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.
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
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
@@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
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😇
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
@@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
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
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
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)?
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
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?
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
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 ?
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
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.
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.
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?
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.
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.
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...
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
@@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?
@@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.
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.
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
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.
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?
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
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.
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
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!! :)
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
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
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! :)
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)
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
@@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?
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.
@@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
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
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.
Dr. Anderson, your presentations are the reasons why I am passing physics now. Please make more!
Mirror?
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
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
I spent half of the tutorial time amazed how you were writing
he wasn't writing in the opposite way but he was writing on the glass board w/ a camera which was recording the text laterally opposite
see 1:50 for proof about the board
Me to
@@reemabansal9594 ohhhhhh
I agree
Dr. Anderson, your presentation is exactly what I needed. Thank you sir.
Great to hear. Keep up with the physics.
Cheers,
Dr. A
@@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
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😇
You're my favorite physics teacher, thank you so much !!!.
You've done such a really really great job, thanks so much sir.
Thanks for this great explained Video Prof Anderson... 👍👍...have a nice weekend
Any Indian here who is watching this and got amazed by seeing how he is writing😉😉😉😱😱😱😱.
He made that board himself if you can believe it.
yes
Surinder Singh bro just write on the glass normally and record from other side and flip the video
You are right bro this is what the guy did
Wow....well a genius idea....if u think about that...isn't. It?
Truly one of the moment of all time
What kind of moment? Moment of inertia?
Thanks, appreciate it.
Cheers,
Dr. A
This was the most interesting presentation on centripetal acceleration I’ve ever seen. Never seen it taught this way
Great to hear. I'm definitely not the first, though. Many textbooks treat it this way.
Cheers,
Dr. A
How do you write backwards??
Answer revealed here: www.learning.glass
Cheers,
Dr. A
thanks
The video image is flipped horizontally
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
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
Mirac,
Thanks for the feedback, I really appreciate it. And keep following, I'll try to add more.
Cheers,
Dr. A
@@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
His voice is like COOPER in INTERSTELLER .
Am I correct?
Understanding the threshold of the most important knowledge th-cam.com/video/j89sJGy3S5U/w-d-xo.html
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
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
You are my dream teacher.thank you
AWESOME 🎉🎉🎉🎉
Naw I understand a question thats been driving me mad for YEARS ❤
I understood everything, but why are the 2 thetas the same??
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)?
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
Your lectures are quite amazing...
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?
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
GREAT PHYSICS LECTURES !!! :))
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 ?
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
Simple and to the point,Thank you so much!
The power of editing!
Cheers,
Dr. A
7:00 hiw are the triangles similar. By which test of similarity ?
really pleased by the way you explained this topic. thank you sir.....
Rahul,
Excellent, this is music to my ears. Keep on learning!
Cheers,
Dr. A
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.
Yo that opened my eyes so damn wide
Excellent. Now put on some shades.
Cheers,
Dr. A
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.
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.
very greatful to ur vedio it amazing explanation
Thomala,
Thanks for the comment. I've had a lot of fun making these videos. More on the way.
Cheers,
Dr. A
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?
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.
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
You explain in a good way.
Awesome dethklok shirt
Why oh why have I not been watching these videos
You are great sir amazing
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.
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...
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.
If the centripetal force direction is radially inwards, then why the mass leaves tangentially when the centripetal force becomes zero?
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
@@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?
@@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.
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.
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
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.
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?
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
Your explanation is excellent.....
great explanation dr, thank you :)
do you fast forward when you wipe off the marker or do u wipe that fast
Busted! Yep, that would be the miracle of editing.
Thanks for noticing.
Cheers,
Dr. A
Thank you sir for making it so simple to understand. 💗
Clear explanation.I like your lecture.
Excellent!
Cheers,
Dr. A
awsome video , really i like it.
+Deepak Aggarwal Thanks Deepak, the sound was a bit wonky on this one, but otherwise it was okay. Thanks for watching!
Cheers,
Dr. A
professor, were you actually writing the other way? and yes, i love your explanation. thank you sir : )
no he wasn't
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.
Why is the velocity always perpendicular to the radius?
Because otherwise it is not moving in a circle.
Cheers,
Dr. A
Preety good. He can taught without hesitation after practice.
rip grammar
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?
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
Thank you very much for the clear explanation sir.
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!! :)
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
really liked his teaching way. nice.
Like to hear that. Keep up with the physics.
Cheers,
Dr. A
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
Thanks for the comment. This derivation was slightly hand-wavy. Maybe next time I'll be more explicit.
Cheers,
Dr. A
"Trapped on Earth" story of my life...
Shit, amazing,how does he do that?
That dethklok shirt though
What does the professor mean about adding the 90 degrees? at 6:29?
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.
deducing uniform circular motion! Made it simple.
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! :)
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)
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
@@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?
Thank u Dr. Anderson
the guy w the dethklok shirt \m/ XD
so beneficial
Appreciate that. Glad to be of help.
Cheers,
Dr. A
How does your board work? It's cool
John,
Thanks. All secrets are revealed here: www.learning.glass
Cheers,
Dr. A
Awesome explanation.
Thanks for the comment. Keep up with the physics.
Cheers,
Dr. A
very amazed with the board!!!!!!!! thank you sir! it helps a lot!
Thanks! I think it turned out pretty good. Our more recent videos look even better, in my opinion.
Cheers,
Dr. A
I... clap!
Thanks for the useful video and cool board. Really like your efforts to make it interesting!
Thanks. I find it interesting, so I'm hoping I can convince others as well.
Cheers,
Dr. A
Distracted by how he;s writting in mirrored?
I enjoyed session and i wanna see mor3
There's plenty more:
th-cam.com/users/yoprofmatt
Cheers,
Dr. A
is this a serway reference?
How are you able to Wright in reverse, On the other side.
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.
Is it just me or is this guy writing in reverse
Why is velocity tangent to the circular path
Pls help professor
@@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
Nice one, anderson sir really know how to teach, but ∆r/∆t isn't it ∆v instead of v?
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
Are there any other methods besides similar triangle method?
If you break the problem down, trig functions would work as well. But I like the simplicity here.
Cheers,
Dr. A
Derivatives definitely make it easier, but I think in this course I was trying to not use calculus.
Cheers,
Dr. A
Thanks a lot, was quite helpful:)
Good to hear. Have a great day.
Cheers,
Dr. A
what does Δv/v mean? i didn't understand why he did that. what triangle property it is?
I was using the property of similar triangles.
Cheers,
Dr. A
@@yoprofmatt thanks
am I the only one who wants to know how he writes from the other side of the board???
thank you
Yeah we are accelerating right now because of rotating Earth.
7|13.1.'98
M
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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.
I kind of love it. Might be at least one typo.
Cheers,
Dr. A
why the speed constant while there is acceleration (it become UARM)
2 years late, but, its because the acceleration is inwards, not in the direction the object is traveling
Because the net force on the object is perpendicular to its velocity. The net work being done on the object is zero.
You should've used ucm to erase the board!
is he writing backwards
i cant with the marker sound
cool
Well iust say .. thats a cool board😆
Thanks much. Check it out here: www.learning.glass
Cheers,
Dr. A
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
How are you writing backward
ooooohhhhj so thats where this formula came from :o
i dont get the angle thing
Great video😊😊😊😊😊😣😣😣😣😅😄😄😄😄😰😮😮
Muito boa aula!
Obrigado pela sua presença.
Felicidades,
Dr. A
Are these secondary students?
First year college students mostly.
Cheers,
Dr. A
Thank you Professor,
will you explain coriolis acceleration in one of your videos?
Good idea. I'll work on it.
Cheers,
Dr. A
Thank you
Thanks for your lecture
You are welcome. Keep up with the physics!
Cheers,
Dr. A
excellent
Thanks. Keep up with the physics.
Cheers,
Dr. A
is he.. writing backwards?
Thank you sir