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Please show all the steps to understandA uniform thin rod of length 0.795 m is hung from a horizontalnail passing through a small hole in the rod located 0.054 m fr...

Question

Please show all the steps to understandA uniform thin rod of length 0.795 m is hung from a horizontalnail passing through a small hole in the rod located 0.054 m fromthe rod's end. When the rod is set swinging about the nail at smallamplitude, what is the period 𝑇 of oscillation?

please show all the steps to understand A uniform thin rod of length 0.795 m is hung from a horizontal nail passing through a small hole in the rod located 0.054 m from the rod's end. When the rod is set swinging about the nail at small amplitude, what is the period 𝑇 of oscillation?



Answers

A thin, straight, uniform rod of length $\ell=1.00 \mathrm{m}$ and mass $m=215 \mathrm{g}$ hangs from a pivot at one end. $(a)$ What is its period for small-amplitude oscillations? (b) What is the length of a simple pendulum that will have the same period?

Well, Omega for physical panda LEM is given by Omega is equal to square Who toe mt Times l divided by I here G is acceleration due to gravity. L is length and eyes the rotational inertia here and further the time p a. T is equal to two by divided by omega and rotational inertia. I is equal to one divided by three m l square. So with these equations, um, Omega, which is anger frequency is equal to square root off m times g times L, uh, divided by, um do I Siegel to do I right? And now Time PDT is equal to to buy divided by Omega. And we have omega right to therefore but the sequel to do by Divided by the square Root off mgl divided by two times to multiply by I well, here I is one divided by three ml square, right, and therefore diabetes D is equal to Time PDT Nichols Dubai into square root off to l divided by three Gene nylex Plugging values well by plugging values, Time PDT is equal to to buy into the square root off two times L, which is doing to one meter divided by three, multiplied by 9.8, and therefore Time PDT is a call to 1.639 seconds, 1.63 line seconds.

Hello students in this question we have a uniform role of length, capital L. It is nailed at opposed so that the two third length is below the nail. So suppose we can draw this diagram. So I suppose this is a roll of total length L. And uh the nail is here so that the two third length is below the nail such like this. And this will be held by three. Okay, and this is the nail at which this is suspended. Okay, So we can calculate the center of mass will be here. So this will be the center of mass point and this is the hinge point. Okay, changed point. So the distance between them is supposed small edge. So small, it will be equal to the small edge. This can be calculated. Uh this resistance is small. And so from up to here these two L by three minus this restaurant center of Mass will be held by two minus L by two. So this is equals two L by six. Okay so mom entrepreneurs here about this point of suspension can be calculated. So I will be equal to the center of mass plus M. H. Square where smaller mr mass of this role. So center of mass. Movement of nausea will be M. L. E squared divided by 12 plus M. Replaced by H. S. L. By six. And this holy square. So moment of nausea comes out to be Emily square by nine. Okay so now we can write that the time 30. This will be equals 2 to 5. And the root of I divided by M. G. H. So we can substitute the value so to five. And the root of all which is M. L. Squared divided by nine divided by MG. More player B. H. Which is held by six. Okay so from here after solving the time period T. This will be obtained as to buy under root off To well developed by 3G. So this will be the period of oscillation for the road. Okay? Thank you.

The moment of inertia about is given by I is equal to hi center of gravity plus MhZ square which is equal to Emily squared by two world plus M L. E squared by food Which is equal to Emily Square by three. I can write developed is equal to two pi route under al by mg. Multiplication L by two which is equal to two pi route under two MLB square bye three MG L. On further simplification. Finally I can write developed is equal to two pies or to tender. Well bye treaty. Now let the time period he is equal to the time period of simple pendulum or plant X like. And I. D. Value of T is equal to two point a route under X by G. So here the value of well by three. G is equal to X by G. Or X. Is equal to Well by three. So the length of the pendulum media well by three. This is the answer.

Hello students in this question we have given a circular hope of demeter D. Hanging on a nail. So we have to determine the period of oscillation for the small amplitude. Okay so suppose the radius of the who pieces small art and the hope is hanging from this center. So this is the hope and this is the hinge point and this is also a center of mass point. Okay so moment of inertia about the point of suspension. This can be uh sorry this is hinged from here. So this is the hinge point about which it is hinged. Okay so moment of inertia about this hinge point. I this will be equal to the center of mass plus M. R. Square. Where M. Is the mass of the school? Okay so center of mass, movement of nausea will be emery square plus again emery square. So we get to M. R. Square. Okay. And the height edge above which between center of mass and this hinge point is edge which is equal to this radius. So we can write the period of oscillation T. It is equal to the root of Mhm. I divided by M G H. Okay, So I here is to M R. E square divided by M G M B H. Is our. Okay? So this become equals two are divided by G. And demeter is D. And radius is taken are so D. Will be equal to poor. So this to work and be replaced by D. So D by G. And this under. So this will be the time period of oscillation for this pool. Okay? Thank you.


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