parts F. And there

Like Newton's second law to Mars. Embrun. A plan Newton sick in law. Do most Angwin Do you want minus and one G Sign off. Could be 71 Theodore getting equals toe. I'm 18 now. There were 51 if he pulls through 1 50 kilograms. 15 10 kilograms magically. 9.8 meter, 9.8. Sign off. 30 zone victory. Bless. I don't know, Major Burns. Think couldn't quit. We get our include 11 H. Nugent. Now Marcel m do I'm too is opposed to and g minus. Do you do He wants to and do a now Do you do with the girls too? And do you minus eight? Is it closed toe to indicate program, multiply 9.8 meters. Four second floor minus 21 to 2 meters per second. Where is it? Was 2156 neuter now in part B Be happy. Do you to minus D one into? Uh, is it worth toe? I l for which is opposed to I e a one kep it and are and I use it was to do you do my nurse d one? You took our squared divided right dealing we could devolution after putting the values t two with month of toothpicks, Newton minus 111 h Newton multiply with little want to find video. Meters were divided by a, which is to ban fetal meters per second squared. Richard W. I. Is equal to 1.17 He looked around, meet their squid.

In this problem we have to determine the exploration of this system. So Newton's second law is and the net force is mass into the excavation. So we have from the figure the net force acting on the mass. M one is T one minus M one G. This is a question one. And the net force yeah acting on mass. Everyone is everyone. Mm. This is the equation to So using both the equations we can say Using Creation one and two We can say and one a. is equal to 51- and one G. So from this T one is equal to M. One A plus M one gee. This is a question three. And then that force acting on the mosque two is I am to G -62. This is a question for and the net force acting on them as two is um to a Now using equation Using equation four and 5 we have em too is a call to M two G minus T two. So T two is M two G-. And to desist equation six. And the equation for net torque on the police half mm are square multiplied by a upon. Hi this is equation seven. And the equation for net torque is this is the equation for net torque on literally. Yeah. And the expression for net torque is um RT to minus RT one minus style. This is equation eight. So using equation seven and eight. Mhm. Uh huh. Rt to minus R. T. One minus so is equal to have an eye square multiplied by upon our. So from here the two minus the one minus how are is equal to half M. This is equation nine. So now you think equation three, six and nine we have M two G minus and one g minus tao upon us, he recalled two and one plus M two plus half. I am might be fired by a So from here is a Dorito. I am to G- and one key minus towel upon our of born and one I am too. Let's have Yeah, so We have M to 8.8 Kg. She is 9.8 metaphor 2nd sq my nurse and one is 10.4 kg multiplied by 9.8 metre for second square. My style is point 35 new 10 m upon. It is .15 upon and one is 10.8 K drip plus M two is point for Fiji. Let's have M. S 0.2 G. So from here is 1.2 m for a second square. Therefore the isolation of the system is 1.2 metaphor second square.

Mhm. All right. So for part a of this when we're looking for the angular acceleration because it's starting from restaurant on that delicate data, we can use this equation here, Delta data is omega T plus one half alpha T squared where it this omega T. Is going to be a zero. So we're going to have an equation for alpha in terms of feta time, where alpha is going to be able to to fada over T squared. Next. We need to find the acceleration of the blocks and we can use that angular acceleration here to find that because we can say that A is equal to alpha times radius of the polling. And so we already have an equation for alpha in terms of the T. So we can just plug that in. So we're gonna have to pay to our of her T squared like self next. We need to find the tension in the upper and lower halves of the string. So let's start with the lower half. We can do with some of the forces acting on this mass. So we're going to have um the wait for us, our mom, let's do this. We've got the net force acting on it. It's going to be equal to the weight force. OMG minus the tension force. Or rather T. one. And so we can solve for T. one it's going to be I. M g minus or M times G minus A. And we can substitute in our value for it here. So we got to peter are over T squared as our value for a and That's our attention one. Now to find tension to it's going to be slightly trickier since we don't know the whether or not there's friction here. But we can actually use this pulley and do the some of the forces on the pulley instead. So we can use the forces the equilibrium forces around the pulley. We saw 42 there you have T. Two is going to be equal to T. One minus I. Alpha or are R. Is the radius of that police. We can substitute in 41 with this value right here. We can substitute in for alpha using this right here. So we're going to get T. Two is equal to M. G minus tooth data are over T squared. Uh huh minus I over our times data which is to paint a horror over T squared. And the ours are going to cancel. And so will give me luck with this. Oh excuse me. I substituted an A. Instead of alpha. So there should still be an are down here. We don't cancel that. Um It's two fatal over T squared. Um Times I overall. Excuse me sorry about that.


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