Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

5: Uniform Circular Motion

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Today we will discuss what we call "uniform circular motion." What is uniform circular motion? An object goes around in a circle, has radius r and the object is here.This is the velocity.It's a vector, perpendicular. And later in time when the object is here the velocity has changed, but the speed has not changed. We introduce T, what we call the period-- of course it's in seconds-- which is the time to go around once.

We introduce the frequency, f, which we call the frequency which is the number of rotations per second. And so the units are either seconds minus one or, as most physicists will call it, "hertz" and so frequency is one divided by T. We also introduce angular velocity, omega which we call angular velocity.

Angular velocity means not how many meters per second but how many radians per second. So since there are two pi radians in one circumference-- in one full circle-- and it takes T seconds to go around once it is immediately obvious that omega equals two pi divided by T. This is something that I would like you to remember. Omega equals two pi divided by T-- two pi radians in capital T seconds.

The speed, v, is, of course, the circumference two pi r divided by the time to go around once but since two pi divided by T is omega you can also write for this "omega r." And this is also something that I want you to remember.

These two things you really want to remember. The speed is not changing, but the velocity vector is changing. Therefore there must be an acceleration. That is non-negotiable. You can derive what that acceleration must be in terms of magnitude and in terms of direction. It's about a five, six minutes derivation. You'll find it in your book. I have decided to give you the results so that you read up on the book so that we can more talk about the physics rather than on the derivation.

This acceleration that is necessary to make the change in the velocity vector is always pointing towards the center of the circle. We call it "centripetal acceleration." Centripetal, pointing towards the center. And here, also pointing towards the center. It's a vector.

And the magnitude of the centripetal acceleration equals v squared divided by r, which is this v and therefore it's also omega squared r. And so now we have three equations and those are the only three you really would like to remember. We can have a simple example.

Let's have a vacuum cleaner, which has a rotor inside which scoops the air out or in, whichever way you look at it. And let's assume that the vacuum cleaner these scoops have a radius r of about ten centimeters and that it goes around 600 revolutions per minute, 600 rpm.

600 rpm would translate into a frequency, f, of 10 Hz so it would translate into a period going around in one-tenth of a second. So omega, angular velocity, which is two pi divided by T is then approximately 63 radians per second and the speed, v, equals omega r is then roughly 6.3 meters per second.

The centripetal acceleration--

and that's really my goal--

the centripetal acceleration would be omega squared r or if you prefer, you can take v squared over r. You will get the same answer, of course, and you will find that that is about 400 meters per second squared.

And that is huge.

That is 40 times the acceleration due to gravity. It's a phenomenal acceleration, the simple vacuum cleaner. Notice that the acceleration, the centripetal acceleration is linear in r. Don't think that it is inversely proportional with r. That's a mistake, because v itself is a function of r.

If you were sitting here then your velocity would be lower.

Since omega is the same for the entire motion you really have to look at this equation and you see that the centripetal acceleration is proportional with r.

Therefore, if you were...

if this were a disc which was rotating and you were at the center of the disc the centripetal acceleration would be zero. And as you were to walk out, further out, it would increase. Now, the acceleration must be caused by something.

There is no such thing as a free lunch.

There is something that must be responsible for the change in this velocity and that something I will call either a pull or I will call it a push. In our next lecture, when we deal with Newton's laws we will introduce the word "force." Today we will only deal with the words "pull" and "push." So there must be a pull or a push. Imagine that this is a turntable and you are sitting here on the turntable on a chair. It's going around with angular velocity omega and your distance to the center, let's say, is little r.

You're sitting on this chair and you must experience--

that is non-negotiable--

centripetal acceleration A of c, which is omega squared times r.

Where do you get it from? Well, if your seat is bolted to the turntable then you will feel a push in your back so you're sitting on this thing, you're going around and you will feel that the seat is pushing you in your back and so you feel a push, and that gives the push out.

Yeah, I can give this a red color for now.

So you feel a push in your back.

That push, apparently, is necessary for the acceleration.

Alternatively, suppose you had in front of you a stick.

You're not sitting on a chair.

You don't get a push from your back.

But you hold onto the stick and now you can go around by holding onto the stick.

Now the stick is pulling on you in this same direction.

So now you would say, aha, someone is pulling on you.

Whether it is the pull or whether it is the push one of... either one of the two is necessary for you to go around in that circle on that turntable with that constant speed.

Now, the classic question comes up, which we often ask to people who have no scientific background.

If you were to go around like this and something is either pushing on you or is pulling on you to make this possible suppose you took that push out, all of a sudden.

The pull is gone.

[makes whooshing sound]

What is now the motion of the person who is sitting on the turntable? And many non-scientists say, "Well, it will do like this." That's sort of what your intuition says.

You go around in a circle, and all of a sudden you no longer have the pull or the push and you go around in a spiral and obviously, that is not the case.

What will happen is, if you have, at this moment in time a velocity in this direction and you take the pull or the push out you will start flying off in that direction and depending upon whether there is gravity or no gravity there may be a change, but if this were...

if there were no gravity you would just continue to go along that line and you would not make this crazy spiral motion.

I have here a disc, which we will rotate and at the end...

the edge of the disc here we have a little ball.

And the ball is attached to that disc with string.

So now this is vertical, and so this is going to go around with angular velocity omega.

And we have a string here and the string is attached to this ball and the whole thing is going around and so at one moment in time this has a velocity, like so.

And therefore there must be non-negotiable centripetal acceleration which in magnitude is omega squared r or, if you want to, v squared divided by r.

Now I cut it and that's like taking away the push and the pull. The string that you have here is providing the pull on this ball. This ball is feeling a pull from the string and that provides it with the centripetal acceleration. Cut the string and the pull is gone and the object will take off. And if there were gravity here, as there is in 26.100 it would become a parabola and it would end up here. If, however, I cut the ball exactly when it is here--

not the ball, but I cut the string--

then, of course, it would fly straight up gravity would act on it, it would come to a halt and it would come back. So it really would then go along a straight line. But you would clearly see, then that it's not going to do what many people think-- that it would start to swirl around.

It would just go...

[makes whooshing sound]

and comes back. Let's look at that. We have that here. So here is that ball. The string is behind here; you cannot see the string. I will rotate it, wait for it to pick up a little speed and the knife, that you can't see either, is behind here and when I push the knife in, I do it exactly here.

It cuts the string and it goes up. You ready for this? You sure you're ready? Three, two, one, zero. Wow! That was very high. So you see, it's nothing like this. It simply continued on in the direction that it was going. It wasn't going into a parabola because I was shooting it straight up. The string forms the connection between the rotating disc and the ball and therefore, the pull is responsible for the centripetal acceleration.

Let's now think about planets.

Planets go around the sun.

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

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Fisika untuk Universitas

Fisika untuk Universitas

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4: The Motion of Projectiles

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Today, we're going to take it quite easy.

I also have to take it a little easy because my voice may be petering out, if I'm not careful.

We're going to apply today what we have learned, so there is nothing new but its applications.

And that's important--

things that... you can let it sink in.

We have here a trajectory of a golf ball or a tennis ball in 26.100.

We shoot it up at an angle alpha.

The horizontal component in the x direction is v zero cosine alpha and the vertical component is v zero sine alpha.

It reaches the highest point at P and it returns to the ground at point S.

This is the increasing y direction and this is the increasing x direction.

We're going to use, very heavily, the equations that you see here that are so familiar with us.

These are the one-dimensional equations in x direction where there is no acceleration and the one-dimensional equations in the y direction where there is acceleration.

In order to use these equations we need all these constants--

x zero, v zero x and v zero y.

We have seen those last time.

I choose for x zero...

I choose zero arbitrarily.

Also for y zero.

The velocity in the x direction will never change.

This v zero x will always remain v zero cosine alpha.

The velocity in the y direction, however, in the beginning at t equals zero is v zero sine alpha.

And that one will change, because there is here this t and that's why the velocity is going to change.

This t will do it.

And the acceleration in the y direction--

since this is increasing value of y--

is going to be negative 9.8.

Since I call always 9.8 plus...

since I always call g "plus 9.8," this is minus g.

I now want to ask first the question that you may never have seen answered: what is the shape of this? Well, we can go to equation number three there and we can write down this equation number three: That y, as a function of time, equals v zero yt so it is v zero sine alpha times t minus one-half gt squared.

That's the equation in y.

I go to equation number one and I write down x--

at any moment in time--

equals v zero z times t so that is v zero cosine alpha times t.

Now I eliminate t, and the best way to do that is to do it here--

to write for t, x divided by v zero cosine alpha.

Now I can drop all subindexes t because we're now going to see x versus y.

We're going to eliminate t.

So this time here, I'm going to substitute in here and in there and so I'm going to get y equals...

There's a v zero here and there's a v zero there that cancels.

There's a sine alpha here and a cosine alpha there that makes it a tangent of alpha.

And then I have here the x and I get minus one-half g times this squared--

x squared divided by v zero cosine alpha squared.

And now look very carefully.

Y is a constant times x minus another constant times x squared.

That is a parabola.

It's a second-order equation in x, and is a parabola and a parabola has this shape.

So you so see now, by eliminating the time that we have here a parabola.

Now I want to massage this quite a bit further today.

I would like to know at what time the object here comes to a halt to its highest point.

It comes to a halt in the y direction.

It comes to a highest point and I want to know how high that is.

Well, the best way to do is to go to equation four and you say, to equation four, "When are you zero?" Because that is the moment that the velocity in the y direction becomes zero.

It must be at its highest point, then.

So in order to find, for us, the position of the highest point P we first ask ourselves the question from equation number four: when is the velocity in the y direction zero? And that then becomes v zero y which is v zero sine alpha minus gt and out pops that t at point P is going to be v zero sine alpha divided by g.

That's the time that it takes for the object to reach the highest point.

Where is it, then? What is the highest point above the ground? Well, now we have to go to equation number three and you have to substitute this time in there so that highest point h, which is y at the time t of P equals v zero yt--

that is v zero times the sine of alpha.

But you have to multiply it by this time and so I get another v zero sine alpha and I get a g here minus gt mi...

oh, no, no, this equa...

minus half dt squared minus one-half g times this one squared which is v zero sine alpha squared, divided by g...

divided by g squared because there is a g here, you see? So you square the whole thing if it's t squared.

You lose one g and you will find, then, that the highest point--

let's write it down here so that we don't block that blackboard--

the highest point in the sky equals v zero sine alpha squared divided by 2g.

That is the highest point.

Let's give that some color because we may want to keep that.

Is it reasonable that the point, the highest point in the sky gets higher when v zero is higher? Of course.

If I shoot it up at a higher speed of course it will get higher.

So that's completely intuitive that v zero is upstairs.

If I increase the angle from a small angle to larger and larger and larger is it reasonable that it will get higher? Of course.

You all feel in your stomach that the highest possible value you can get is when you make alpha 90 degrees for a given velocity.

That's the highest it will go in the sky.

So clearly, this is also very pleasing.

If you did the experiment on the moon with the same initial speed, it will go much higher so you are also happy to see that this g here is downstairs.

So that makes sense.

At what time will the object be at point S? Now, there are two ways that you can do that.

You either go to this equation, number three and you ask equation number three, "When are you zero?" It will give you two answers.

It will say, "I am zero here at this time" and "I am zero at that time." And those are the two times that you want and this is the one you pick.

That's perfectly fine.

I think there's a faster way to do it, and that's the following.

This is a parabola, so it's completely symmetric about the vertical, about P.

So to climb up from O to P must take the same amount of time as to go down from P to S and so I claim that the time to reach point S must be twice the time to reach point P and therefore it's going to be two v zero sine alpha divided by g.

But now we want to look again whether the v zeros and the sine alphas have the right place.

Indeed, if I increase the speed, I would expect it to take longer before it reaches S.

If I give it a larger speed, it will come out farther and obviously, the time will take longer.

If I do it at a higher angle, it will also take longer and if I do it on the moon, it will also take longer.

So this makes sense--

these equations are pleasing in terms of the rate that v zero and sine alpha appear in the equations.

But now comes an important point which I am going to use throughout this lecture.

I want to know what OS is.

The distance OS... I shoot it up and it hits the floor again What is that distance that it travels? Well, for that, I need equation number one.

It is v zero x times the time and v zero x is v zero cosine alpha.

We got v zero cosine alpha times the time to hit it--

that is two v zero sine alpha.

So I get a two here, I get a sine alpha, and I get a g here and I have another v zero there, and so the answer is a v zero squared times the sine of the double angle--

remember, two cosine alpha sine alpha is the sine of two alpha--

divided by g.

And this is OS, and I'm going to need this a lot.

This reminds me not to remove it.

Now, I sort of wonder, and you should too why is it that the highest point in the sky has a v zero squared and why is the farthest point also...

why does it also have a v zero squared? There must be a way that you can reason that.

Why is it not just v zero? Why is it v zero squared? Well, I'll let you argue about the highest points, and I'll give you a good reason for the distance, OS.

Don't look at the equations.

You simply...

Think for a change.

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

Belajar itu Asyik

Berikut ini adalah tips dan triks yang dapat menjadi masukan berharga dalam mempersiapkan diri dalam menghadapi ulangan atau ujian :

1. Belajar Kelompok

Belajar kelompok dapat menjadi kegiatan belajar menjadi lebih menyenangkan karena ditemani oleh teman dan berada di rumah sendiri sehingga dapat lebih santai. Namun sebaiknya tetap didampingi oleh orang dewasa seperti kakak, paman, bibi atau orang tua agar belajar tidak berubah menjadi bermain. Belajar kelompok ada baiknya mengajak teman yang pandai dan rajin belajar agar yang tidak pandai jadi ketularan pintar. Dalam belajar kelompok kegiatannya adalah membahas pelajaran yang belum dipahami oleh semua atau sebagian kelompok belajar baik yang sudah dijelaskan guru maupun belum dijelaskan guru.

2. Rajin Membuat Catatan Intisari Pelajaran

Bagian-bagian penting dari pelajaran sebaiknya dibuat catatan di kertas atau buku kecil yang dapat dibawa kemana-mana sehingga dapat dibaca di mana pun kita berada. Namun catatan tersebut jangan dijadikan media mencontek karena dapat merugikan kita sendiri.

3. Membuat Perencanaan Yang Baik

Untuk mencapai suatu tujuan biasanya diiringi oleh rencana yang baik. Oleh karena itu ada baiknya kita membuat rencana belajar dan rencana pencapaian nilai untuk mengetahui apakah kegiatan belajar yang kita lakukan telah maksimal atau perlu ditingkatkan. Sesuaikan target pencapaian dengan kemampuan yang kita miliki. Jangan menargetkan yang yang nomor satu jika saat ini kita masih di luar 10 besar di kelas. Buat rencana belajar yang diprioritaskan pada mata pelajaran yang lemah. Buatlah jadwal belajar yang baik.

4. Disiplin Dalam Belajar

Apabila kita telah membuat jadwal belajar maka harus dijalankan dengan baik. Contohnya seperti belajar tepat waktu dan serius tidak sambil main-main dengan konsentrasi penuh. Jika waktu makan, mandi, ibadah, dan sebagainya telah tiba maka jangan ditunda-tunda lagi. Lanjutkan belajar setelah melakukan kegiatan tersebut jika waktu belajar belum usai. Bermain dengan teman atau game dapat merusak konsentrasi belajar. Sebaiknya kegiatan bermain juga dijadwalkan dengan waktu yang cukup panjang namun tidak melelahkan jika dilakukan sebelum waktu belajar. Jika bermain video game sebaiknya pilih game yang mendidik dan tidak menimbulkan rasa penasaran yang tinggi ataupun rasa kekesalan yang tinggi jika kalah.

5. Menjadi Aktif Bertanya dan Ditanya

Jika ada hal yang belum jelas, maka tanyakan kepada guru, teman atau orang tua. Jika kita bertanya biasanya kita akan ingat jawabannya. Jika bertanya, bertanyalah secukupnya dan jangan bersifat menguji orang yang kita tanya. Tawarkanlah pada teman untuk bertanya kepada kita hal-hal yang belum dia pahami. Semakin banyak ditanya maka kita dapat semakin ingat dengan jawaban dan apabila kita juga tidak tahu jawaban yang benar, maka kita dapat membahasnya bersama-sama dengan teman. Selain itu

6. Belajar Dengan Serius dan Tekun

Ketika belajar di kelas dengarkan dan catat apa yang guru jelaskan. Catat yang penting karena bisa saja hal tersebut tidak ada di buku dan nanti akan keluar saat ulangan atau ujian. Ketika waktu luang baca kembali catatan yang telah dibuat tadi dan hapalkan sambil dimengerti. Jika kita sudah merasa mantap dengan suatu pelajaran maka ujilah diri sendiri dengan soal-soal. Setelah soal dikerjakan periksa jawaban dengan kunci jawaban. Pelajari kembali soal-soal yang salah dijawab.

7. Hindari Belajar Berlebihan

Jika waktu ujian atau ulangan sudah dekat biasanya kita akan panik jika belum siap. Jalan pintas yang sering dilakukan oleh pelajar yang belum siap adalah dengan belajar hingga larut malam / begadang atau membuat contekan. Sebaiknya ketika akan ujian tetap tidur tepat waktu karena jika bergadang semalaman akan membawa dampak yang buruk bagi kesehatan, terutama bagi anak-anak.

8. Jujur Dalam Mengerjakan Ulangan Dan Ujian

Hindari mencontek ketika sedang mengerjakan soal ulangan atau ujian. Mencontek dapat membuat sifat kita curang dan pembohong. Kebohongan bagaimanapun juga tidak dapat ditutup-tutupi terus-menerus dan cenderung untuk melakukan kebohongan selanjutnya untuk menutupi kebohongan selanjutnya. Anggaplah dengan nyontek pasti akan ketahuan guru dan memiliki masa depan sebagai penjahat apabila kita melakukan kecurangan.

Semoga tips cara belajar yang benar ini dapat memberikan manfaat untuk kita semua, amin.

Sources: Godam