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A long; thin, uniform rod with a pivot at its top acts as a physical pendulum with a small angular amplitude. If this pendulum has a period of 1.4 seconds, find the...

Question

A long; thin, uniform rod with a pivot at its top acts as a physical pendulum with a small angular amplitude. If this pendulum has a period of 1.4 seconds, find the length of the rod;

A long; thin, uniform rod with a pivot at its top acts as a physical pendulum with a small angular amplitude. If this pendulum has a period of 1.4 seconds, find the length of the rod;



Answers

A uniform rod of length $l$ is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the rod.

We know that. Pity it off. A pendulum is given by to buy Route L over G Where a list event of the pendulum Angie's The gravitational acceleration. Putting the values eed to bias. Explain to eight, then off. The pendulum is given by 1.4 meter on the graph. Additional acceleration is 9.8 meter for a second square. We have to do the math now, so this will come out to the 6.28 one point 4/9 1.0.8 is 1/7. So divided by squared off seven. That will help me punching it in the calculator. Six point to a divided by a squared off seven. This is Trooper and 37 Seconds.

We have given ah, physical pendulum and we have to find the time period which is equals toe five a little I which is moment off, Garcia The right way I m g and Toby D is the distance between Hinge and the center off. Mass off the pendulum. Suppose this is the pendulum Here, Here This is Lord and this is must m Yeah and mass off road is also M. This is Hinge Point Onda Total length off rod is given L which is equal to 1.35 meeting Now we have to find the moment off Energia, about point over here So I about oh will be called toe must m l e squared plus moment of jealousy off role which is equal to one by three aml square So this will be equal to around after putting the value am is 1.1 kg Ellis 1.35 putting the values four by three The mass is one kg l is 1.35 m So the moment of energy about hinge will be called toe 2.43 kg meter squared. This is the moment of inertia about Hinge Now we have to find the distance between center of gravity and hinge Point. We know that the center of mass off the road is between here. Add distance zero point 675 Now the center of gravity off This pendulum will be equal to here between this distance. So this distance will be called toe zero point 675 Divide way too. With this total distance from Hinge will we call toe be Will we call toe? 0.675 plus 0.675 divided by two. It will be equal to 1.1 to 5 m. So now I put the value off I nd in the equation, Steve will be called to pie on. That would help. Sorry, I with just 2.43 years mass total bust This two K g g s 9.81 and B is 1.125 The time period is a call to 2.1. Line 75 or equal to two point 20 seconds. Okay,

Position of center of masks from that point of violation. Actually, I am an expert over ever present id everyone into and by two I'm going to minus l V two. I am meticulous and groups experiment. Moment of energy, of the system. And we're going to invite to the square mhm and 22 minus and bite me sweat. So I am one plus m tool to tell you. Yeah. Yeah, the time period off or situation would be to buy. How do you golf i upon and GDP substitutes available? I is a month plus and two into any square. Mm, yeah, yeah. I am much less ample. G into X and X. We have major. Huge. Yeah. You there? Yeah. Okay. Mhm. Yeah. Mhm. No. So I'm better. You will get Yeah. Dubai the root of two and mhm. A month plus and two upon g m. One. My lesson. Yeah, that's it.

So here to find the period of a physical pendulum. This is going to equal to pi times that the moment of inertia divided by MGH So this will equal to pi times the square root of 1/3. Um, I'll squared and this is gonna be divided by M g times 1/2 of l. So this is gonna equal to pi times to AL over three g and we can solve. So the period is gonna be equal to two pi times the square root of two times one meter being the length divided by three times 9.8 meters per second squared being the acceleration due to gravity. And we find that the period of this pendulum is going to be 1.64 seconds. So this would be your final answer for a party and then for part B, we're going to relate the formula above to a simple pendulum to find length. So t is equaling two pi times the square root of all over G. This would be the formula for the This would be the formula for these simple pendulum, and this is gonna equal to pi times to our over three g. So at this point, we know that the length of the simple pendulum would be equal to two out over three. Or we could just say 0.667 So two times one, remember three. So l simple, but vehicle 2.667 Yeah, meters, This'd be our final answer. That is the end of the solution. Thank you for watching.


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