Zak's Lab
Zak's Lab
  • Видео 1 469
  • Просмотров 2 851 089
How to make a phase space plot for the simple harmonic oscillator.
Given the mass, spring constant and amplitude of a simple harmonic oscillator, we construct a phase space plot for the simple harmonic oscillator.
The phase space plot for a harmonic oscillator is a graph of velocity vs. position, and the oscillator traces out a trajectory in phase space (ordered pairs consisting of position and velocity) as it evolves in time. Taking a look at the expression for the total energy of a harmonic oscillator: 1/2mv^2+1/2kx^2=E, we see that the phase space trajectory is an ellipse.
To plot this ellipse in phase space, we only need to find the extreme values of position and velocity. The extreme values of position correspond to the amplitude of the oscillator an...
Просмотров: 435

Видео

Find the speed of a harmonic oscillator at a given position.
Просмотров 129Месяц назад
Given the mass, spring constant and amplitude of a simple harmonic oscillator, we find the speed of the oscillator at a specific position of x=10cm. This is an energy conservation problem, where the given spring constant and amplitude allow us to find the total initial energy as 1/2kA^2. The final total energy of the oscillator has both kinetic and potential energy components, 1/2kx^2 and 1/2mv...
Find the amplitude and maximum acceleration of a simple harmonic oscillator.
Просмотров 76Месяц назад
Given the speed of a simple harmonic oscillator as it passes through equilibrium, we find the amplitude and maximum acceleration of the harmonic oscillator. Finding the amplitude of the oscillator is an energy conservation problem: in the initial state, all the energy is kinetic because the spring is at its relaxed length in the initial state of the system. In the final state, all the energy is...
Find the maximum speed of a pendulum, the period and the time to get to maximum speed.
Просмотров 204Месяц назад
Given the length of a simple pendulum released from rest at a given small angle, we compute the time it takes for the pendulum bob to reach its maximum speed, then we find the maximum speed of the pendulum bob. The maximum speed of the pendulum occurs at the lowest point on its path which takes a time of one quarter period, so we use the period formula for a pendulum: T=2pi*sqrt(L/g) and divide...
Find the spring constant given the position time graph for a simple harmonic oscillator.
Просмотров 100Месяц назад
Given the position time graph for a simple harmonic oscillator and given the mass of the oscillator, we calculate the spring constant by using the period formula for a spring and mass system T=2pi*sqrt(m/k). Along the way, we use the position time graph for the simple harmonic oscillator to determine the period and frequency of oscillations and the amplitude of oscillations, then we use the kno...
Converting frequency to period for a sound wave (middle C or C4).
Просмотров 74Месяц назад
Given the frequency of a musical note (middle C or C4 in this case), we compute the period of oscillation of the ear drum when the sound reaches our ears. The relationship between period and frequency is that they are reciprocals of each other (frequency has units of oscillations per second or 1/s or Hz, while period has units of seconds per oscillation or s). So, T=1/f, and we sub in the frequ...
Period of a physical pendulum: uniform disk drilled off-center.
Просмотров 356Месяц назад
We compute the period of oscillation of a uniform disk drilled off-center and used as a physical pendulum. To begin, we bring in three results from previous videos: the period of oscillation of a physical pendulum, the parallel axis theorem, and the moment of inertia of a disk rotating about its center of mass (axis perpendicular to the plane of the disk). We put it all together to find a symbo...
Period of oscillation for a thin ring physical pendulum.
Просмотров 218Месяц назад
Given a ring of mass M and radius R, we use our previous result for the period of oscillation of a physical pendulum (T=2pi*sqrt(I/mgR)) and the parallel axis theorem (I_parallel = I_cm MR^2) to calculate the period of oscillation for a thin ring physical pendulum. The moment of inertia of the thin ring about the edge is found using the parallel axis theorem, so I_parallel=I_cm MR^2=2MR^2 when ...
Atwood machine with a heavy pulley: find acceleration and angular displacement.
Просмотров 622Месяц назад
Given an Atwood machine with a heavy pully, we find the acceleration by using a force and torque approach to the problem. So, we make a detailed force diagram for the mass and pulley system, including the force of gravity and tension on each hanging mass, and the force exerted by each side of the string on the heavy disk pulley. We apply Newton's second law F=ma and the rotational analog of New...
Solve elastic collisions using the center of mass frame trick!
Просмотров 478Месяц назад
Two carts initially head toward each other and collide in a perfectly elastic one-dimensional collision, meaning both momentum and energy are conserved. This time, we approach the problem by switching to the center of mass reference frame. First, we find the center of mass velocity of the system by using a theorem of classical mechanics: p_net=Mv_cm. We solve for v_cm and calculate the net mome...
Find the Laplace transform of the sine function (Laplace transform of sin(t)).
Просмотров 608Месяц назад
To calculate the Laplace transform of sin(t), we take the integral on 0 to infinity of e^-st*sin(t), where s is a constant with respect to the t integral. This Laplace transform combines improper integrals with the integration by parts looping trick, which made it a perfect bonus problem for my last Calculus II exam! To proceed with the integral, we apply integration by parts, letting u=e^-st a...
Gravitational self-energy of a uniform sphere.
Просмотров 231Месяц назад
Given the mass M and radius R for a uniform sphere, we calculate the gravitational self-energy of a uniform sphere by building up the sphere from the center out using layers of incremental mass dm and incremental thickness dr. We start with a quick review of gravitational potential energy between two masses: U(r)=-GMm/r. Now we build our sphere from the center out, viewing an intermediate momen...
Period of oscillation for a rectangular slab physical pendulum.
Просмотров 293Месяц назад
We bring together several recent results in order to compute the period of oscillation of a uniform rectangular slab drilled near the corner and oscillating as a physical pendulum. The key prior results are: 1. derivation of the period for a physical pendulum: ruclips.net/video/JiG991EnaIo/видео.html 2. the parallel axis theorem: ruclips.net/video/_vjUXz4d1lc/видео.html 3. moment of inertia for...
Moment of inertia of a rectangular slab with rotation axis perpendicular to the slab.
Просмотров 196Месяц назад
We compute the moment of inertia of a uniform rectangular slab of mass M, width a and height b by slicing it into thin rods. This calculation requires the formula we previously derived for the moment of inertia of a thin rod about the center of mass, in addition to applying the parallel axis theorem to find the moment of inertia contribution of each thin slice of the rectangle. We use an integr...
Integral of sin^n(x) reduction formula derived using the integration by parts "looping trick".
Просмотров 629Месяц назад
Integral of sin^n(x) reduction formula derived using the integration by parts "looping trick".
u-substitution combined with integration by parts: integrate cos(sqrt(x))!
Просмотров 176Месяц назад
u-substitution combined with integration by parts: integrate cos(sqrt(x))!
Find the integral of sin^3(x)/sqrt(cos(x)).
Просмотров 343Месяц назад
Find the integral of sin^3(x)/sqrt(cos(x)).
Integral of cos^2(pi*x) on [0,2] + an average value integral trick.
Просмотров 185Месяц назад
Integral of cos^2(pi*x) on [0,2] an average value integral trick.
How to study physics: what my 'A' students said.
Просмотров 698Месяц назад
How to study physics: what my 'A' students said.
Find the spring constant given the maximum compression of a spring.
Просмотров 376Месяц назад
Find the spring constant given the maximum compression of a spring.
Find the final speed using a work integral with F(x)=3sin^2(pi*x).
Просмотров 233Месяц назад
Find the final speed using a work integral with F(x)=3sin^2(pi*x).
This trapezoid sum is undefined - but we can fix it! (integral of sinx/x on [0,pi/2])
Просмотров 289Месяц назад
This trapezoid sum is undefined - but we can fix it! (integral of sinx/x on [0,pi/2])
Partial fractions integral with a cubic denominator using factoring by grouping.
Просмотров 430Месяц назад
Partial fractions integral with a cubic denominator using factoring by grouping.
Rotational kinetic energy of the Earth - is Wikipedia wrong?
Просмотров 328Месяц назад
Rotational kinetic energy of the Earth - is Wikipedia wrong?
Compressed spring between two masses: find the final velocities when the system is released.
Просмотров 343Месяц назад
Compressed spring between two masses: find the final velocities when the system is released.
Maximum speed for a block shot up a rough incline (physics challenge problem!)
Просмотров 4512 месяца назад
Maximum speed for a block shot up a rough incline (physics challenge problem!)
Find the energy required to pump all the water out of a tank + center of mass shortcut.
Просмотров 1732 месяца назад
Find the energy required to pump all the water out of a tank center of mass shortcut.
How to find the minimum escape velocity from a planet, and finding the escape velocity from Earth.
Просмотров 2362 месяца назад
How to find the minimum escape velocity from a planet, and finding the escape velocity from Earth.
Integral with two power functions: integrate (2x^2+3sqrt(x))/x^3.
Просмотров 2322 месяца назад
Integral with two power functions: integrate (2x^2 3sqrt(x))/x^3.
Chain rule backwards vs. u substitution: integral of sec^2(x)/(1+tan(x)) on [0,pi/4]
Просмотров 2172 месяца назад
Chain rule backwards vs. u substitution: integral of sec^2(x)/(1 tan(x)) on [0,pi/4]

Комментарии

  • @tverdyioleksandr7010
    @tverdyioleksandr7010 2 дня назад

    You are amazing! I don't understand why profs in university can't explain all of this that simple as you

    • @ZaksLab
      @ZaksLab 2 дня назад

      Thanks! I do have the benefit of editing, but I've also been teaching for 20 years! z

  • @RealEverythingComputers
    @RealEverythingComputers 4 дня назад

    I had a similar question to this, except that the wedge moved in the opposite direction as the mass slid down (but the mass stuck onto it).... Thus, like your video theta was the wedge's angle of inclination, but due to the wedge going the other way simultaneously, the "true course" of the mass sliding down would have a steeper angle. Is it necessary that the steeper angle and theta must add up to 90? It almost feels intuitively correct, but I'm having difficulty proving it...

    • @ZaksLab
      @ZaksLab 2 дня назад

      I haven't solved that problem in a long time, and I'm not even confident the block would slide on a straight line trajectory during the motion . . . could it be a curved path? If you're just looking for final speed at the bottom, you should be able to get it with energy conservation and conservation of momentum (x direction). I have a note on my desk right now to make a video on that problem, but it will probably have to wait until the next time I teach mechanics (fall '25). z

    • @RealEverythingComputers
      @RealEverythingComputers День назад

      @@ZaksLab Thanks I'll stay tuned - my physics 1 final was just today - would you believe fall '25 is exactly the semester I'll be taking 2nd year mechanics!

  • @worldwidearmy4235
    @worldwidearmy4235 6 дней назад

    Thank you

    • @ZaksLab
      @ZaksLab 2 дня назад

      you're welcome! -z

  • @RajKumartop
    @RajKumartop 7 дней назад

    Tamil la solra

  • @mahsadarvishi
    @mahsadarvishi 10 дней назад

    Thank you so much, I understand everything but I don't really understand why should we derivate acceleration respect to theta, how could this give us the angle that we want? Would you please explain it.

    • @ZaksLab
      @ZaksLab 10 дней назад

      finding minimum or maximum of a function: take the derivative and set it equal to zero. That's going to be the location of a max or min because it's the top of a hill or bottom of a valley where the slope is zero (calculus I optimization stuff), and you can usually infer which based on physical context. z

  • @copernicus6420
    @copernicus6420 10 дней назад

    Wonderful explanation

  • @adrianajaquelingarciapined8036
    @adrianajaquelingarciapined8036 11 дней назад

    Spanish

  • @jdogjones6398
    @jdogjones6398 11 дней назад

    very helpful

  • @friedhegg8874
    @friedhegg8874 12 дней назад

    idk how you dont have more views. thank you for your videos ! subscribed

    • @ZaksLab
      @ZaksLab 12 дней назад

      thanks for your support! -Zak

  • @og-moadjinate5912
    @og-moadjinate5912 12 дней назад

    Thanks for the vid bro

    • @ZaksLab
      @ZaksLab 12 дней назад

      you're welcome! z

  • @Niglnws
    @Niglnws 15 дней назад

    Very nice video. Thanks a lot. I spent much time thinking alone about the kinetics or dynamics of rolling and rolling after rest and rolling after slipping. Nearly no one talk about it or about its intuition. I am trying to figure something in case of rolling after slipping as in your case but there is a motor which keeps applying a force. If the force applied is greater than the max static friction, then it will slip or keep slipping and there will be acceleration at the c.m. so, it keeps rotating faster due to (f.app - f.kinetic) and also the c.m keep going faster due to the (f.app - f.kinetic) so the point of contact is never 0 because the tangential velocity due to rotation is larger than the velocity of c.m. Does that mean that the wheel im this case will not roll. Chatgpt say that after some time, it will roll because angular velocity will decrease, but how will it decrease although the there is torque applied! It say kinetic friction will reduce the angular velocity, but konetic friction is constant. Can you help me?

    • @ZaksLab
      @ZaksLab 14 дней назад

      I think you're right . . . chat gpt still makes a lot of mistakes with unusual physics problems. I write a ton of new problems for my classes every semester, and I try to push the boundaries into problems that are fairly original. If I check my solution against ChatGPT, it often makes several mistakes and needs to be "coached" through the solution, apologizing every time I point out an error. It's still a useful tool, but just try pointing out its mistakes to work toward a solution. z

  • @OctureX
    @OctureX 15 дней назад

    Can you make a video studying the values for a to make the series 1/(ln n)^a converge?

    • @ZaksLab
      @ZaksLab 14 дней назад

      that's an interesting idea! z

  • @entropycore
    @entropycore 17 дней назад

    Sir got me feel sad about nodes

  • @alexacosta7659
    @alexacosta7659 21 день назад

    Thank you, I really liked how you explained what you were thinking and why you made your choices.

    • @ZaksLab
      @ZaksLab 21 день назад

      Glad I could help! z

  • @hungrythundr4892
    @hungrythundr4892 21 день назад

    You just saved me dude. Cramming for an exam and had no idea what to do but this helped a ton. Great video

    • @ZaksLab
      @ZaksLab 21 день назад

      glad I could help! note that if you are interested in creating math animations, you now have the secret to making polar curves spin: just plot r(theta-t), where t is the animation parameter, and you'll get a CCW rotation of the curve. z

  • @subspace1830
    @subspace1830 21 день назад

    Thank you for the wonderful video, you explained this much better than my professor did in about half an hour <3

    • @ZaksLab
      @ZaksLab 21 день назад

      thanks! z

  • @heromask3432
    @heromask3432 22 дня назад

    Conteudo top mano (I know how to speak english btw, nice video man!)

    • @ZaksLab
      @ZaksLab 21 день назад

      thanks! z

  • @Jason-o5s
    @Jason-o5s 23 дня назад

    Cheer~~~situated or extending across something(of a line)😅

  • @haileykk1035
    @haileykk1035 23 дня назад

    Why do u assume the thickness to be equal to arc length instead of straight line in the middle

    • @ZaksLab
      @ZaksLab 23 дня назад

      This is a really good question, and one that bugged me for a long time. I know there are more rigorous ways to argue this, but let's simplify the problem and compare two simpler problems from calculus: finding the volume of a sphere using thin disks vs. finding the surface area of a sphere using thin ribbons. In the volume case, you can get away with using dz for the thickness of each disk, because the tiny amount you are cutting off the curved edge is negligible compared to the volume of the disk (volume of the disk is like pi*r^2dz, but the tiny angled ring we neglect is like 2pir*1/2*dx*dz, which has a product of *two* infinitesimals in it, making it of negligible order). In the surface area using ribbons case, we don't get so lucky: the surface area of a ribbon using dz for thickness is 2pi*r*dz, and the surface area using arc increment is 2pi*r*ds or 2pi*r*sqrt(dx^2+dz^2). This time the error committed is on the same order as the quantity being calculated (one infinitesimal), and that will lead to a finite cumulative error. With experience you just kind of develop a gut instinct for how detailed you need to be in order to not create a finite error -- generally speaking, surface integrals require the extra detail of using ds, and volume integrals can be treated more coursely, using dz for thickness. Try using just a dz for thickness in the surface area of a sphere calculation, and you'll see that it underestimates the total surface area compared to using ds. Hope this helps. z

    • @haileykk1035
      @haileykk1035 22 дня назад

      @ZaksLab yes, it helps. Thank you!!!

  • @MingyaDing
    @MingyaDing 23 дня назад

    thank you so much

    • @ZaksLab
      @ZaksLab 23 дня назад

      you're welcome! z

  • @jeremyred6343
    @jeremyred6343 26 дней назад

    Lol big "D"

  • @ouimonreuf
    @ouimonreuf 26 дней назад

    merci le frère 🔥🔥🔥

  • @JohnVKaravitis
    @JohnVKaravitis 27 дней назад

    The convention for torque is that a counterclockwise motion is a positive torque. And so on.

  • @marcosvanzulli1041
    @marcosvanzulli1041 Месяц назад

    Thanks for the explanation, nice and simple

    • @ZaksLab
      @ZaksLab Месяц назад

      you're welcome! z

  • @user-dr7co8hw4l
    @user-dr7co8hw4l Месяц назад

    I'm an italian guy who loves physics. I've found the best place in the world! Very good explainations

    • @ZaksLab
      @ZaksLab Месяц назад

      thank you! z

  • @ZianLu
    @ZianLu Месяц назад

    good

  • @MUGABO-parfait
    @MUGABO-parfait Месяц назад

    Thanks sir, helpful video

  • @huyviethungnguyen7788
    @huyviethungnguyen7788 Месяц назад

    I'm wondering why can we use conservation of energy in this problem. Isn't a part of them will be taken away by friction?

    • @ZaksLab
      @ZaksLab Месяц назад

      This may not be a very satisfying answer, but friction is not doing net work on the system here. The reason why is that the contact point with the ramp is always stationary, so there is no mechanism for dissipating energy through slipping. Instead, the friction force is always a static friction force. But this force *is* exerting a torque on the system, and in fact it's critical to have this static friction force in order to enforce the rolling without slipping condition. The static friction force is the key to converting the rotational KE of the ball into additional height here: it supplies a torque slowing the rotation down, but simultaneously supplies a force pointing up the ramp. It is a weird thought to have friction acting as a conservative mechanism to convert one type of energy into another, but if you look at the force/torque solution of the problem, it does corroborate the answer given by energy conservation.

    • @huyviethungnguyen7788
      @huyviethungnguyen7788 Месяц назад

      Thank you for your response! My another question is how would we solve the problem if the ball is allowed to both slip and roll?

    • @ZaksLab
      @ZaksLab Месяц назад

      @@huyviethungnguyen7788 that will depend on the angle of incline and the coefficient of kinetic friction. You'd have to do a force/torque approach to the problem instead of the energy approach.

    • @huyviethungnguyen7788
      @huyviethungnguyen7788 Месяц назад

      @@ZaksLab Thanks a lot!

  • @plushy7424
    @plushy7424 Месяц назад

    Is the distance traveled on this for part b only x-distance or total distance on the hypotenuse

    • @ZaksLab
      @ZaksLab Месяц назад

      total distance! We found a_parallel, so we're describing motion along the hypotenuse.

  • @angel6418
    @angel6418 Месяц назад

    Thank you.

    • @ZaksLab
      @ZaksLab Месяц назад

      you're welcome! z

  • @lolsandgiggles111
    @lolsandgiggles111 Месяц назад

    thank you so much you DON"T understand how much this helped me

    • @ZaksLab
      @ZaksLab Месяц назад

      Glad I could help! By the way, I recently updated this kind of problem with better production quality: ruclips.net/video/qTm3LyVVYI8/видео.html

  • @iwishthatiwasjustacloud6105
    @iwishthatiwasjustacloud6105 Месяц назад

    why cant we plug in the initial velocity for the kinetic energy and equate that to the spring energy? shouldnt they be the same since energy is conserved?

    • @ZaksLab
      @ZaksLab Месяц назад

      because in the final state (the moment of max compression) the masses are still moving, so some of the energy will be in the translational kinetic energy of the masses; i.e., not all the initial energy is in the compression of the spring. z

  • @mushrooom3783
    @mushrooom3783 Месяц назад

    YOU'RE AWESOME! been looking up and down for an explaination I can understand. Your teaching logic is extremely helpful❤♥️. Thank you!

    • @ZaksLab
      @ZaksLab Месяц назад

      Glad I could help! z

  • @bloggerboltipublic
    @bloggerboltipublic Месяц назад

    Nice approach thank you

    • @ZaksLab
      @ZaksLab Месяц назад

      You're welcome! z

  • @VarnikaReddyGelella
    @VarnikaReddyGelella Месяц назад

    I searched for nearly 30 min for this kind of video and finally found it thanks a lot

    • @ZaksLab
      @ZaksLab Месяц назад

      Cool! I've got a huge playlist of MOI integrals now because I'm dedicating the last week of my mechanics course purely to moment of inertia integrals and application to the physical pendulum! z

  • @aru-qt9ju
    @aru-qt9ju Месяц назад

    thank you for the clear explanation

    • @ZaksLab
      @ZaksLab Месяц назад

      you're welcome! z

  • @hopeofworld59
    @hopeofworld59 Месяц назад

    How can work be done in a negative since area can never be in negative!???

    • @ZaksLab
      @ZaksLab Месяц назад

      If you're thinking about work as an integral of force over displacement - we can interpret integrals as area bounded by the curve, but when we say 'area' we really mean 'signed area'; i.e, area under the displacement axis counts as negative. Going back to the fundamentals on integrals from first semester calculus, you can imagine slicing the area bounded by a F-x graph into thin vertical strips with height F(x) and width dx (or delta x). Then each incremental contribution to area is just the area of a thin rectangle: F(x)dx or F(x)*delta-x. When F(x) is negative, these areas get a minus sign (they count as negative).

  • @ethanlee3002
    @ethanlee3002 Месяц назад

    You had said in the video that the attachment point of the string was at 80cm in discovering the relevant angle but then proceeded to find the arctan(40/60)?

    • @ZaksLab
      @ZaksLab Месяц назад

      This is a mistake because I was going insane in Fall '20. You're right, and it should be 40/80, or theta = arctan(1/2). I'll add this video to my replace asap list. Thanks. BTW, I spent the last few hours replacing my garbage disposal (third thing to break this week if you don't count me), so that homework thing we talked about is going to take another day. - z

  • @thehorizontries4759
    @thehorizontries4759 Месяц назад

    This is an AWESOME model that explains exactly how it works. The “layers” of the galaxy are roughly elliptical and the interior ellipses spin faster than the outer ones due to keplers laws and then the edges of the ellipses run into each other and this is where gas and dust are pushed together triggering star formation and overdensities of the gas and dust which appear as bulky arms

  • @6AxisSage
    @6AxisSage Месяц назад

    This IS a neat trick and gives me all kinds of new visualiser ideas, I wish i had a fraction of your teaching ability 😅

    • @ZaksLab
      @ZaksLab Месяц назад

      thanks! z

  • @EvanBulatek
    @EvanBulatek Месяц назад

    Thank you for this.

    • @ZaksLab
      @ZaksLab Месяц назад

      you're welcome! quadratic drag next . . .

  • @ToM-ee2ln
    @ToM-ee2ln Месяц назад

    Thank you bro, really appreciate it!

    • @ZaksLab
      @ZaksLab Месяц назад

      you're welcome!

  • @chinmay_0518
    @chinmay_0518 Месяц назад

    😮😮😮😮😮😮

  • @vmened
    @vmened Месяц назад

    Thanks a lot, Zak it was interesting. I have two may be stupid questions. 1) how to determine from which distance from our planet asteroids will be captured by gravity of our planet with final whan occurs landing asteroid on the planet? 2) Lets imagine same situation between two small size black holes (1cm in diameter) in the air of Earth which moves in some random direction with random speed at not crossing routes. Air has visconsity and if i understand physics correctly they will merged together at any distance because gravitation force has no limitation on the distance and their starting speed will be dissipated into gas heating by air drag forces. It that task i suppose ignore gravitation field of Earth for avoid problem of interaction between 3 body. Thanks in advance if you have time to answer thees questions!

  • @LukeyTaylor
    @LukeyTaylor Месяц назад

    If you get the function's height with respect to the horizontal distance traveled (the actual cross-sectional view), then solve the integral for its arc length, then I believe you can generalize the solution to its path length: v^2[sin(θ)+cos^2(θ)ln({1+sin(θ)}/cos(θ))]/g If you want to include an initial height, then it becomes a bit more complicated: {sqrt(v^2sin^2(θ)+2gh)sqrt(v^2+2gh)+v^2[sin(θ)+cos^2(θ)ln({sqrt(v^2sin^2(θ)+2gh)+sqrt(v^2+2gh)}/{v(1-sin(θ))})]}/{2g}

  • @faredoondastur8114
    @faredoondastur8114 Месяц назад

    Thank you! Awesome derivations as always! One question - is the --(i+j) configuration in the acceleration vector function indicative of the centripetal acceleration pointing toward the center of the circle (as opposed to i+j or i-j etc.)?

    • @ZaksLab
      @ZaksLab Месяц назад

      you're welcome -- glad I could help! z

    • @ZaksLab
      @ZaksLab Месяц назад

      by the way -- I updated this video a couple years ago, so here's the good version: ruclips.net/video/iBUnGqWFeJ4/видео.html

    • @faredoondastur8114
      @faredoondastur8114 Месяц назад

      @@ZaksLab Thanks!

  • @davidperezdelafuente1910
    @davidperezdelafuente1910 Месяц назад

    For my case, I have a cylinder (with a hole inside) that is surrounded by a coil, and then I have a smaller cylinder that is made of magnets. The cylinder with magnets is going to move with a continuous veocity. The induced voltage desire is 2.5 V. I will like to know a formula or formulas that could relate the dimensions of the cylinder, the number of turns of the coil, and maybe the properties of the magnet, and obvisuly the velocity of the movement. I have a formula, but i do not know if this formula is correct: E = (μ0·4π ·2M)∙πr∙v∙cos(θ) μ0 = permeability of free space ( 4π∙10−7T∙mA). M = magnetic moment of the magnet (T∙m3 or Amp∙ m2 ) r = radius of the coil( since is a cylinder) ( m ). v = linear velocity ( m/s ) E = induced voltage ( Volts ). In this case a minimum of 2.5 V. θ = perpendicular angle to magnetic field ( rad ).

  • @davidperezdelafuente1910
    @davidperezdelafuente1910 Месяц назад

    Hello. I am looking for some information about electromagnetic induction. In this case "Magnet Moving Through a Coil of Wire". I would like to know what type of formula should I use for this scenario.

  • @GachPawlowicz
    @GachPawlowicz Месяц назад

    Very shitty explanation. If you want a good one try khan academy. Type inline skier