+Alex Zilio Thank you Alex for your kind words. I am delighted that my lectures are "extremely" ,helpful for you.

I completely agree.

@@frankdimeglio8216 in your way

My pleasure!

Yes, you are right.

Excellent!

Hi

How your jee exam went? Can i prefer his lectures for my jee prep?

Lectures by Walter Lewin. They will make you ♥ Physics. sir, the link is down. please check.

You're very welcome!

Which university do you study

Thank you for your kind words.

For the love of physics

class 7 e ami physics er namE jantam na 🙄

You are very welcome

👍🏻

Best of luck

thank you Mark!

:)

You are most welcome

Wonderful!

SUPERRRRR en.wikipedia.org/wiki/Five_hundred_meter_Aperture_Spherical_Telescope

If you have finished 8.01, 8.02 and 8.03 then go to MIT OCW and watch 8.04 and after that 8.05. 8.05 is lectured by Barton Zwieback. He is a very good lecturer, I have attended several of his lectures at MIT.

she did not get the Nobel Prize instead she got a $3.5 million prize which is way more than a Nobel Prize. I have congratulated her!

Thank you so much for correcting me, Walter: I won't make that mistake again. She *deserves* the recognition and more all the same, but the Nobel snub still rankles. What amazes me is that (as you and others have described her) she's humble, gracious and generous enough to let all that slide *and* offer up her huge cash prize to support others. Yeah... a woman can be a mensch. :-) :-) :-)

Best of luck!

If an object of mass m is rotating about its com with angular momentum A and if it is also translating with velocity V, then the angular momentum relative to any point P is A + m*dXV, d is the position vector from P to the com. X indicates cross product.

Spin angular momentum of the Earth is an intrinsic property it's omega*I. Orbital angular momentum of the Earth is ONLY conserved relative to the Sun, NOT relative to any other point. If you choose a point somewhere on the orbit, when the Earth is at that point the orbital angular momentum relative to that point is ZERO.

yes, that is correct torque=I*alpha. alpha is dw/dt

If its rotational inertia remains unchanged (individual rigid body), it requires a net torque to act upon it, in order to have an angular acceleration. An object can have angular acceleration without torque, while changing its rotational inertia. Unlike linear inertia (i.e. mass), that requires changing the identity of the body to change its mass (like a rocket loosing its propellant mass, or farm equipment collecting a harvest), rotational inertia can change just by redistributing the mass. As you can see with the turntable demonstration in this lecture.

Manish Singh You are welcome

Manish Singh you are welcome. This website "Lectures by Walter Lewin. THey will make you love Physics" is the ONLY complete site. It has all my lectures. enjoy! \\/\///////////

+Lectures by Walter Lewin. They will make you ♥ Physics. Would you ever consider doing another complete lecture series? These are so fantastic.

the answer is in my 8.01 lectures. WATCH them or use google

Glad to hear that

angular momentum is r cross p r and p are vectors

The moment of inertia of a hoop is Mr^2, and the Cylinder is 1\2mr^2.

I sent you a revised version of my earlier answer. Plse read it.

total energy (KE + PE) is conserved. When the star implodes the PE decreases (the smaller R the less PE) thus KE goes up.

@@lecturesbywalterlewin.they9259 so to work out the loss in PE, work out the increase in rotational KE?

I cover all this in my 8.01 lectures

yes it increased. As I pulled my arms in, I had to do work.

*This channel was created in Febr 2015 by my Dutch friend Daniel Dekkers.* It has become way more popular than "For the Allure of Physics" (created in Dec 2014) which also carries my 94 MIT course lectures + my Farewell Lecture at MIT "For the Love of Physics" of May 16, 2011. That lecture alone has been viewed by more than 6 million people.

Sir, you have always been an inspiration to me and your lectures always motivate me to pursue my dream of getting into research and studying astronomy.....only the fact that you replied makes me feel so special. Believe it or not but this is like a blessing to me. -Regards

watch my 8.01 lecture where I cover linear momentum

physics.stackexchange.com/questions/26515/what-is-exactly-the-density-of-a-black-hole-and-how-can-it-be-calculated

:)

:)

use google, simple math

use google

@@lecturesbywalterlewin.they9259 sir only if you have written neutron star in the title of the video google will give me the result. I checked all the titles and nowhere stars were mentioned

please learn how to use google en.wikipedia.org/wiki/Neutron_star section 2.3

omega^2*r is not angular acceleration. That is centripetal acceleration, which is unrelated to alpha.

+Harish Yadav downward on earth, but toward the larger body in space. I would watch earlier lectures for this (gravity is actually attractive force between all objects

derivating a position vector (r) with respect to time gives a velocity :)

Bhanu Sharma Thank you Bhanu

really from dumb to pro .... credit goes to sir walter lewin ... n to ... thnxx for uploading whole course ...good work

Bhanu Sharma click on "playlist" you view wayyyyyyy more than my course lectures alone!

yup...i saw ur playlist ...but the problem is the all have japanese title...nd i don't know japanese language...all i know is ...."watashi wa bhanu te imashu"...:)

all have Japanese titles ???????????? For me all have English titles. Why are your Japanese? This is absurd.

if an object is in pure translation then there is angular momentum relative to all points except points on the "straight" line of the movement. Depending on the problem you need to solve this angular momentum can be CRUCIAL and can not be ignored.

+Srujan Govindu Physics Hans C. Ohanian 2nd edition W.W. Norton & Company ISBN 0-393-95748-9

x X y = z that's a righthanded coordinate system. PERIOD it does not matter how you draw it as long as x X y = z.

@@lecturesbywalterlewin.they9259 okay, thank you

1st year college

17 years appx age I guess?

@@sadhgurusfunniestandwittie3620 Around 19 or 20

use google

How much are you scoring in Physics in JEE Advanced Mock Tests per paper of around 66 marks , and which coaching institute?

:)

In the absence of external torque, angular momentum is conserved. Watch my "ice skater" demo. I*omega is conserved. If I increase I omega goes down and vice versa.

kzfaq.info/get/bejne/eLaZrbR1rsm1Z2Q.html

Good question. To understand what happens to the rotational kinetic energy when his arms extend and his body slows down when rotating, consider a linear motion example. Suppose there is a 1 kg bottle of water on top of a 1 meter table above the floor. You pick it up, and carry it to set it gently on the floor. It starts at rest, and ends at rest, so something must've happened to the 9.8 Joules of potential energy in this example, as that energy cannot be in the bottle's kinetic energy. Your muscles generate thermal energy when they act as brakes for objects in motion, absorbing the work done on them, and dissipating the energy as heat. That's what happens to his rotational kinetic energy as well, when he extends his arms while rotating on that platform. The kinetic energy becomes work done on his muscles, and when work is done on a person's muscles, heat is dissipated. Since our bodies aren't built for regenerative braking of objects in motion, we also get tired from doing negative work on objects in motion, just as we would get tired from doing positive work to put objects in motion.

It's my pleasure

It's the result of Newton's 3 laws which have all been verified 10^99 times. *In the absence of an external torque angular momentum is conserved.*

Most welcome

It's NOT 10^46 W it's J. It's the change in gravitational potential energy.

Sorry! Slip of the mind. But in any case where can you begin in order to find this number?

Take a spherical mass M of radius R2. Then change the radius of the star to R1 (R1

Thank you sir!

I cannot add to the clarity of my lecture in which I define Grav PE (zero at infinity). --MmG/r (r being the distance of m to the center of M). Watch my lecture again or use google

they evolve in a natural way from Newton's three laws. If you view my 8.01 lectures you should be able to see that for yourself.

The work (negative or positive) is done by the skater when she changes her moment of inertia. When pulling her arms in she will have to do positive work.

Thank you so much for answering!!

The masses are being treated as point masses, rather than a uniform cylinder about the axis of rotation. Even if you did treat the masses in his hands as solid uniform cylinders, the parallel axis theorem would render their own size insignificant, such that they might as well be treated as point masses.

Half mr2 is for a disk bt here in the case we consider a single inst. object

+pajaczekps3 Let's take a satellite in circular orbit around the Earth (speed V). That satellite has a total energy which is negative. If it were positive it would not be a bound orbit. The total energy is the sum of KE (which is >0) and PE (which is

+Lectures by Walter Lewin. They will make you ♥ Physics. Thank you Professor :)

yes, the stability of a bike depends on angular momentum of the wheels. However, that not the only thing. Use google

Lectures by Walter Lewin. They will make you ♥ Physics. Does static friction force matter?

Lectures by Walter Lewin. They will make you ♥ Physics. I think delta U should be negative because W gravity = delta K = -delta U => gravitational potential energy converted to kinetic energy. Did I misunderstand something?