whoever does the teaching in this video is really good, he needs a raise
@doni11416 жыл бұрын
Radioactive Music 4 U I’ve been studying for days for my physics midterm and i finally realized how to use fnet=ma properly because of this video. He’s so good
@tek16453 жыл бұрын
Non profit organization. Donate to them if you want to help
@frankdimeglio82163 жыл бұрын
@@doni1141 Consider the man who IS standing on what is the Earth/ground. Touch AND feeling BLEND, AS ELECTROMAGNETISM/energy is gravity. The Earth is blue, AND the sky is blue. SO, overlay THE EYE in BALANCED RELATION to/WITH what is the Earth ! (The BLUE SKY is ALSO translucent !) GREAT. Objects fall at the SAME RATE (neglecting air resistance, of course), AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. NOW, we proceed to the next step. The stars AND PLANETS are POINTS in the night sky. Invisible AND VISIBLE SPACE in fundamental equilibrium and BALANCE IS the MIDDLE DISTANCE in/of SPACE in fundamental RELATION to the universal fact that E=mc2 is F=ma. The Earth (A PLANET) involves BALANCED electromagnetic/gravitational force/ENERGY, AS E=MC2 IS F=MA. The Earth AND the Sun are CLEARLY E=mc2 AND F=ma IN BALANCE pursuant to what is the BALANCED MIDDLE DISTANCE in/of SPACE !!! Consider what is the speed of light (c), AND consider WHAT IS THE SUN. NOW, think about what is the Earth. Importantly, outer "space" involves full inertia; AND it is fully invisible AND black ! Great. Accordingly, the Earth is in BALANCE with what is the Sun, AS the orange Sun represents what is LAVA ON BALANCE !!! Great. (Notice the role and relation of what is the EYElid.) The MIDDLE DISTANCE in/of SPACE is universally balanced to/with what is THE SUN AND the speed of light (c), AS E=MC2 IS F=MA !!! Gravity IS ELECTROMAGNETISM/energy. Notice the black space of WHAT IS THE EYE, as the stars AND PLANETS are POINTS in the night sky IN BALANCED RELATION to/WITH what is the MIDDLE DISTANCE in/of SPACE; AS E=MC2 IS F=MA !!! The FULL DISTANCE in/of SPACE is truly linked AND BALANCED to/with what is THE MIDDLE DISTANCE in/of SPACE, AS E=mc2 IS F=ma IN BALANCE !!!; AS ELECTROMAGNETISM/energy is gravity !!!!!! GREAT !!!! Notice that the viscosity of lava IS BETWEEN that of WATER AND the Earth/ground ! The orange Sun is the same size as THE EYE. Outstanding. E=mc2 IS F=ma as what is the BALANCED MIDDLE DISTANCE in/of SPACE !!!! GREAT !!! LOOK at the progression AND THE BALANCE regarding FULL DISTANCE in/of SPACE, MIDDLE DISTANCE in/of SPACE !!!, AND A POINT (ALSO c) !!! E=mc2 IS F=ma !!!, AS ELECTROMAGNETISM/energy is gravity ON BALANCE !!! GREAT !!! It ALL CLEARLY makes perfect sense. BALANCE AND completeness go hand in hand. E=mc2 is F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity !!! Gravity IS ELECTROMAGNETISM/energy ON BALANCE !!! TIME is NECESSARILY possible/potential AND actual IN BALANCE, AS E=mc2 is F=ma; AS ELECTROMAGNETISM/energy is gravity. INDEED, TIME DILATION ultimately proves ON BALANCE that ELECTROMAGNETISM/energy is gravity; AS E=MC2 IS F=ma !!!! INSTANTANEITY is thus FUNDAMENTAL to what is the FULL and proper UNDERSTANDING of physics/physical experience, AS E=MC2 IS F=MA ON BALANCE !!!; AS ELECTROMAGNETISM/ENERGY IS GRAVITY !!! GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=mc2 IS F=MA; AS ELECTROMAGNETISM/energy is gravity. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution !!! Notice what is the fully illuminated (AND setting) Moon in DIRECT comparison with what is the orange Sun !!! Again, the stars AND PLANETS are POINTS in the night sky. A given PLANET (including WHAT IS THE EARTH) sweeps out EQUAL AREAS in equal times consistent WITH/AS F=ma, E=mc2, AND what is perpetual motion, AS ELECTROMAGNETISM/energy is gravity; AS E=MC2 IS F=MA ON BALANCE !!!!!! GREAT. Gravity IS ELECTROMAGNETISM/energy, AS E=MC2 IS F=ma !!! By Frank DiMeglio
@kaimenobeng87927 жыл бұрын
Amazing teaching skills, very well explained.
@anessa7105 жыл бұрын
David....I love you so much.....this concept has been hurting my brain for so long lol. but you helped me understand it! thank you! and thank you Sal!!!
@nicholasbraud19867 жыл бұрын
Thanks, I go back to this example quite frequently when I forget how to do these.
@weenewman81236 жыл бұрын
I love this guy so much.Amzing teaching skill. that next generation of sal has born.
@LL-kn1de6 жыл бұрын
Thank you David, you are awesome! One more question, can you make video for the question below: A playground slide is in the form of an arc of a circle that has a radius of 12 m. The maximum height of the slide is h = 4.0 m, and the ground is tangent to the circle as shown below in the figure. A 25 kg child starts from rest at the top of the slide and has a speed of 6.2 m/sec at the bottom. (a) What is the length of the slide? (b) What average frictional force acts on the child over this distance? (c) What is the normal force on the child when the child is the bottom of the slide? THANKS!!!
@tomiwajohnson34565 жыл бұрын
years later and this is still SO helpful, thank you so much!
@hysteriodx6 жыл бұрын
ENLIGHTENED!!!!!!!
@pdriaodra4 жыл бұрын
영어로 설명해주셔서 이해하기 조금 어려웠지만 엄청 유익하고 도움이 됩니다 감사합니다 !^^
@SuperMeatball12345 жыл бұрын
Saved my GPA, thank you!
@grantwood29557 жыл бұрын
What if your theta is unknown and you only know that the object in motion is travelling constantly, its mass, and the radius of the circle
@davisjohn-d6h6 жыл бұрын
Grant Wood 69 degrees to the z axis
@xavi87803 жыл бұрын
decompose Tension vectors and use arctan or instead use arcsin(radius/string length) to get the angle as well. I know this is 3 years late, but for those who are reading the comments right now lost I hope this helps.
@hrn89353 жыл бұрын
@@xavi8780 thanks i was also wondering about that
@pasindushavinda80664 жыл бұрын
Really helped to understand the concept. ✌️✌️
@heej84615 жыл бұрын
Hey man, i love you
@duncanram6 жыл бұрын
how would you deal with a situation where the connecting arm/rod had a mass (kg)?
@samsonridge45264 жыл бұрын
OOOOOOHHHHHHHHHH MY GOD.I JUST GOT SAVED.THANKS SIR
@duncanram6 жыл бұрын
A conical pendulum consists of a uniform straight bar (L) long and of mass (M), with a bob of mass (m) at its lower end. it rotates at (N rev/min) about a vertical axis through the upper end of the bar. How would you find the radius of rotation of mass (m)
@seth854694 жыл бұрын
this dudes a boss
@AlaizaF3 жыл бұрын
Thankyou so much. T T The way you explain is just so smooth
@alevelsdemystified34106 жыл бұрын
Awesome video.
@praisemugodi47333 жыл бұрын
I feel like im on top of a cloud.....☺☺thank you
@zeinabfarhat19202 жыл бұрын
saved my life!!
@vasunith96825 жыл бұрын
this question is the most basic question in India for anyone who is preparing for iit jee.......no offence and the teaching is really great and spectacular(CU)
@laurachiriac13366 жыл бұрын
GOOD TEACHERA AND TUTORIAL BUT HOW DO I DO FOR TWO MASSES SWINGING IN THE SAME HORIZONTAL CIRCLE.. I NEED HELP (AT DIFFRENT ANGLES)
@rochaksinghthapakaji83196 жыл бұрын
amzingly good
@ryanlang29215 жыл бұрын
You helped me so much!
@AlaizaF3 жыл бұрын
Thankyou so much!!!
@chrislee50444 жыл бұрын
You just saved my life.
@sers65615 жыл бұрын
I love you soo much. So So So So SO So SO SO SOmuch
@dzunimthombeni92186 жыл бұрын
well done,,I salute
@Cricketlover-sc8fm3 жыл бұрын
Thanks ☺️
@sababatamanna21175 жыл бұрын
Thank you!
@inthebackwiththerabbish5 жыл бұрын
I'm mesmerized 😂
@yash11525 жыл бұрын
Which software is used? And is a stylus & graphic tablet used too?
@subschallenge-nh4xp7 жыл бұрын
one year and you dont have any dislike you should be GOD
@weenewman81236 жыл бұрын
maybe because of your comment now it have 4 dislikes lol
@calvinmuchemwa38805 жыл бұрын
your comment triggered alot of shit
@sgpaints92574 жыл бұрын
On my practice exam they have this problem exactly, but they ask you to express Newton's second law in the radial and tangential directions. They end up with Fr = T - mgcos(theta) = m^2/r and Ft = -mgsin(theta) = mat I don't understand how the radial direction is not towards the center of the circle where the equation would be Fr = -Tsin(theta).
@bleuemoone87106 жыл бұрын
How do you solve for the angle if you are given the velocity???
@alidarwish4086 жыл бұрын
Great video
@graceb24047 жыл бұрын
Nice example~
@ainfatini28213 жыл бұрын
Hye can u do vertical circle for angled centripedal force?
@prod.hxrford38965 жыл бұрын
but what do you call the vertical component of the string tension?
@spartenx87435 жыл бұрын
Thank you💐💐💐💐💐
@yash11525 жыл бұрын
6:47 can theta be never equal to 90°?? Intro: I was stuck on a much more complex problem, and I thought of this case as it is the most simple case, but with angle theta = 90°. So, the next question was, what is balancing the mass here (when theta is 90)?? On a little bit searching, I found this: physics.stackexchange.com/a/277016 But this answer is voted -1 as of 2019.12.08 . So.... Also, just a quickie: this case is called a conical pendulum (for sake of a shorter & more precise search keyword 😃).
@tunauzel2 жыл бұрын
If theta equals to 90° ,the y-coordinate of tension equals zero.Thus there won't be any forces that makes total force of x coordinate zero. So it is not possible.
@johnnywong51506 жыл бұрын
@ 5:44, if we know two angles, you can determine the 3rd angle if it’s a right triangle 🥴
@NightLapseStudios3 жыл бұрын
what if you dont know the angle, but you know L and the velocity and g? is it impossible? to know the angle
@anc40026 жыл бұрын
How would you find the length of the string...
@EdwinFairchild3 жыл бұрын
ww@5:50 we do know that missing angle, 180-30-90 =60
@eolvra7 жыл бұрын
excellent
@prashanttmg54327 жыл бұрын
its ..awesome
@aldrichdias90866 жыл бұрын
if I had to find the time can I use Vc=2πr/t to find it or is there another way
@raselmahamudrobin41623 жыл бұрын
Nice
@Alexander0501016 жыл бұрын
Do you know how to calculate the extension of the string if you knew the time period and radius too?
@tunauzel2 жыл бұрын
If you know time period and radius you can calculate speed. And ıf you know speed you can calculate the x coordinate of tension. But if you want to find total tension force ,you also need to know the angle between it.
@mrchedda7 жыл бұрын
How to solve for the angle if given velocity?
@bleuemoone87106 жыл бұрын
how to do this?
@lordpinochetuttp38196 жыл бұрын
The method depends on what variables you have to work with. If you have the velocity, radius and mass, do the following: Calculate horizontal force by rearraging the velocity-centrifugal force equation. v^2/r = Fx/m , Fx = mv^2/r. Calculate the upward force by using classic F=ma. Fy=9.81m. Lastly, calculate the angle by using the tangent function. Divide the Fx by Fy and plug the result into the inverse tangent function. The answer is the angle which the rope makes with the vertical.
@CompositeAcoustic6 жыл бұрын
What happen to the yellow m in the first equation?
@mayankgoyal64956 жыл бұрын
thanku sir
@prankie7274 жыл бұрын
What happen if angle is 90 degree. In this scenario which force balance the weight (mg) of the body
@jayjalloh39125 жыл бұрын
Shouldn't T (y-axis) be sin30?
@bryanguzman63334 жыл бұрын
what if I wanted to find the angle given the speed, mass, and length of string
@tunauzel2 жыл бұрын
Firstly, you can find tension. Let us call them Ty and Tx. If you want to find the angle between them you can use Arctan(Tx/Ty) .
@mirzabaig55837 жыл бұрын
nice video
@arthurmorgan44605 жыл бұрын
Got an exam tomorrow and im only half way through what i should know. It's 6pm, looks like an all nighter. *nervous laugh
@Ashleenaccarato5 жыл бұрын
He forgot the extra 'm' at 6:40. You get (m^2*g)/cos(30). And I got 101.8 N
@josephsoo53185 жыл бұрын
Get good
@deekshasingh88827 жыл бұрын
how to solve for tension when angle is not given but mass length nd velocity is given
@scotty150027 жыл бұрын
deeksha singh You can find the X-component of the tension from the velocity and solve the angle measure from there.
@bleuemoone87106 жыл бұрын
Can you elaborate on this? I'm not quite sure how you'd go about doing that. I have the same setup as deeksha.
@joesiu49725 жыл бұрын
IM HERE FROM MY BOI AUSTIN
@lasokl37182 жыл бұрын
how do you find V if you dont know L or R?
@jaimed.campos21468 жыл бұрын
why isnt there a normal force on the FB diagram?
@radhab54087 жыл бұрын
If the object is in contact with the surface then only we will have normal force acting perpendicular to the surface
@arunimachakraborty45963 жыл бұрын
"Zeros are wonderful. They make calculation easier."😂😂😂
@andredubbs48544 жыл бұрын
you just saved MY ASS thank you
@danbeale68185 жыл бұрын
My Professor got this wrong in class today
@taykeith37954 жыл бұрын
6:48 is probably the reason you clicked this video. The rest is waffle
@taykeith37954 жыл бұрын
I never knew this video was gonna include a waffle recipe. Not grateful