Problem #4 - Pendulum For Advanced Viewers

Problem #4 - Pendulum For Advanced Viewers

[@pedramamani]

Professor Lewin, I would like to thank your tremendous effort in creating such videos. They truly are amazing and I am a prominent fan. My answer to this question would be: x is a fraction of the total length such that 7*numerator + 4*denominator = 19

[@lecturesbywalterlewin.they9259]

x=AL what is A?

[@harshshah9003]

This means I have solved the problem. Yes I am so much excited that I did a tough one as quoted by you sir

[@raytheboss4650]

x=AL where A is 1/3 solution, let the distance of the point(s) from the centre of mass of the rod be x, the force of gravity on that half is M/L* x *g, similarly on the bottom half it will be M/L (L-x) * g. now imagine an angle b just like in your diagram, the perpendicular components of the forces are the forces times sinb which is approximately b, now write torque about the hinge point (distance x from the centre). torque will be clockwise torque - anti clockwise torque = moment of inertia about hinge * alpha, moment of inertia about the hinge can be found using parallel axis theorum which is ml^2/12 + mx^2. now rearrange for alpha and compare it with alpha in the case of the case where its hinged about its end and solve the equation you will get the answer as x = L/3. I just finished my 11th grade and im preparing for jee so i am happy i solved this as it was the 2nd last chapter of 11th grade jee physics

[@sciwizShreyasKulkarni]

We will write I at x via parallel Axis theorem , then T=√I/MgL , we will equate T on both sides and find x

[@amirawadalla5866]

Hello prof. I'm 16 years old and have only a decent understanding of physics. I know I'm most likely not going to be able to solve it but still, I'm trying. Your way of teaching things is absolutely top notch. Again, I probably won't get this right but still, I'm tempted to try.

[@lecturesbywalterlewin.they9259]

:)

[@icefreezer7]

i believe this is the same question as asking what the period is of a shorter/longer pendulum, so i think the period would stay the same since the kinetic/potential energies would balance across any length of pendulum

[@lecturesbywalterlewin.they9259]

>>> i think the period would stay the same since the kinetic/potential energies would balance across any length of pendulum>>> I suggest you watch my solution to this problem

[@bios546]

Awesome! Sir you look so great in this video! I can feel the excitement from this side of the screen! my day has become so much more better :D :D :D :D :D !!! This is a lovely problem!!!! I did get the solution... but I'll just say : SPOILER ALERT!!!! Answer: from the point where you held the rod, there will be three other points away from the point, distanced: 1) L/3 2) 2L/3 3) L (for (2) and (3), the experiment must be repeated by turning the rod upside down) ((1) and (2) divide the rod into three similar sections!!! woaw!!! never noticed that... ) My explaination: Intuitively, we can say that the time period values, given the distance of the axis of rotation from the end of the rod gives values that are continuously changing with different distances... Now, for a point very far away (in theory only, as g varies) and for the point coinciding with the centre of mass, the time period is large enough that its value will tend to infinity... as we move the axis closer in from the indefinitely far point or away from the centre, there will be a decrease in time period values... so, as the values of time period vary in a continuous fashion, and are decreasing when the axis shifts from centre outward and from a far away point inward, there must be a minima somewhere... which also means, as long as the point is not a minima, and the time period corresponding to such a point is less than, or equal to the time period value obtained at the end, there will be four points on the rod for which the time periods will be equal. This will consist of two pairs of points which are equidistant from the centre of mass. Also, from the math, the points which lie within L/3 and 2L/3 away from the initial axis point, the other points where the time periods will be equal will lie outside the rod. The final result for time period came out to be (for a rigid pendulum with uniform mass distribution) T = 2 pi sqrt( ( l + k^2/l )/g ) where, l = distance of centre of mass from the axis of rotation, k = moment of inertia about the centre of mass, parallel to the axis of rotation / mass of the body. mathematically, the term: l + k^2/l must equal 2L/3 this will give a quadratic equation whose roots will be L/2 and L/6... which means, the axis of rotation must be L/2 and L/6 away from the centre. L/2 corresponds to one end of the rod and the other end, L/6 corresponds to points L/3 and 2L/3 away from that end.

[@bgdg323]

The oscillation speed would be the same at all points above the center of gravity because the length and mass ratio relative to the pivot point would be the same.

[@kankagoswami4267]

Respected Professor, The time period will remain same for all values of 'x'

[@lecturesbywalterlewin.they9259]

>>>The time period will remain same for all values of 'x' >> incorrect

[@benjaminrousseaux806]

Hi Walter. I've come to an answer to your problem and here's how I did: We have to calculate both the moment of inertia I and the torque tau with respect to point P, with an integral over the lengths of the two segments on each side of P. Also for this we need the mass per unit length rho = M/L. I find I = (M/3L)*(x^3 + (L - x)^3) ; tau = (Mg/2L)*(x^2 + (L - x)^2)*sin(b). Using I*alpha + tau = 0 and sin(b) approx b, we then find the eqn of motion on the angle b: d^2b/dt^2 + C*b = 0. Then the period appears: T = 2*pi/sqrt(C) = 2*pi*sqrt[(2/3g)*(x^3 + (L - x)^3)/(x^2 + (L - x)^2)]. If we choose x = 0 or x = L, we find the solution to the previous problem. I don't know if that's correct, but that was fun to do ! Thank you :)

[@lecturesbywalterlewin.they9259]

If you decided that there is no value of x between 0 and L/2 which has the same period as at x=0. that's not correct

[@leonardoandreae6143]

Using angular momentum principle w.r.t. point P we get: I_P*\ddot(b) + Mg*(L/2-x)sin(b) = 0 where I_P is the moment of inertia of the bar w.r.t. point P: I_P = 1/12 ML^2 + M(L/2-x)^2 = M*(L^2/3 - Lx+x^2) using small angles approx we can rewrite the first equation as: \ddot(b) + g*(L/2-x)/(L^2/3 - Lx+x^2)*b = 0 This is an harmonic oscillator with period: T_new =2 \pi * \sqrt(L^2/3 - Lx+x^2/(g*(L/2-x))) equating this to the old period T we get a quadratic equation in x: x^2+x*L-2/3L^2 = 0 The only solution that satisfies the requirement x>0, x

[@lecturesbywalterlewin.they9259]

INCORRECT

[@MrTheAlaka]

There are no points on the rod where the period is the same, though I'm curious what the negative solution means. As before, we can create an equation of the form d^2b/dt^2 +C*b = 0. By equating C_old = C_new we get an equation in x (or x/L) that gives us the points where the period is the same. Starting from I*alfa + tau = 0 we need to find the new I and tau for this situation. I can be calculated from the parallell axis theorom as I_new = I_old + M*x^2, so the new I is M*(L^2/3 + x^2). The torque tau can be found as (L-x)/2*Mg - x/2*Mg. By rearranging the equation to find C_new and setting C_new = C_old we get the equation 3*(x/L)^2 = -2(x/L). The solutions to this equation is x/L = 0 and x/L = -2/3. x/L = -2/3 is not a point on the rod itself. Would a rod suspended by a massless string of length 2L/3 have the same period then? The x/L = 0 is of course the same situation as before, but not a new point on the rod

[@lecturesbywalterlewin.they9259]

incorrect answer

[@lizkambas7332]

Im totally in love with your exceptional physics questions...and your style of dress!

[@lecturesbywalterlewin.they9259]

:)

[@shivangisrivastava3152]

love you love you love you soooooooo.... much sir. you are my idol. what I was going to do in physics if I were not there to help. 👏👏👏👏👏👏👏

[@lecturesbywalterlewin.they9259]

:)

[@mathematicalworld7337]

sir thanks for your kind advice, you are the best physics teacher I have encountered so far, you are my idol and I assure you that one day I will make you proud when I will become theoretical physicist

[@lecturesbywalterlewin.they9259]

:)

[@fadiakbik2742]

The point would be at the centre of mass which is 1/12ML^2. It will give the same period

[@lecturesbywalterlewin.they9259]

incorrect

[@srihariification]

at x = 0.45 * length of the rod, the time period of oscillation will be same as that of the one when the rod is made to oscillate at its end.

[@lecturesbywalterlewin.they9259]

incorrect

[@chang.stanley]

There is no point from the end where the period is the same, unless X=L. The period approaches infinity as the point of rotation approaches the center and oscillation stops. So, I assume the period would only increase with distance from the end to the center.

[@lecturesbywalterlewin.they9259]

incorrect

[@chang.stanley]

D:

[@chang.stanley]

Is it, x = (3/8)L ?? I also get L/3, but that's apparenty incorrect.

[@chang.stanley]

Also, I have a bit of confusion I can't figure out. When an electron of an atom absorbs a photon of a specific wave, it can re-emit that photon at the same wavelength. But due to entropy, there should be some inefficiency and loss of energy as heat, since a real reaction is thermodynamically irreversible. Reversible reactions are theoretical and don't actually occur. Photons have no mass and always travel at the same speed so, where is the loss of energy? How come it can re-emit the photon at the same wavelength, seemingly at 100% efficiency. A friend studying physics suggested that entropy is a statistical phenomenon and as such, does not apply to individual fundamental particles. But, he was unsure and drunk at the time. Is this true?

[@chaitanyagiri76]

l/3 is correct answer

[@kushagramisra4429]

Sir.in this problem the point is L/4. solving the differential equation of S.H.M for some value of length[ y],finding moment of inertia by parallel axis theorem & then equating values of time period according to the question we will get y=L/4. THANK YOU! for the problem.

[@lecturesbywalterlewin.they9259]

incorrect

[@kushagramisra4429]

sorry sir there was a calculation mistake in my solution and now I got to the conclusion that there exists no two points on the rod for which time periods are equal as T is proportional to 1/square root of separation between point of axis of rotation and point of center of mass.

[@lecturesbywalterlewin.they9259]

there is a point x>0 where the period is identical to the one at x=0

[@kushagramisra4429]

Sir i again tried to solve the problem and get the equation X=anti log{0.46 log [on the base 10] L} which led to the point that if L=1 m then point where time period=that of previous problem is x=0 and if you increase L then value of x also increases towards C.O.M

[@kot0564]

Is it warm yet in Boston? Do you really need the puffy jacket?

[@SoumilSahu]

I got last week's problem right, so I hope this is correct. Since you mentioned that posting the process would be no fun for others, i will only post the answer: T = 2π √(2l² + 6x² - 6lx) / g(3l - 6x) P. S: I used the same method that was applied for the previous problem along with the parallel axes theorem

[@lecturesbywalterlewin.they9259]

answer my question. x=AL what is A

[@SoumilSahu]

Professor, i think A can be anywhere between 0 and 1/2. If it's greater than 1/2, the motion will not be simple harmonic , since if the rod is turned slightly, the net torque will not be restoring and if displaced a little, the rod will rotate and "fall" (I could be wrong, but this is what I think)

[@jakesky100]

Hi Walter Lewin, I don't have a solution yet but I was wondering: out of all the lectures posted you have done, which one is YOUR favorite? Just curious :D

[@tejved5531]

Sir, please provide some selective problems from book IE IRODOV. It is a wonderful book with some highly complex problems.

[@nishanttiwari1013]

Sir MOI about axis passing through a distance of L/6 from center of mass is Ml^2/9 and by the same formula we will get same time period so answer will be x=L/3

[@mathematicalworld7337]

sir, can the principle of conservation of energy be violated at atomic level as per the Heisenberg principle?.

[@lecturesbywalterlewin.they9259]

HUP: delta(t) x delta(E) > h_bar/2. "In classical mechanics (in the sense of non-quantum) physics, there is no mechanism to allow for non-conservation for energy. Since classicalmechanics is pretty exact for macroscopic objects, there is never any macroscopic violation of conservation of energy.However, for small enough (quantum) systems, we have the Heisenberg uncertainty relation."delta(t) x delta(E) > h_bar/2. "Which means that the uncertainty in energy times the uncertainty in time is greater than some very small number. Interpreted correctly, this means that it is possible to violate conservation of energy given you do it for a very short time, in other words, it is possible to "borrow" energy E from "nowhere", given you return it in a time period t, where t is given byt ~ h/EThis "borrowing" of energy is so small that again by the uncertainty principle, it can never be directly observed. In other words, violation of conservation of energy can occur if and only if the violation can not be observed due to the uncertainty principle."

[@015-n1c]

I forgot all the physics I've learned... but let me try. Let x denote the ratio between the length of the indicated piece, and L. So x is a real number in [0,0.5). On the right of P, mark another segment of length xL, so that the left over has length yL, as in the following: I--(1)--P--(2)--I---(3)---I. Part 1,2 has length xL, 3 has length yL. With 2x+y=1. The the forces on 1 and 2 cancels. We can remove 1, replace 2 by a massless string and obtain an equivalent problem. After applying formulas, we end up solving the equation (x+y)^2-x^2=(x+y)^3-x^3, together with previous constraints, gives x=0. I actually have a question I would like to ask. Suppose the sun disappears. Do we fly off in a straight line? Or do we keep orbiting around for another 8 minutes before we fly off?

[@lecturesbywalterlewin.they9259]

q1 you did not answer my question. q2 If we were the ONLY planet and if there were NO moons, we would fly off in a straight line tangential to our orbital motion.

[@015-n1c]

Lectures by Walter Lewin. They will make you ♥ Physics. Sorry, q1 I meant the only possible answer is 0 so there isn't any values >0. q2 I meant, the earth "knows" what happens 8 light minutes away instantly? Seems very interesting to me. I guess I'll google. Thanks for replying.

[@lecturesbywalterlewin.they9259]

>>>> the only possible answer is 0 so there isn't any values >0>>> INCORRECT

[@lecturesbywalterlewin.they9259]

you are right. the ice cube would have a radial velocity probably much much smaller than its velocity in orbit, but it means that the ice cube will not "escape" the tube tangentially. But it will move in a straight line which will be very very close to tangential.

[@younggunshot5425]

Im having a hard time believing that perpetual motion to make energy isnt possible, whats the best thing someone has came up with

[@aetrew5499]

Hi, Mr. Lewin I enjoy your lecture greatly and they do "make me love physics". Do you have any recommendations for a non-calculus based book of problems in physics? I ask as I am in a highschool/college physics class, however I find the exam problems are always particularly challenging. So I figure I should probably try to practice more of these kinds of problems, however I guess I am just not finding problems on the same level of difficulty present in my exams and wondered if you had any resource recommendations. P.S. You're a great teacher and I always visit your lectures if I ever cannot understand the concept. If you have no advice just tell me to use google haha.

[@icystrangers5482]

Prof, I am curious... serious question: do you "dry run" your videos to make them go smoothly, as you did your lectures? (And how did you manage 50 years in the US and not get an American accent?)

[@lecturesbywalterlewin.they9259]

I tape them sometimes 3-4 times before I am happy with them. The longer they are the more often as I have to tape them as I do not know how to edit a video. I taped the 20 min soln of today's problem 3 times. I will post it next week. Today I taped the next problem which was only 3 min. I accepted my 2 nd version.

[@lecturesbywalterlewin.they9259]

preparation of the writing on paper which I paste on the wall for this week's soln took me many hours as I changed my mind several times to improve on clarity.

[@icystrangers5482]

Thank you, Prof. Your diligence is an inspiration.

[@shaktirajdaudra7285]

Sir, would you like that if I also start uploading a solution video from your next problems on my channel and attached a link in the comment.

[@rsn9394]

Hope it's okay to post the solution publicly. Inertia about the pivot is (mL^2)/12 + m((L/2) - x)^2 use it in the equation as done in the previous problem and find an expression for frequency. equate to freq in previous problem and you get x=5L/6 or x=L/6

[@lecturesbywalterlewin.they9259]

incorrect

[@Mahesh_Shenoy]

Sir you haven't replied to my solution? Is it correct or incorrect? :(

[@mathematicalworld7337]

sir,i think orbital angular momentum is conserved about the Centre of sun as position vector of earth about sun and gravitational force are anti parallel so torque is 0 and hence L is conserved only about Centre of sun but when sun disappears and earth moves with constant velocity L is conserved about every point in space , is it correct?

[@lecturesbywalterlewin.they9259]

STOP this! do your homework on what is angular momentum and when it is conserved. I'll help you a little. 1. why is the orbital angular momentum of the Earth ONLY conserved relative to the sun and not to any other point? 2. If an object is moving with constant speed along a straight line, are there points, P, in space about which the angular momentum of the moving object is conserved. This is my last msg on this topic.

[@abhishekrevinipati4873]

You are wonderful sir!

[@lecturesbywalterlewin.they9259]

:)

[@kot0564]

Is there practical application for this problem?

[@Mr_nuke99]

PROFESSOR I GOT T= 2π [ 2(L^2 + 3x^2 - 3Lx ) / 3Lg ]^1/2 . ONLY FOR x=0 Time period will be same as the previous simple problem. Also x can't be negative.For increasing value of x time period will increase too. Hope this time my expression and intution are correct ,this one is really tough :)YOURS STUDENT :)

[@lecturesbywalterlewin.they9259]

You did not answer my question.

[@mathematicalworld7337]

sir, can you please suggest me some books for classical mechanics undergraduate course

[@lecturesbywalterlewin.they9259]

search the web. At MIT I used 8.01 Physics Hans C. Ohanian 2nd edition W.W. Norton & Company ISBN 0-393-95748-9 8.02 Physics for Scientists & Engineers by Douglas C. Giancoli. Prentice Hall ISBN 0-13-021517-18 8.03 Vibrations and Waves by Anthony French CRC Press ISBN 9780748744473 8.03 Electromagnetic Vibrations, Waves and Radiation by Bekefi and Barrett. The MIT Press ISBN 0-262-52047-8

[@utkarshraswant8394]

When u roll a cone on an inclined plane then the cone turns toward the point of least radius. This can be explained using centre of mass.but can we explain it by imagining that the atoms in the cone are in form of circular disks the radius of which keeps decreasing. Now the no of rotations of each disk remains same.So the velocity of each disk is also different but a force is present that keeps the different disks together and thus a net force vector is produced. Is this model right? I am extremely sorry for putting this here.i know its ot related but didn't know how to reach u.

[@sujithatesh2471]

i have a doubt sir, can gravity attract electricity??

[@lecturesbywalterlewin.they9259]

ask google

[@daanhoogers2000]

there is no point x, for 0

[@lecturesbywalterlewin.they9259]

incorrect

[@russellbrooks5073]

I wish my uni followed the same structure that you teach! keep the problems coming! :)

[@harshshah9003]

there is no such position between end of the rod and centre of mass for which the period will be the same . Because when I solved it properly I got x=0 and x=L which means it only happens at end points of the rod.

[@lecturesbywalterlewin.they9259]

INCORRECT

[@mathematicalworld7337]

sir, if during rolling motion, point in contact with ground has zero speed instantaneouly and friction does no work then why in daily life rolling object come to a halt?

[@HimanshuKumar-dv6nb]

sir will there be any magnetic force on two charges moving with same velocity if yes then what is the formula of that force

[@lecturesbywalterlewin.they9259]

use google

[@chotirawee]

sir, this is a tough one for me. how difficult is this compare to 8.01 exam at MIT?

[@lecturesbywalterlewin.they9259]

this is covered during the first month for freshman when they enter MIT.

[@lecturesbywalterlewin.they9259]

This could be one problem (out of 5 problems) in one of my FOUR exams. My 8.01 exams (+ solutions) are posted as pdf files below the video thumb nails.

[@chotirawee]

When I watched your solution video I just noticed that I forgot the minus sign between (L/2) and (x) !!! I'll be more carefull next problem. Hahaha :D

[@mathematicalworld7337]

sir, does iota has real significance in physics or it is just mathematical construct.

[@ashutoshsahu9979]

sir how many correct answers you got till now??

[@lecturesbywalterlewin.they9259]

about 35 correct about 125 incorrect

[@yath3681]

In what ?

[@mathematicalworld7337]

thank you sir

[@shreyanshpratapsingh8572]

Sir, I have a question in my mind, If we consider a Bohr-hydrogen atom then it's only electron has quantised energy levels in which electron can exist and that's why it can not go into the nucleus by coulomb interaction, the electron has to be in any one of the principle quantum number and no where in between. Bohr's theory well explained the atomic spectrum , i am also satisfied by his theory upto 90% but the question is that: in quantum mechanical model of atom,we have orbitals ,let us assume a p(x) orbital then according to it the electron can be anywhere in that dum bell shaped p orbital (with a high probability) , so why is the electron not falling into the nucleus and why it keeps on moving in p orbital regardless of any force except the coulombic force ( say at t=7sec, we made the measurement of electron's velocity in particular direction and its position) OR the electron is just like a human being moving by itself without any external force

[@lecturesbywalterlewin.they9259]

use google

[@armybrat6915]

Heisenberg uncertainty principle tells us that speed of electron will be higher that light in nucleus but this not possible so electron cannot fall into the nucleus

[@lecturesbywalterlewin.they9259]

>>>>Heisenberg uncertainty principle tells us that speed of electron will be higher that light>>>> that is incorrect. The size of an atom is about 10^(-10)m. If you use HUP with delta_x of about 10^(-10) m, you will find an approx value of delta_p. Since you know the mass of the electron you can then get a handle on the speed of an electron in an atom. Using It's much smaller than c.

[@armybrat6915]

but radius of nucleus is of the order of 10^(-15) so if it is confined to nucleus than speed comes out to be approximately 2.8* 10^10 which is greater than c.

[@lecturesbywalterlewin.they9259]

electrons are NOT confined to the nucleus. They are confined to the size of the atom.

[@abhishekchopane3996]

Hellooo sirr ! Greetings from India ! :) Well I think there is no solution for x as the time period of the oscillation will be constantly increasing as one increases x ! at x=l/2 the net torque acting on the rod becomes 0 hence making the time period infinite ! As one increases x the net torque decreases so thinking in this manner I came to the conclusion that the time period must be unique at every value of x! Did try solving using equations but got stuck with x^2 term ! Then the aha moment struck and the above thought came to my mind ! I hope I was right ! Eager to hear from you professor ! :)

[@lecturesbywalterlewin.they9259]

>>>>Well I think there is no solution for x as the time period of the oscillation will be constantly increasing as one increases x >>>>> INCORRECT

[@sharmajalandhar5194]

@@lecturesbywalterlewin.they9259 u

[@hseverins]

Excuus mijn domheid, maar mij lijkt het logisch dat jouw vorige uitleg gespiegeld is aan de bovenkant . Misschien een waardeloze uitleg , maar ik heb helaas maar LTS, MTS gedaan 😘🤔😉..... Ps gezien dat we wat contact hadden , zeg ik dat mijn dochter vandaag 10 jaar is geworden ....... Het mooiste wat ik heb ! Fijne avond Walter

[@lecturesbywalterlewin.they9259]

feliciteer je dochter namens mij

[@mastersubhadeep]

Do we need to worry about elliptic functions for solving this problem?

[@siddharthamishra1999]

no

[@AMotoVlogger]

There is absolutely no way that the period of oscillation can equate the same amount of time as in the first problem. UNLESS you are skilled in the art of mind jitsu. I>. ANY given point on the rod is a point of symmetry (Sym. A and Sym. B, respectively). Since we are dealing with time of oscillation of SIMPLE pendulums, the most efficient point (ie: the point which will convert the MOST potential energy into kinetic energy) is Sym. A-- from the first experiment. 2>. Any change in symmetry demands a change in angle B, so that potential energy can be converted (and maintained) into highly efficient kinetic energy. Yes, you could do it, but you would have to find the angle proportional to the first angle used. That is some serious math that I, for one, would scarcely endeavor to take on. . . or you could just turn the rod upside down since mass is constant. duh

[@lecturesbywalterlewin.they9259]

>>>>There is absolutely no way that the period of oscillation can equate the same amount of time as in the first problem.>>>> INCORRECT

[@jacobprince3573]

Professor, I have been thinking for years about a rotational problem that you might care to comment upon. If a piece of ice (ie near frictionless) is placed in a straight tube and swung in a horizontal circle it will emerge radially from the end of the tube. Since it was a frictionless operation there cannot have been any radial reaction on the tube, so where does the centripetal force come from to make the ice move outwards?

[@lecturesbywalterlewin.they9259]

In the rotating frame of the ice cube there is a centrifugal force (it's called a ghost force). This force will move the object radially with an acceleration V^2/R. R will increase as the object moves away from the center. When it reaches the end of the tube it will fly out tangentially (NOT radially).

[@jacobprince3573]

Thank you for the reply. The surprise for me there is that it will come out tangentially. Unlike the situation where a centripetal force has confined an object in a circular path and is suddenly let go to carry on with the speed and direction it last had, in this case the ice also has a radial velocity when it comes out. Perhaps I should try to perform an experiment to video the situation and see what happens?

[@baldpolnareff7224]

Is the Pivot frictionless? That would really simplify the problem!

[@lecturesbywalterlewin.they9259]

yes that is assumed

[@baldpolnareff7224]

Thanks!

[@operatorenabla8398]

First of all, we want to find the formula for the period with the new "pin". J is the torque I is the moment of inertia at a distance x from the top of the rod with a ' I mean a time derivative (so '' is a second time derivative) w is b' I_CM is the moment of inertia of the rod spinning with an axis passing through the center of mass and ortogonal to the rod ^ is the cross product I'll write vectors in line but think of them as if they were written in column My origin will be the spin point p is pi We have J=Ib'' First of all, let's find the new I. We can use the parallel axis theorem: I = I_CM + Mx^2 = M(1/12*L^2 + x^2) (I'll keep calling it I for a while for simplicity) The Torque will be r ^ F = ((L/2-x)sin(b) ; (-L/2+x)cos(b) ; 0) ^ (0 ; -Mg ; 0), which leads to J = (0 ; 0 ; Mg(-L/2+x)sinb). Let's drop the vector notation to solve J=Ib'' Mg(-L/2+x)sinb - Ib'' = 0 -> using small angles approximation: Mg(L/2-x)b + Ib'' = 0 which has the form of an oscillator's equation, infact: b'' + (Mg(L/2-x)/I)b = 0 which implies w^2 = Mg(L/2-x)/I -> T = 2p*sqrt[I/(Mg(L/2-x))] -> T = 2p*sqrt[(L^2+12x^2)/(6g*(L-2x)] Let's now equal it to the period of last week's problem, which was 2p*sqrt[2L/(3g)] we know from doing so that x =/= L/2 in order for the solution to be acceptable (which makes sense since the torque would equal 0 in that case) solving it, we end up with a second order equation: 12x^2 + 8Lx - 3L^2 = 0. Now we can solve for x, we end up with two solutions: x_1= L*(sqrt(13)-2)/6 and x_2 = -L*(sqrt(13)+2) Hope I got it right

[@lecturesbywalterlewin.they9259]

incorrect

[@operatorenabla8398]

Is the new answer correct?

[@mathematicalworld7337]

I am not able to understand ,why it is called a transfer of resistor

[@mathematicalworld7337]

sir,as liquid like water can't bear shearing stress then how surface waves(transverse in nature - requiring shearing property of medium) are possible on surface of water?

[@lecturesbywalterlewin.they9259]

use google

[@kaustuvmohapatra114]

would it be (sqrt(13)-1)L/6 ??

[@lecturesbywalterlewin.they9259]

incorrect

[@googavo1d]

Would it be wrong to ask for measurable parameters of that rod and conduct measuring experiment like 10 times then just give an answer , then perhaps do the same thing for 100 more rods and make my own equation. If i have the brains i will come up with next best thing after the book equation if i dont il end up in static results like 2.4 sec doh. How hard can it be to measure time using my iPhone, perhaps an app with proximity sensor(which probably uses equations alike) and all that cool stuff that Newtons didnt have back there.

[@howtosport425]

Sir isn't it that the distance of x has the same amount of mass on either side of the pendulum rather than a pendulum balanced at the top is uneven and unweighted too. Therefore the inertia and centre of mass is then uneven and therefore oscillates at t and a more consistent rate. Then when it is halved it has a 50/50 mass on both sides of the pendulum therefore the centre or mass ad inertia is uneven and bobs. But so long as the angle plus the momentum and force of the swing creates the same outcome as it is swung at the top. The constant motion of force given creating momentum and centre offs circling the pendulum causing a force of Lmg 10T 26sec 10T (Middle mass) 25seconds the halving of the pendulum splits the mass and therefore the force and inertia creating a movement swinging slower as mass is halved (I could've done the equation and solution but didn't have time)

[@lecturesbywalterlewin.they9259]

what is your answer to the problem?

[@howtosport425]

+Lectures by Walter Lewin. They will make you ♥ Physics. My answer to the problem is that either way only depending on the even weight and the angle it swings from it will oscillate the exact same if those attributes are not met than it will cause the frictional force caused by amplitude to bob. Plus the fact that wherever p is placed is vital for it to oscillate PLUS!!! The amount of centre of mass coming from that push is also vital too. Like I said the angle in which it is pushed from, without a certain force added to it, it will oscillate perfectly. I could've done the equation but have to go to school and didn't have enough time. PS I'm only 12 turning 13 in a fee weeks time

[@howtosport425]

Lectures by Walter Lewin. They will make you ♥ Physics. also I am 12 years old too

[@affanelahi6163]

Could you be so kind as to explain why a top spins as it does

[@lecturesbywalterlewin.they9259]

use google

[@affanelahi6163]

Thanx

[@mathematicalworld7337]

sir, why we use complex wave function ,not a sinusoidal real function in quantum mechanics?

[@lecturesbywalterlewin.they9259]

ask google

[@mathematicalworld7337]

sir, is it correct to say that magnetic force act as a transferring agent in transferring mechanical energy to heat in resistor when we produce motional emf?

[@lecturesbywalterlewin.they9259]

I would not put it that way.

[@shivangisrivastava3152]

Sir I'm very weak at calculations.... PlZ suggest me something to cope with it.. and this problem has created a barrier between me and Physics ??????

[@lecturesbywalterlewin.they9259]

Watch my 8.01 lectures as a start. They will probably remove the barrier.

[@shivangisrivastava3152]

definitely I'm gonna try it... thank you so much sir. 😊

[@lecturesbywalterlewin.they9259]

most of what you need is covered in lect #21 of 8.01. But unless you have a good grip on torques and moments of inertia, that lecture alone is not enough.

[@shivangisrivastava3152]

Lectures by Walter Lewin. They will make you ♥ Physics. ......okkkk. then first of all I'll try the former one.. then I'll go for the rest.

[@lecturesbywalterlewin.they9259]

Yes do watch the solution to the first MUCH easier problem that will help. But that alone may not be enough. good luck!

[@mathematicalworld7337]

thank you sir for such a good explanation

[@safihalim3747]

by any chance does it happen to be correct? x=[(2/3)L]÷[1-(2/3)L]

[@lecturesbywalterlewin.they9259]

incorrect also dimensionally incorrect

[@mathematicalworld7337]

sir, is the magnetic field does negative work in motional emf concept?

[@lecturesbywalterlewin.they9259]

magnetic fields can do NO work. Use google.

[@psobocik]

hello, on my logic each body hanged above its center of mass should work as physic pendulum. from previous problem we can build universal equation for the body, then compare the parts l/Mx=2/3L then we get L/2 >=I3/2LM>0 And so we see that in 0 which is the center of mass there is no solution couse there is no pendulum no more and the period is the same for x=[L/2÷0). am i right Sir?

[@lecturesbywalterlewin.they9259]

not correct

[@sandeepmahuja]

Dear sir , I need one advise . I am currently studying in 11th grade in India .I started reading physics after qualifying for INMO(Indian National Mathematical Olympiad, which is third round for IMO) in class 8th .I have completed the entire high school physics course upto the level of IPHO (International Physics Olympiad) and have read many books which are easily available in my country including H.C.Verma ,University physics , David Morin , book by Klepner and Kolenkow and electrodynamics book by Griffiths .I am also a fan of your videos without which I think moving forward in physics would have been impossible . My current interests are Quantum physics (I am reading Griffiths for it) and relativistic mechanics. Many summer camps will taking place this summer and I want to join one of them . Can you suggest me one that can cater to my interests .I tried to apply in RSI (conducted by MIT) but could not , because this year there were no sponsors from India for the course , I belong to a middle class family and could not afford to pay the entire course fee. You may reply at my e-mail address: [e-mail] or [e-mail] With Deep Regards Rishit

[@lecturesbywalterlewin.they9259]

I am unfamiliar with summer camps

[@mathematicalworld7337]

sir, when a stationary bomb explodes its Centre of mass does not move as no external force acts on it but kinetic energy changes , but according to work energy theorem this implies there is an external force so there is a contradiction, please solve this problem

[@lecturesbywalterlewin.they9259]

no contradiction. work energy theorem applies ONLY to external forces. Since there are NONE the KE of CM does not change.

[@mathematicalworld7337]

sir , which is your favorite physics book

[@lecturesbywalterlewin.they9259]

"For the Love of Physics"

[@mathematicalworld7337]

sir, if magnetic field can do no work then it must be a conservative force ,as per the definition of conservative force,but line integral of magnetic field is not zero always?

[@lecturesbywalterlewin.they9259]

The magnetic field itself is neither conservative nor non-conservative. Magnetic field lines do go in closed paths but that's not the definition of conservative. Rather, a field is conservative when the force on a test particle moving around any closed path does no net work.

[@mathematicalworld7337]

thank you sir for your precise explanation.

[@Mr_nuke99]

Professor I think that there are no points b/w end ade center of mass and end for which the time period is same as previous problem.

[@lecturesbywalterlewin.they9259]

incorrect

[@Yasinzaii]

love you sir 💓

[@lecturesbywalterlewin.they9259]

:)

[@mathematicalworld7337]

thank you, sir

[@ScientyProf]

Thanks you fot this. J'espère rester comme vous dans mes cours :)

[@sangeetadarpel5153]

no such point is possible .....the quadratic equation that is formed has imaginary roots.

[@lecturesbywalterlewin.they9259]

incorrect

[@shauryajaggi1951]

x=(2/3)L...I've solved it on a paper ..how do I show you the solution? ?

[@lecturesbywalterlewin.they9259]

INCORRECT

[@SupathiRatnayake]

Sir, I have seen a physics toy, Fan blade is pinned on a edge of a stick (like a chopstick) And one side of the stick is corrugated. It is rubbed up and down with another stick. The fan blade always rotates anticlockwise due to vibration energy. Do you know that toy? If it is What's the name of it?

[@lecturesbywalterlewin.they9259]

use google

[@SupathiRatnayake]

But how to google all that information? I tried.

[@Gruuvin1]

google "physics toy rub stick with ridges"

[@Gruuvin1]

by the way, it's a "Gee-haw whammy diddle" doesn't everyone know that? (kidding)

[@SupathiRatnayake]

Thanks!

[@avisingh8830]

Two points at x=0 and x=L\3

[@mathematicalworld7337]

sir , I love beauty mathematics as much as physics so can you please advice some books on mathematics for undergraduate course.

[@lecturesbywalterlewin.they9259]

search the web

[@mathematicalworld7337]

sir, I have problem in which a circular region of radius R is given ,in which B field is increasing at a constant rate and we have to plot E vs r graph so I applied Faraday's law and found that E increases linearly up to R but I can't find E field for r greater than R , sir,please suggest a solution .

[@lecturesbywalterlewin.they9259]

watch my 8.02 lectures.

[@Csky1988]

this is way beyond me but i love your responses in the comments. you're having a hay day with this one. haha

[@lecturesbywalterlewin.they9259]

:)

[@mathematicalworld7337]

sir, are you always available here to take queries of your students

[@lecturesbywalterlewin.they9259]

yes - one question at a time and make them BRIEF

[@mathematicalworld7337]

sir, if I open the door of my refrigerator then does the overall temp. of surrounding increase or decrease ,i am struggling with this question since 2 weeks, sir, please suggest a suitable explanation.

[@lecturesbywalterlewin.they9259]

a refrigerator needs power. That means it produces heat. A refrigerator heats up your home. You cannot ONLY lower the temperature inside the refrigerator. That would be a violation of a the fundamentals of Thermodynamics. Reply ·

[@ganymedkallisto5561]

Another interesting question: for which x does the pendulum have the smallest period time?

[@lecturesbywalterlewin.they9259]

have you found x for which T is the smallest value? And what is T_min? I have already taped it as part of my solution which I will post next week.

[@ganymedkallisto5561]

No, i haven't yet. Today i only solved for the x which gives the same period as the "simple" pendulum (hope that it was correct). Perhaps i will try to find the minimum period and the corresponding x tomorrow. Best Regards.

[@ganymedkallisto5561]

i found a solution for the smallest period: T_min = 2π * sqrt(L/(g*sqrt(3))) and the corresponding x is L/2 * (1 - 1/sqrt(3)) Best Regards.

[@lecturesbywalterlewin.they9259]

correct

[@ganymedkallisto5561]

Thank you. :-) Another interesting problem with this pendulum is the following: Imagine a rigid stopper, which is located somewhere vertically under the pivot point. The pivot point is at the very end again (simple physical pendulum). Now you can lift the pendulum and let it go. It will swing until it hits the stopper hard (an eccentric collision occours). During the collision there will be a horizontal force acting on the pivot point of the pendulum. The question is: are there any special locations for the stopper, so that there is no horizontal force acting on the pivot point? If so, where are these locations measured verticaly down from the pivot point? Best Regards.

[@rachydamine9458]

Is there any way to contact you with my answer since I can't type the whole thing here ? Thanks :)

[@rachydamine9458]

I can send you my demonstration professor if you want it

[@armybrat6915]

Sir, how should I send you my solution?

[@lecturesbywalterlewin.they9259]

write it here

[@armybrat6915]

okay sir

[@shaktirajdaudra7285]

Sir, I want to buy a book "For the love of Physics" but when I checked online it is the one with light brown, or nearly about yellowish color cover. So, is that original ? (I am asking this because I had seen one with black cover)

[@lecturesbywalterlewin.they9259]

I can sell you a copy of my book for $13 + postage. postage will depend on where you live. Where do you live? If you transfer the money to my Paypal account and let me know your full name and address I will send you a signed copy with a special note for you written in the book. I will also include a signed colored picture of me swinging on the pendulum (with your name written on it). Keep in mind that Paypal charges 6% for every currency transition. pay pal: [e-mail]

[@shaktirajdaudra7285]

Prof. I live in India (Gujarat), will you please give me a Rough Figure of Money including postage. Then , I will ask to my dad to have one

[@lecturesbywalterlewin.they9259]

INDIA I can sell you a copy of my book for $13 + $30 postage =$43. If you transfer the money to my Paypal account and let me know your full name and address I will send you a signed copy with a special note for you written in the book. I will also include a signed colored picture of me swinging on the pendulum (with your name written on it). Keep in mind that Paypal charges 6% for every currency transition. 1.06*43=$46 pay pal: [e-mail]

[@shaktirajdaudra7285]

Sorry Professor, I need some time to convince my parents, I will soon buy it on my 17th birthday. I am sure they will allow me. So will you please wait for few months, I am really interested to read your book.

[@mathematicalworld7337]

sir, is string theory is the theory of everything

[@lecturesbywalterlewin.they9259]

use google

[@jatinbhatt7826]

professor i got very complicated answer. if u consider its- -(1-3/2L)+or- sqrt of {(1-3/2L)^2+4L (1-L). It's the final expression i couldn't find more simple form of this :(

[@lecturesbywalterlewin.they9259]

I need the know what x is X=A*L. Tell me what A is.

[@lecturesbywalterlewin.they9259]

your new answer for A is correct

[@jatinbhatt7826]

+Lectures by Walter Lewin. They will make you ♥ Physics. thanks a lot sir.you have no idea how happy i am right now. At first i miss a term that's the reason i got some giant answer. but then i cross check then finally i found that. love u professor, all credit goes to you :-*

[@mathematicalworld7337]

sir, when a pendulum of length L and with bob of mass M exhibit shm then how can I find the maximum tension in string during its motion, sir, please give me some hints.

[@lecturesbywalterlewin.they9259]

watch my 8.01 Lectures. !! You will then find the answer.

[@michabronicki1367]

Sir, can I email you the answer? Is there a chance that you will reply if it is correct?

[@lecturesbywalterlewin.they9259]

Send it here. I will not post it if it is correct

[@baldpolnareff7224]

I would love to send you the solution I found, but I can't attach the file into KZbin's private message, is there a mail I can send it to?

[@lecturesbywalterlewin.they9259]

you can ONLY post it here. If your answer is correct I will not post it thus no one will see it.

[@mathematicalworld7337]

sir, can you explain the working of a transistor

[@lecturesbywalterlewin.they9259]

use google

[@baldpolnareff7224]

Where can I post the solution?

[@lecturesbywalterlewin.they9259]

send it to me. If your soln is correct I will not post it.

[@tahamimarlar2141]

sir are the questions in university exams comes like this or harder maybe ?

[@lecturesbywalterlewin.they9259]

any freshman at any University who is taking an introductory course in classical mechanics should be able to solve this.

[@mathematicalworld7337]

sir, why is it so that every element has different properties but nucleus of these element have same density?

[@lecturesbywalterlewin.they9259]

the number of protons in the nucleus determine what element they are. 2 protons He, 3 p LI, 4 Be, 5 B etc

[@yash29210]

How would you rate it on a scale of 0 to 10?

[@lecturesbywalterlewin.they9259]

rate what?

[@yash29210]

this physics problem.....

[@lecturesbywalterlewin.they9259]

it's a great problem. So far my best. grade 9

[@mathematicalworld7337]

sir, you told in your lecture video that L is conserved only about Centre of sun so I think L will remain conserved about this point only.

[@lecturesbywalterlewin.they9259]

you did not understand why the orbital angular momentum of the Earth ONLY relative to the Sun is conserved. GO over that again. When the sun is gone, the Earth (if there are no other planets and moons) will "take off" in a straight line with constant speed. So your question is: "If an object moves with constant speen along a straight line is angular momentum conserved and is the answer is YES about which points is it conserved? ANSWER THAT ONE!

[@mathematicalworld7337]

sir,if sun were to disappear then L will remain conserved or not?

[@lecturesbywalterlewin.they9259]

L about which point?