Uniform Circular Motion

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  • čas přidán 8. 09. 2024

Komentáře • 265

  • @Ender.wigginn
    @Ender.wigginn Před 10 lety +169

    Dr. Anderson, firstly I would like to say how impressed I am with the setup of your online lectures. Most instructors wouldn't think of using a mirror and camera, illuminated glass and a small audience for the creation of online lectures. This setup of yours is extremely engaging and interesting. I almost feel like I'm in the classroom.
    Moreover, I've found these lectures extremely helpful. Sometimes my instructors are not always clear with their explanations, it's always great to have another instructor, with a different explanation, to illustrate the concepts in a way that may help me make sense of the material.

    • @sherryvoss1566
      @sherryvoss1566 Před 6 lety +2

      Dr. Anderson, your presentations are the reasons why I am passing physics now. Please make more!

    • @shrinivasyadav1798
      @shrinivasyadav1798 Před 5 lety

      Mirror?

    • @mehekahmed3152
      @mehekahmed3152 Před 5 lety +2

      I couldnt agree more. My professor speaks faster than the speed of light; this professor explains every little step and writes out things and goes slowly too

    • @EmpyreanLightASMR
      @EmpyreanLightASMR Před 11 měsíci

      I mean, I feel bad that the kids don't have a surface to write on except their knees. But I'm guessing this is an extracurricular lecture, not an actual classroom lecture

  • @siame9250
    @siame9250 Před 4 lety +180

    I spent half of the tutorial time amazed how you were writing

  • @jameserayburn
    @jameserayburn Před 7 lety +37

    Dr. Anderson, your presentation is exactly what I needed. Thank you sir.

    • @yoprofmatt
      @yoprofmatt  Před 7 lety +11

      Great to hear. Keep up with the physics.
      Cheers,
      Dr. A

    • @frankdimeglio8216
      @frankdimeglio8216 Před 2 lety

      @@yoprofmatt THE CLEAR, TOP DOWN, SIMPLE, AND BALANCED MATHEMATICAL PROOF OF THE FACT THAT E=MC2 IS F=MA:
      E=MC2 IS F=ma ON BALANCE. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE; AS ELECTROMAGNETISM ENERGY IS GRAVITY !!! Gravity IS ELECTROMAGNETISM/energy. TIME is NECESSARILY possible/potential AND actual IN BALANCE, AS E=MC2 IS F=ma ON BALANCE; AS ELECTROMAGNETISM/energy is gravity !!! INSTANTANEITY is thus fundamental to what is the FULL and proper UNDERSTANDING of physics/physical experience, AS E=MC2 IS F=ma IN BALANCE; AS the stars AND PLANETS are POINTS in the night sky; AS ELECTROMAGNETISM/energy is gravity. THE SUN AND what is THE EARTH/ground are E=MC2 AND F=ma IN BALANCE. TIME DILATION ultimately proves ON BALANCE that E=MC2 IS F=ma ON BALANCE, AS ELECTROMAGNETISM/energy is gravity. (The sky is blue, AND THE EARTH is ALSO BLUE. CAREFULLY consider what is THE EYE.) Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=ma ON BALANCE; AS ELECTROMAGNETISM/energy is gravity !!! (THEREFORE, the rotation of WHAT IS THE MOON matches it's revolution.) "Mass"/ENERGY involves BALANCED inertia/INERTIAL RESISTANCE consistent with/as what is BALANCED electromagnetic/gravitational force/ENERGY, AS ELECTROMAGNETISM/energy is gravity; AS E=MC2 IS F=ma. GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=ma ON BALANCE; AS ELECTROMAGNETISM/energy is gravity !!! Energy has/involves GRAVITY, AND ENERGY has/involves inertia/INERTIAL RESISTANCE. ("Mass"/ENERGY IS GRAVITY. ELECTROMAGNETISM/energy is gravity. E=MC2 IS F=ma. Carefully consider what is THE EYE.) Objects (AND what is the FALLING MAN) fall at the SAME RATE (neglecting air resistance, of course), AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. Again, carefully consider that the stars AND PLANETS are POINTS in the night sky !!! (Very importantly, outer "space" involves full inertia; AND it is fully invisible AND black.) It ALL CLEARLY makes perfect sense. BALANCE AND completeness go hand in hand. SO, carefully consider what are the ORANGE SUN AND the fully illuminated and setting MOON ! Both are the size of THE EYE. Think LAVA !!! The Moon is ALSO BLUE on balance. Therefore, E=MC2 IS F=ma IN BALANCE !! It all CLEARLY makes perfect sense !!! Carefully consider THE MAN who IS standing on what is THE EARTH/ground !!! Great !!! E=MC2 IS F=ma ON BALANCE !!!!
      By Frank DiMeglio

  • @jittususanmathew468
    @jittususanmathew468 Před 5 lety +11

    Thanks a lot professor....
    I had a two page derivation which made no sense but ultimately by watching your video i was able to solve the derivation within half a page....
    Great job... Keep going😇

  • @mariela2462
    @mariela2462 Před 3 lety +6

    You're my favorite physics teacher, thank you so much !!!.

  • @user-wr5kj3ug8i
    @user-wr5kj3ug8i Před rokem +2

    Truly one of the moment of all time

    • @yoprofmatt
      @yoprofmatt  Před rokem

      What kind of moment? Moment of inertia?
      Thanks, appreciate it.
      Cheers,
      Dr. A

  • @abbydl6545
    @abbydl6545 Před 7 lety +11

    You've done such a really really great job, thanks so much sir.

  • @steffenleo5997
    @steffenleo5997 Před rokem +1

    Thanks for this great explained Video Prof Anderson... 👍👍...have a nice weekend

  • @yoprofmatt
    @yoprofmatt  Před 10 lety +19

    Daniel, thanks very much for the feedback. That was indeed the goal, to make students feel like I was talking directly to them. Glad you're enjoying it.
    Here's a nice little story about the Learning Glass: newscenter.sdsu.edu/sdsu_newscenter/news.aspx?s=75004
    Cheers,
    Dr. A

  • @shahrahman7337
    @shahrahman7337 Před 2 lety

    This was the most interesting presentation on centripetal acceleration I’ve ever seen. Never seen it taught this way

    • @yoprofmatt
      @yoprofmatt  Před 2 lety +3

      Great to hear. I'm definitely not the first, though. Many textbooks treat it this way.
      Cheers,
      Dr. A

  • @JasonHannah-jf6ry
    @JasonHannah-jf6ry Před měsícem

    AWESOME 🎉🎉🎉🎉
    Naw I understand a question thats been driving me mad for YEARS ❤

  • @akifmertturk3657
    @akifmertturk3657 Před 7 lety +6

    Waov. You are excellent.
    Thanks for sharing your knowledge with us in such a great way.
    We keep following you and your lectures
    Thanks,
    Miraç Akif

    • @yoprofmatt
      @yoprofmatt  Před 7 lety +1

      Mirac,
      Thanks for the feedback, I really appreciate it. And keep following, I'll try to add more.
      Cheers,
      Dr. A

    • @frankdimeglio8216
      @frankdimeglio8216 Před 2 lety

      @@yoprofmatt The ultimate unification and understanding of physics/physical experience combines, BALANCES, AND INCLUDES opposites, AS E=MC2 is CLEARLY manifest as F=ma ON BALANCE; AS ELECTROMAGNETISM/energy is gravity. Here's the proof. This also explains why objects (including WHAT IS THE FALLING MAN) fall at the SAME RATE (neglecting air resistance, of course), AS E=MC2 IS CLEARLY F=MA ON BALANCE; AS ELECTROMAGNETISM/energy is gravity. TIME dilation ULTIMATELY proves (ON BALANCE) that E=MC2 IS clearly and necessarily F=ma ON BALANCE, AS ELECTROMAGNETISM/energy is gravity. Gravity is ELECTROMAGNETISM/energy ON BALANCE.
      ON THE CLEAR, EXTENSIVE, SENSIBLE, BALANCED, THEORETICAL, AND UNIVERSAL PROOF THAT ELECTROMAGNETISM/energy is gravity, AS E=MC2 IS clearly PROVEN TO BE F=MA ON BALANCE:
      Balanced inertia/INERTIAL RESISTANCE is fundamental, as ELECTROMAGNETISM/energy is gravity. The stars AND PLANETS are POINTS in the night sky. Energy has/involves GRAVITY, AND ENERGY has/involves inertia/INERTIAL RESISTANCE; AS gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites; AS ELECTROMAGNETISM/energy is gravity; AS E=MC2 is CLEARLY proven to be F=ma ON BALANCE. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE. Indeed, A PHOTON may be placed at the center of what is THE SUN (as A POINT, of course); AS the reduction of SPACE is offset by (or BALANCED with) the speed of light (c); AS ELECTROMAGNETISM/energy is gravity ON BALANCE; AS E=MC2 is CLEARLY F=ma IN BALANCE !!! Gravity is ELECTROMAGNETISM/energy ON BALANCE, AS E=MC2 is CLEARLY proven to be F=ma IN BALANCE. TIME dilation ULTIMATELY proves (ON BALANCE) that ELECTROMAGNETISM/energy is gravity, AS E=MC2 is CLEARLY F=ma ON BALANCE. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS ELECTROMAGNETISM/energy is gravity; AS E=MC2 is CLEARLY F=ma ON BALANCE; AS the stars AND PLANETS are POINTS in the night sky. Accordingly, ON BALANCE, the rotation of WHAT IS THE MOON matches it's revolution. TIME is NECESSARILY possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity; AS E=MC2 is CLEARLY F=ma ON BALANCE. The stars AND PLANETS are POINTS in the night sky. Accordingly, ON BALANCE, it makes perfect sense that THE PLANETS (including WHAT IS THE EARTH) will move away very, very, very slightly in relation to what is THE SUN !!! ELECTROMAGNETISM/energy is gravity, AS E=MC2 is CLEARLY F=ma ON BALANCE. Inertia/INERTIAL RESISTANCE is proportional to (or BALANCED with/as) GRAVITATIONAL force/ENERGY, as this balances gravity AND inertia; AS E=MC2 is CLEARLY F=ma ON BALANCE; AS ELECTROMAGNETISM/energy is gravity. GREAT. I have explained the cosmological redshift AND the supergiant stars. Stellar clustering ALSO proves ON BALANCE that ELECTROMAGNETISM/energy is gravity, AS E=MC2 is CLEARLY F=ma IN BALANCE !!!
      By Frank DiMeglio

  • @QianBing
    @QianBing Před 7 lety +37

    How do you write backwards??

    • @yoprofmatt
      @yoprofmatt  Před 7 lety +14

      Answer revealed here: www.learning.glass
      Cheers,
      Dr. A

    • @QianBing
      @QianBing Před 6 lety

      thanks

    • @timjay6279
      @timjay6279 Před 4 lety +2

      The video image is flipped horizontally

  • @SurinderSingh-gl1dj
    @SurinderSingh-gl1dj Před 6 lety +104

    Any Indian here who is watching this and got amazed by seeing how he is writing😉😉😉😱😱😱😱.

  • @p-brane8358
    @p-brane8358 Před 2 lety

    What makes the Earth's oceans, that are moving at hundreds of mph, turn with the sphere Earth, staying put at their latitudes and not moving in a straight line, (as water wants to do), to the sphere-Earth's equator? What "tethers" the waters to the axis at any given latitude and what causes the water to make a left turn, (for instance, in the northern hemisphere)?

  • @VSShiva
    @VSShiva Před 5 lety +2

    Simple and to the point,Thank you so much!

    • @yoprofmatt
      @yoprofmatt  Před 5 lety +1

      The power of editing!
      Cheers,
      Dr. A

  • @rajkr7j
    @rajkr7j Před 4 lety +1

    His voice is like COOPER in INTERSTELLER .
    Am I correct?

    • @sbsanni
      @sbsanni Před 4 lety

      Understanding the threshold of the most important knowledge czcams.com/video/j89sJGy3S5U/video.html

    • @yoprofmatt
      @yoprofmatt  Před 4 lety

      raj kumar,
      No idea. but thanks for the comment, and keep up with the physics!
      You might also like my new website: www.universityphysics.education
      Cheers,
      Dr. A

  • @victorrono9158
    @victorrono9158 Před 7 lety +3

    Your lectures are quite amazing...

  • @user-pf1sm5rh8c
    @user-pf1sm5rh8c Před 8 měsíci

    You are my dream teacher.thank you

  • @alexdilling8778
    @alexdilling8778 Před 7 lety +3

    Yo that opened my eyes so damn wide

    • @yoprofmatt
      @yoprofmatt  Před 7 lety +5

      Excellent. Now put on some shades.
      Cheers,
      Dr. A

  • @bradenmartin1382
    @bradenmartin1382 Před 4 lety

    Why oh why have I not been watching these videos

  • @onspot361
    @onspot361 Před 3 lety

    You explain in a good way.

  • @dentistrystudent2225
    @dentistrystudent2225 Před 4 lety +1

    I understood everything, but why are the 2 thetas the same??

  • @manuboker1
    @manuboker1 Před 3 lety +1

    GREAT PHYSICS LECTURES !!! :))

  • @bignut6269
    @bignut6269 Před 3 lety

    Awesome dethklok shirt

  • @user-qj3rv2mo1b
    @user-qj3rv2mo1b Před 11 měsíci

    One equal angle between the two triangles.
    Angle between r1 r2 is theta where r1 =r2 = r
    is the same as angle between v1 & v2, where v1 = v2.
    One Side equal between the two triangles.
    v2-v1 is equal to r
    Two sides of one triangle are at 90 degree to the corresponding side of the other triangle.
    One side and one angle of the two triangles are equal, how does it prove the two triangles are similar

  • @sairahul3934
    @sairahul3934 Před 8 lety +1

    really pleased by the way you explained this topic. thank you sir.....

    • @yoprofmatt
      @yoprofmatt  Před 8 lety +1

      Rahul,
      Excellent, this is music to my ears. Keep on learning!
      Cheers,
      Dr. A

  • @idrisjalal7887
    @idrisjalal7887 Před 2 lety

    Hi dear professor Anderson.. Your lecture was so amazing it was a very quite lesson and I understood more than my prediction.. Well done and keep your best doing
    Regards. Edris from Kurdistan region of Iraq

  • @shubhamguptashubhamgupta7880

    You are great sir amazing

  • @uzairahmed706
    @uzairahmed706 Před měsícem

    7:00 hiw are the triangles similar. By which test of similarity ?

  • @chadholdaway2089
    @chadholdaway2089 Před 3 lety

    That dethklok shirt though

  • @thomasashish9782
    @thomasashish9782 Před 7 lety +2

    very greatful to ur vedio it amazing explanation

    • @yoprofmatt
      @yoprofmatt  Před 7 lety +1

      Thomala,
      Thanks for the comment. I've had a lot of fun making these videos. More on the way.
      Cheers,
      Dr. A

  • @salamdraidi5915
    @salamdraidi5915 Před 7 lety +1

    great explanation dr, thank you :)

  • @nileshrathod3153
    @nileshrathod3153 Před 4 lety

    I think centripetal force is a necessary condition just to sustain the circular motion. But to start the circular motion, we need to provide torque. Only centripetal force can't start a circular motion. Could you please comment on it?

    • @yoprofmatt
      @yoprofmatt  Před 4 lety

      nilesh,
      You are absolutely correct. Which is why rockets don't just go straight up.
      Thanks for the comment, and keep up with the physics!
      You might also like my new website: www.universityphysics.education
      Cheers,
      Dr. A

  • @fathimafarah767
    @fathimafarah767 Před 4 lety

    Thank u Dr. Anderson

  • @ThehouseofcricketVNR24

    Your explanation is excellent.....

  • @Z78798
    @Z78798 Před 5 lety

    Preety good. He can taught without hesitation after practice.

  • @davesflights6256
    @davesflights6256 Před 4 lety +1

    Hi. For someone revisiting Physics 40 years after school (to help my daughter..!) I am enjoying these immensely. Could I ask one thing though....this lecture starts with asking the students if they are accelerating at that moment, and returns to ask the same question at the end, but doesn't answer or make any statement to summarise it... Is that on another video ?

    • @yoprofmatt
      @yoprofmatt  Před 4 lety +1

      Daves,
      Not sure, exactly. Didn't mean to leave you hanging.
      Thanks for the comment, and keep up with the physics!
      You might also like my new website: www.universityphysics.education
      Cheers,
      Dr. A

  • @deepakaggarwal333
    @deepakaggarwal333 Před 8 lety +2

    awsome video , really i like it.

    • @yoprofmatt
      @yoprofmatt  Před 8 lety +2

      +Deepak Aggarwal Thanks Deepak, the sound was a bit wonky on this one, but otherwise it was okay. Thanks for watching!
      Cheers,
      Dr. A

  • @kelvinreyes5587
    @kelvinreyes5587 Před 4 lety

    the guy w the dethklok shirt \m/ XD

  • @glyphiclive15
    @glyphiclive15 Před 2 lety

    "Trapped on Earth" story of my life...

  • @yogeshwarsingh7975
    @yogeshwarsingh7975 Před 3 lety

    doubt : at 7.30; when you removed the vector sign and wrote in magnitude form, to write that (delta)a = (delta)v/(delta)t; how can that make sense? Instead it should be = | ( delta of vector)v/(delta)t |, and the modulus can not go "inside" the delta, because if (delta)a = (delta)v/(delta)t were to be true; then in a case where change in speed were 0, then it should imply that the magnitude of acceleration is also zero, which is obviously not true.

  • @nomatterthefact
    @nomatterthefact Před 5 lety

    really liked his teaching way. nice.

    • @yoprofmatt
      @yoprofmatt  Před 5 lety

      Like to hear that. Keep up with the physics.
      Cheers,
      Dr. A

  • @gauravkamath9569
    @gauravkamath9569 Před 6 lety +1

    Im in 9th grade, in Alg 2 this year is it possible for me to take Ap physics 1 outside school. Is there any way I can do it to get thorough. I really love physics, and enjoy your videos. what can I do?

  • @gheoffricare4520
    @gheoffricare4520 Před 3 lety

    I... clap!

  • @fattin8961
    @fattin8961 Před 8 lety +1

    very amazed with the board!!!!!!!! thank you sir! it helps a lot!

    • @yoprofmatt
      @yoprofmatt  Před 7 lety

      Thanks! I think it turned out pretty good. Our more recent videos look even better, in my opinion.
      Cheers,
      Dr. A

  • @FakeAccount
    @FakeAccount Před 4 lety +1

    doesn't the first question depends on the adopted referential? We're moving relative to the sun, but we're still relative to your wall, for example.

    • @carultch
      @carultch Před 3 lety +1

      Velocity is relative, but acceleration is not relative. There is such a thing as absolute acceleration. You can infer that the Earth is rotating by tracking the passage of stars and other celestial bodies, and determine Earth rotates once every 23 hrs & 56 minutes as measured to the distatant stars.
      You can also measure Earth's rotation, without referencing any outside body, by constructing Newton's laws in our rotating reference frame, and keeping track of the drift that results from our rotation. The Focault pendulum was the first experimental evidence of Earth's rotation, that did not reference any external body. You can also measure this with a ring laser gyroscope, that tracks the directional drift of our planet, relative to the inertial reference frame in which light travels.

  • @Vivek-cw3zu
    @Vivek-cw3zu Před 5 lety

    Thank you sir for making it so simple to understand. 💗

  • @AK-di4rv
    @AK-di4rv Před 7 lety

    deducing uniform circular motion! Made it simple.

  • @hmwehmweqhmwe9468
    @hmwehmweqhmwe9468 Před 7 lety

    Clear explanation.I like your lecture.

  • @muhammetkurkcu4756
    @muhammetkurkcu4756 Před 2 lety

    cool

  • @stoian1743
    @stoian1743 Před 7 lety

    Can you please sir make a video about special and general relativity... if i watch it, it will be my happiest experience i've ever had
    ! Thank you Sir in advance! :)

  • @GeorgetteOforiNuamah-jy6ge

    Shit, amazing,how does he do that?

  • @PrashantKumar-nz6ih
    @PrashantKumar-nz6ih Před 7 měsíci

    Professor I'm still confused, if we're really facing centripetal acceleration shouldn't we be moving towards sun. Isn't it being cancelled out by centrifugal acceleration and probably that's why we're stable. Please do reply.

    • @ihavenoenemis
      @ihavenoenemis Před 3 měsíci

      Depends on the frame of reference right if we are looking from outer space and need to analyse the force on our body we should put centrifugal force

  • @danawitephrem6272
    @danawitephrem6272 Před 2 lety

    thank you

  • @yaseminozturk8700
    @yaseminozturk8700 Před 7 lety

    Thanks for the useful video and cool board. Really like your efforts to make it interesting!

    • @yoprofmatt
      @yoprofmatt  Před 7 lety +1

      Thanks. I find it interesting, so I'm hoping I can convince others as well.
      Cheers,
      Dr. A

  • @rajnirawat7310
    @rajnirawat7310 Před 7 lety +2

    professor, were you actually writing the other way? and yes, i love your explanation. thank you sir : )

    • @reemabansal9594
      @reemabansal9594 Před 3 lety

      no he wasn't

    • @carultch
      @carultch Před 2 lety +1

      He uses a mirror between his transparent board and the video camera. You can tell this by which hand has his wedding ring, which is traditionally worn on the left hand. The figure in the video appears to be left-handed and has a wedding ring on the right hand, but it is really mirrored footage.

  • @therealnannygoat1071
    @therealnannygoat1071 Před rokem

    Is it just me or is this guy writing in reverse

  • @arshadqureshi4715
    @arshadqureshi4715 Před 6 lety

    Awesome explanation.

    • @yoprofmatt
      @yoprofmatt  Před 6 lety

      Thanks for the comment. Keep up with the physics.
      Cheers,
      Dr. A

  • @userrr703
    @userrr703 Před rokem

    so beneficial

    • @yoprofmatt
      @yoprofmatt  Před rokem

      Appreciate that. Glad to be of help.
      Cheers,
      Dr. A

  • @roohinkukreja1136
    @roohinkukreja1136 Před 2 lety

    Thanks a lot, was quite helpful:)

    • @yoprofmatt
      @yoprofmatt  Před 2 lety

      Good to hear. Have a great day.
      Cheers,
      Dr. A

  • @nileshrathod3153
    @nileshrathod3153 Před 4 lety

    If the centripetal force direction is radially inwards, then why the mass leaves tangentially when the centripetal force becomes zero?

    • @yoprofmatt
      @yoprofmatt  Před 4 lety +1

      nilesh,
      Newtons' first says "Objects in motion tend to stay in motion." If you remove the force, the projectile follows it's last known velocity in a straight line.
      Thanks for the comment, and keep up with the physics!
      You might also like my new website: www.universityphysics.education
      Cheers,
      Dr. A

    • @nileshrathod3153
      @nileshrathod3153 Před 4 lety

      @@yoprofmatt Thank you ! Dr. A. I have one more question regarding centrifugal force. Even if centrifugal force is not real, then why do we need to design systems taking into consideration the amount of centrifugal force it will experience to avoid failure?

    • @carultch
      @carultch Před 2 lety

      @@nileshrathod3153 Centrifugal (pseudo)-force is an apparent force that you experience as an occupant of a rotating reference frame. It is a function of your immediate environment accelerating radially inward as it moves in a circular path.
      It is a shortcut to account for -m*a (the pseudoforce or the D'Alembert force term) as if it were a force acting upon a structure you were designing for use in an accelerating reference frame, but this really isn't a force because there is no agent object that causes this force upon you. What is really happening, is that the net force needs to add up to the acceleration, and if your environment is accelerating, then so are you. The acceleration isn't immediately obvious to you, without observing your surroundings, and your immediate instinct is to assume an outward force acts upon you, as you experience the inward constraint forces that keep you moving with your reference frame.

  • @jeyjey-ft5zy
    @jeyjey-ft5zy Před rokem

    I enjoyed session and i wanna see mor3

    • @yoprofmatt
      @yoprofmatt  Před rokem

      There's plenty more:
      czcams.com/users/yoprofmatt
      Cheers,
      Dr. A

  • @aytunc5662
    @aytunc5662 Před 6 lety

    Yeah we are accelerating right now because of rotating Earth.

  • @joshuadilao1496
    @joshuadilao1496 Před rokem

    Distracted by how he;s writting in mirrored?

  • @qualquan
    @qualquan Před 6 lety

    v = angular velocity = delta arc/delta t and not delta chord/delta t.
    your delta r is a chord and not an arc.
    So one has to use angle d theta or limit of angle theta to equate arc with chord.
    Only then dr/dt = v
    One cannot use angle delta theta and make delta r/delta t = v

    • @yoprofmatt
      @yoprofmatt  Před 6 lety

      Thanks for the comment. This derivation was slightly hand-wavy. Maybe next time I'll be more explicit.
      Cheers,
      Dr. A

  • @ericportillo8277
    @ericportillo8277 Před 3 lety

    You should've used ucm to erase the board!

  • @ptyptypty3
    @ptyptypty3 Před 7 lety

    how do we know that the acceleration always points to the Center of the Circle?... I can see that the Magnitude of the Acceleration is (V^2)/R ... but how do we know that the Direction of the acceleration Points to the Center? ... in lieu of my question.. YOUR VIDEO WAS VERY GOOD!! THANK YOU!! :)

    • @yoprofmatt
      @yoprofmatt  Před 7 lety +1

      The vector drawing at 5:43 illustrates this. Since Δv is pointing towards the circle's center, this is the direction of the acceleration. As a follow-up, you might ask "where do we draw this on the circle?" And the answer is, halfway in between the vi and vf vectors. As you let Δθ get smaller and smaller, the two vectors vi and vf approach each other, and you can quickly convince yourself that the Δv indeed points towards the circle's center.
      Important note: This only happens when we have constant speed (the magnitude of vi and vf are the same). If we are increasing our speed (or decreasing), the total acceleration no longer points towards circle center.
      Hope this helps. And thanks for the question.
      Cheers,
      Dr. A

  • @Naimka2002
    @Naimka2002 Před 6 lety +2

    Well, I was searching the internet for an explanation, and till now I didn't find a proper video, anyone tell me why these two triangles are similar, why do we use this method from the start, it doesn't seem logical to me.

    • @behnamasid
      @behnamasid Před 5 lety

      If the angle between two triangles is the same, then the triangles are similar. To see why the vector difference angles are the same, then imagine if r_i and r_f form a 90 degree angle. In that case the vectors v_i and V_f also form90 degrees angle. Try 180 degrees and the vectors for both r and v will be opposite, etc...

    • @evam2351
      @evam2351 Před 4 lety

      Similar triangles: if two triangles share the same two angles, then their third is the same. That means that they are proportional to each other.

  • @zeynaviegas
    @zeynaviegas Před 5 lety

    ooooohhhhj so thats where this formula came from :o

  • @nileshrathod3153
    @nileshrathod3153 Před 4 lety

    Hello Prof! I wanted to know exactly why we don't feel the earth's rotation on its own axis? I watched many videos about it, but I didn't find any convincing answer. All videos explain it is because earth is rotating at constant speed ( but accelerating,right!) and because our relative velocity w.r.t earth is zero. But why then we can feel the circular motion on a merry-go-round horizontal or vertical one?

    • @yoprofmatt
      @yoprofmatt  Před 4 lety

      nilesh,
      It's because the earth is big and the rotation is (relatively) slow.
      That makes the centripetal acceleration very small.
      Thanks for the comment, and keep up with the physics!
      You might also like my new website: www.universityphysics.education
      Cheers,
      Dr. A

  • @zhefanfei4660
    @zhefanfei4660 Před 2 lety

    am I the only one who wants to know how he writes from the other side of the board???

  • @saikrishnabiswas8627
    @saikrishnabiswas8627 Před 4 lety

    I have a doubt that how can a body travelling in an uniform circular motion have both same speed throughout and a centripetal acceleration having a MAGNITUDE(ie v^2/r)

    • @yoprofmatt
      @yoprofmatt  Před 4 lety

      Saikrishna Biswas,
      Acceleration is a change in velocity. But velocity has both magnitude (what we call speed) and direction. By changing the direction, you have an acceleration.
      Thanks for the comment, and keep up with the physics!
      You might also like my new website: www.universityphysics.education
      Cheers,
      Dr. A

    • @saikrishnabiswas8627
      @saikrishnabiswas8627 Před 4 lety

      @@yoprofmatt but sir then how can the centripetal acceleration be calculated by formula v^2/r. So what does this formula for centrietal acceleration give us, does it give us the rate in which the DIRECTION is changing?

  • @VamsiKrishna-no5fx
    @VamsiKrishna-no5fx Před 5 lety

    Why is the velocity always perpendicular to the radius?

    • @yoprofmatt
      @yoprofmatt  Před 5 lety

      Because otherwise it is not moving in a circle.
      Cheers,
      Dr. A

  • @ronrexy4683
    @ronrexy4683 Před 5 lety

    Well iust say .. thats a cool board😆

    • @yoprofmatt
      @yoprofmatt  Před 5 lety

      Thanks much. Check it out here: www.learning.glass
      Cheers,
      Dr. A

  • @butternnut
    @butternnut Před 7 lety

    do you fast forward when you wipe off the marker or do u wipe that fast

    • @yoprofmatt
      @yoprofmatt  Před 7 lety +3

      Busted! Yep, that would be the miracle of editing.
      Thanks for noticing.
      Cheers,
      Dr. A

  • @SurinderSingh-gl1dj
    @SurinderSingh-gl1dj Před 6 lety +1

    Great video😊😊😊😊😊😣😣😣😣😅😄😄😄😄😰😮😮

  • @neilmcbain411
    @neilmcbain411 Před 6 lety

    The question here I need to ask Dr Armstrong ? suppose in gravity free space in a free vacuum in a free vacuum ,we have two masses M1 and m2 that are attracted to each other by a centripetal force so tat they spin around their centre of mass.
    when the force disappears m2 leaves the circle at a tangent,then M1 must also leave the the circle at a a tangent but in the opposite direction.
    Suppose that the mass M1 is much greater than m2 that is M2>>>>m2 so that the centre of mass of the system almost coincides with M1 's centre of mass
    if the centripetal force suddenly disappears the smaller mass leaves the circle away from the centre of mass but the larger mass and its centre of mass leaves the circle in the opposite direction away from the common centre of mass of the system
    AM I CORRECT IN ASSUMING THIS?
    the larger mass M1 FEELS NO FORCE ACTING ON IT AS IT LEAVES THE SYSTEM NOR DOES THE MASS m2 IS THIS CORRECT??
    The problem with lectures on circular motion and centripetal force is that you take into account the situation in a gravity free vacuum of space involving two bodies.

    • @yoprofmatt
      @yoprofmatt  Před 6 lety

      Neil,
      Good question. I think you're right in a gravity-free space.
      Here's my analysis:
      Pretend the two objects are tied together by a string and orbiting about the center of mass. If you suddenly cut the string, then both bodies will leave in a straight line tangent to their particular orbit and in opposite directions. This is also at a right angle to the line that joined both bodies and the center of mass when the string was cut.
      After the string is cut, there is no force acting on either body (since we assumed no gravitational force either), and they will move with constant velocity.
      (Put gravity back in and that of course changes the situation.)
      Cheers,
      Dr. A

    • @carultch
      @carultch Před 3 lety

      Objects in orbit do not feel the force that keeps them in orbit. The sensation of the gravitational force is nullified, when the object is free to accelerate according to the gravitational force. This is due to the fact that gravity acts uniformly on every kilogram of an object, such that there are no constraint forces necessary to keep an occupant of an orbiting spacecraft at rest within their immediate environment.

  • @justinyi8766
    @justinyi8766 Před 3 lety

    i cant with the marker sound

  • @videoupload6898
    @videoupload6898 Před 3 lety

    What does the professor mean about adding the 90 degrees? at 6:29?

    • @carultch
      @carultch Před 3 lety

      He's demonstrating that the change in angle between the two velocity vectors, is no different than the change in angle between the two radius vectors.

  • @nurlatifahmohdnor8939
    @nurlatifahmohdnor8939 Před 2 lety

    7|13.1.'98
    M
    I
    N
    T
    '92
    '93 1
    '94 2
    '95 3
    '96 4
    '97 5
    '98 6
    My sibling number 6 was 6 years old. I was at that time Form 5. 18.

    • @yoprofmatt
      @yoprofmatt  Před 2 lety

      I kind of love it. Might be at least one typo.
      Cheers,
      Dr. A

  • @jamesadriantan5272
    @jamesadriantan5272 Před 4 lety

    is this a serway reference?

  • @savinorojas6836
    @savinorojas6836 Před 8 lety

    How does your board work? It's cool

    • @yoprofmatt
      @yoprofmatt  Před 8 lety +1

      John,
      Thanks. All secrets are revealed here: www.learning.glass
      Cheers,
      Dr. A

  • @charlinascimento3225
    @charlinascimento3225 Před 6 lety

    Muito boa aula!

    • @yoprofmatt
      @yoprofmatt  Před 6 lety

      Obrigado pela sua presença.
      Felicidades,
      Dr. A

  • @m.muthupalaniyappan.7842

    Hi Anderson thanks for simple Math for centripetal acceleration, can you please explain this math in most simple intuitive way of physical motion....thought experiments via imagination....for long i am searching...still missing & chasing for some thing

  • @ssss855167
    @ssss855167 Před 2 lety

    Dear sir,
    I have one question. We know the value of acceleration is V^2/r, but how do we know its direction is towards the center?

    • @yoprofmatt
      @yoprofmatt  Před 2 lety

      That's what we're showing with the triangle at 5:43. But also, you know that the acceleration has to be at a right angle to the velocity vector, otherwise the speed would be changing (and since we're assuming uniform circular motion, that means no change in speed).
      Cheers,
      Dr. A

    • @ssss855167
      @ssss855167 Před 2 lety

      Thank you very much for the clear explanation sir.

  • @ysfkyya
    @ysfkyya Před 5 lety

    excellent

    • @yoprofmatt
      @yoprofmatt  Před 5 lety

      Thanks. Keep up with the physics.
      Cheers,
      Dr. A

  • @pranjalverma3501
    @pranjalverma3501 Před 6 lety

    Nice one, anderson sir really know how to teach, but ∆r/∆t isn't it ∆v instead of v?

    • @yoprofmatt
      @yoprofmatt  Před 6 lety

      I simplified this discussion a bit. I think it's okay, but really we should be using derivatives, where v = dx/dt
      Cheers,
      Dr. A

  • @bayarjavkhlanlhagvadorj4937

    How are you writing backward

  • @krasimirronkov17
    @krasimirronkov17 Před 5 lety

    Why is velocity tangent to the circular path

    • @krasimirronkov17
      @krasimirronkov17 Před 5 lety

      Pls help professor

    • @justadreamerforgood69
      @justadreamerforgood69 Před 4 lety

      @@krasimirronkov17
      Because if velocity is dr/dt so imagine ri and rf very close to each other(as shown in this example) then they will be tangent to the circle
      When you divide by dt( where dt tends to zero but not zero) the vector dr is the same except it's magnified because we're dividing by a number less than one and greater than 0

  • @srinivasanp4018
    @srinivasanp4018 Před 4 lety

    Thank you sir

  • @anjanakathuria4194
    @anjanakathuria4194 Před 6 lety

    How are you able to Wright in reverse, On the other side.

    • @carultch
      @carultch Před 3 lety

      He's using a mirror in between the camera and his glass panel. He writes normal, the class who sees him teach in person, will see him writing backward on this board. He shows a live video feed on another monitor, so the class can see normal writing.

  • @lifethrownoutofthewindow
    @lifethrownoutofthewindow Před 7 lety +1

    godtier

    • @yoprofmatt
      @yoprofmatt  Před 7 lety

      Don't know what that means but it's hopefully good?
      Cheers,
      Dr. A

  • @vanessaokoth
    @vanessaokoth Před rokem

    is he writing backwards

  • @cristianmartinez9049
    @cristianmartinez9049 Před 5 lety

    what does Δv/v mean? i didn't understand why he did that. what triangle property it is?

  • @mohammadhousseini3470
    @mohammadhousseini3470 Před 6 lety +1

    why the speed constant while there is acceleration (it become UARM)

    • @zachbarry1992
      @zachbarry1992 Před 3 lety +1

      2 years late, but, its because the acceleration is inwards, not in the direction the object is traveling

    • @carultch
      @carultch Před 2 lety

      Because the net force on the object is perpendicular to its velocity. The net work being done on the object is zero.

  • @trungac2414
    @trungac2414 Před 6 lety

    Thanks for your lecture

    • @yoprofmatt
      @yoprofmatt  Před 6 lety

      You are welcome. Keep up with the physics!
      Cheers,
      Dr. A

  • @aka3673
    @aka3673 Před rokem

    Tnx man

  • @neilmcbain411
    @neilmcbain411 Před 6 lety

    sorry Prof Matt Anderson I typed your name incorrectly below as Armstrong below

    • @yoprofmatt
      @yoprofmatt  Před 6 lety

      No problem. Response below.
      Cheers,
      Dr. A

  • @frootloop1879
    @frootloop1879 Před 5 lety

    i dont get the angle thing