If you were in India a temple would have been built in your name XD. Can't believe how underrated this channel is. One of the best for CONCEPT CLARITY 💯
Even though I used calculus to derive these equations on my own, I still could not intuitively conceive these formulae. Your video was so incredibly helpful! It also helps that you insinuate a classroom setting where each student comes to the answer on their own, so it makes it even more memorable. Thank you for putting these out on the internet!!
Thank you so much for this video! It really helped me understand Rotational Inertia very well! But I have a question: Why does the thin hoop have a larger rotational inertia than the hollow sphere? I keep getting confused by it.
Seams like there was a conceptual leap from the solid cylinder's fraction to the solid sphere's; I would have paused and looked at the differences in the shapes, or how the sphere tucks more of the mass in close. Other than that, great video!
@@itsjoeeeeeeYou ultimately need calculus to determine why these moments of inertia are what they are. The thin cylinder and thin ring are two trivial examples that don't need calculus, since it is immediately obvious that all of the mass is concentrated at the full radius. You can use this as a shortcut, so that your differential mass unit is a series of thin cylinders, that approximate the solid cylinder or solid sphere. You then set up a sum of the corresponding moments of inertia of these differential mass units, which becomes an integral when there are infinitely many of them. Here's the solution for the solid cylinder: Given a cylinder of radius R, length L, and density rho, set up the differential mass element as a thin cylindrical shell at position r and thickness dr. It will also have length L and density rho like the solid cylinder. This differential mass element can unwrap like the label of a can, and will have thickness dr, length L, and width 2*pi*r, consistent with the circumference. Its mass dm is therefore dm=2*pi*rho*L*r*dr. We know the moment of inertia of a thin cylinder is m*R^2, so the differential moment of inertia (di) of our differential mass element cylinder of mass dm and radius r: dI = dm*r^2 Substitute dm: dI = 2*pi*rho*L*r*r^2*dr Simplify: dI = 2*pi*rho*L*r^3*dr We can treat 2*pi*rho*L as a constant. Pull it out in front as we integrate: integral dI = 2*pi*rho*L* integral r^3*dr I = 2*pi*rho*L * integral r^3 dr integral r^3 dr = 1/4*r^4 + C Evaluate the difference from r=R to r=0, noting that the arbitrary constants C cancel: 1/4*R^4 Reconstruct: I = 1/2*pi*rho*L*R^4 Recall that rho is mass/volume, and that the volume of a cylinder is pi*R^2*L: I = 1/2*pi*M/V*L*R^4 Substitute V = pi*R^2*L I = 1/2*pi*M*L*R^4 / (pi*R^2*L) Cancel pi, L, and 2 of the R's. Simplify and we get our solution: I = 1/2*M*R^2
Why it is r2 in moment of inertia ....instead of r from axis of rotation ...I feel one r is enough to make rotation ...then why r square...pls help to understand .... I hope you will explain .... thanks
Just wondering how did you get (1/12)mL^2 for the moment of inertia of a thin rod rotating about its center of mass? Is that something that just needs to be memorized or something we can derive?
I derive that equation in this video: www.flippingphysics.com/thin-rod-rotational-inertia.html Please do not memorize moments of inertia. You need to understand how mass distribution affects rotational inertia and to be able to compare their relative magnitudes like we do in this video, however, you do not need to have the equations memorized.
A point mass has all of the mass concentrated in a volume of space that is negligible. In other words, you can treat the object as if all of its mass is at the same point, and still get the same result as if you considered its distribution. A point mass is usually an approximation to keep the math simple, and not necessarily something that exists in reality. An extended object has a distribution of mass that is important to the problem in question, and we need to consider the location and distribution of mass to get the correct answer. As an example, consider a 100 gram (m) apple hanging on a string of negligible mass with its center 1 meter (L) below the top of the string, with a radius of 3 cm (r). In other words, a simple pendulum. Assume 3 significant digits. If we treat the apple as a point mass, the moment of inertia about the top of the string would be 0.1 kg-m^2, calculated via m*L^2. If we account for the apple's size, and treat it as a uniform solid sphere, we will calculate moment of inertia about the top of the string through 2/5*m*r^2 + m*L^2, and get the moment of inertia to be 0.100036 kg-m^2, which rounds to the same answer we had before. The apple might as well be treated as a point mass, since its size is insignificant compared to the 1 meter string length. Now, let's shorten the string to so that the center of the apple is 10 cm below the top of the string. Treating the apple as a point mass, gets us a moment of inertia of 0.001 kg-m^2. Treating the apple as an extended object, accounting for its shape as a uniform solid sphere, we will instead get 0.001036 kg-m^2, which we'd need to round to 0.00104 kg-m^2 for 3 significant figures. You now see that the 3 cm radius of the apple makes a difference, compared to what it would be if treated it as a point mass.
If you were in India a temple would have been built in your name XD. Can't believe how underrated this channel is. One of the best for CONCEPT CLARITY 💯
Maybe one day I will visit India!
💯%😊
Even though I used calculus to derive these equations on my own, I still could not intuitively conceive these formulae. Your video was so incredibly helpful! It also helps that you insinuate a classroom setting where each student comes to the answer on their own, so it makes it even more memorable. Thank you for putting these out on the internet!!
You are welcome!
i love how the guy in the middle is always high.
Another one of the great video!
Thank you again sir!
Beautifully explained
And for the first time I enjoyed and learnt seriously
Good luck mate
You are the best, Mr P.
I salute the best Physics teacher.
thank you your the best teacher
I can certainly relate with that Beau.
Thanks For your Contribution to our understanding of physics.
Thank you for appreciating it.
I like the way of teaching.
That's very informative, thanks!
You're welcome!
0:59 will you please post calculus based explanation for calculation I
My in-class lectures about this are here :ua-cam.com/video/X36BAPvxWQs/v-deo.html
@@FlippingPhysics thank you
Thank you so much for this video! It really helped me understand Rotational Inertia very well! But I have a question: Why does the thin hoop have a larger rotational inertia than the hollow sphere? I keep getting confused by it.
Nice explanation sir
very cool approach indeed - thank you very much
Really worth watching...great effort
You are an amazing teacher!!!
Seams like there was a conceptual leap from the solid cylinder's fraction to the solid sphere's; I would have paused and looked at the differences in the shapes, or how the sphere tucks more of the mass in close. Other than that, great video!
The video was already way too long and I decided not to dwell on that particular issue, though I agree it is the least obvious of the comparisons.
@@FlippingPhysics Where could I find an explanation for how the shape of sphere and cylinder affects its moment of inertia?
@@itsjoeeeeeeYou ultimately need calculus to determine why these moments of inertia are what they are. The thin cylinder and thin ring are two trivial examples that don't need calculus, since it is immediately obvious that all of the mass is concentrated at the full radius. You can use this as a shortcut, so that your differential mass unit is a series of thin cylinders, that approximate the solid cylinder or solid sphere. You then set up a sum of the corresponding moments of inertia of these differential mass units, which becomes an integral when there are infinitely many of them.
Here's the solution for the solid cylinder:
Given a cylinder of radius R, length L, and density rho, set up the differential mass element as a thin cylindrical shell at position r and thickness dr. It will also have length L and density rho like the solid cylinder.
This differential mass element can unwrap like the label of a can, and will have thickness dr, length L, and width 2*pi*r, consistent with the circumference. Its mass dm is therefore dm=2*pi*rho*L*r*dr.
We know the moment of inertia of a thin cylinder is m*R^2, so the differential moment of inertia (di) of our differential mass element cylinder of mass dm and radius r:
dI = dm*r^2
Substitute dm:
dI = 2*pi*rho*L*r*r^2*dr
Simplify:
dI = 2*pi*rho*L*r^3*dr
We can treat 2*pi*rho*L as a constant. Pull it out in front as we integrate:
integral dI = 2*pi*rho*L* integral r^3*dr
I = 2*pi*rho*L * integral r^3 dr
integral r^3 dr = 1/4*r^4 + C
Evaluate the difference from r=R to r=0, noting that the arbitrary constants C cancel:
1/4*R^4
Reconstruct:
I = 1/2*pi*rho*L*R^4
Recall that rho is mass/volume, and that the volume of a cylinder is pi*R^2*L:
I = 1/2*pi*M/V*L*R^4
Substitute V = pi*R^2*L
I = 1/2*pi*M*L*R^4 / (pi*R^2*L)
Cancel pi, L, and 2 of the R's. Simplify and we get our solution:
I = 1/2*M*R^2
Very good
In my life, first tine I understood rotational physics.🥰🥰🥰🥰
Very nice 😀 explanation sir
Thank you.
Thanks sir
You are welcome.
Thank you so much! My final is around the corner
wow, that's a great video!
Glad to help!
Thank you so much sir..
It was excellent.thank u so much 👍👌
Why it is r2 in moment of inertia ....instead of r from axis of rotation ...I feel one r is enough to make rotation ...then why r square...pls help to understand .... I hope you will explain .... thanks
I explain it in this video. ua-cam.com/video/gdUkXvBWdcs/v-deo.html
THIS VIDEO SAVED MY LIFE THANKYOU!!!!!!!
Hyperbole appreciated.
You are welcome.
O really !!!
I know this is 5 years late but I rlly like ur shirt
Thanks you are the best.
As always. I appreciate your support.
Bo is my favourite.
This tells me a lot about you. 🙃
Thank u sir❤️
Just wondering how did you get (1/12)mL^2 for the moment of inertia of a thin rod rotating about its center of mass? Is that something that just needs to be memorized or something we can derive?
I derive that equation in this video:
www.flippingphysics.com/thin-rod-rotational-inertia.html
Please do not memorize moments of inertia. You need to understand how mass distribution affects rotational inertia and to be able to compare their relative magnitudes like we do in this video, however, you do not need to have the equations memorized.
@@FlippingPhysics thank you!
great
thanks
you are cool my friend
super fun :))))
flipping useful.
Pls help me
How you make difference between point mass
And extended object
A point mass has all of the mass concentrated in a volume of space that is negligible. In other words, you can treat the object as if all of its mass is at the same point, and still get the same result as if you considered its distribution. A point mass is usually an approximation to keep the math simple, and not necessarily something that exists in reality.
An extended object has a distribution of mass that is important to the problem in question, and we need to consider the location and distribution of mass to get the correct answer.
As an example, consider a 100 gram (m) apple hanging on a string of negligible mass with its center 1 meter (L) below the top of the string, with a radius of 3 cm (r). In other words, a simple pendulum. Assume 3 significant digits. If we treat the apple as a point mass, the moment of inertia about the top of the string would be 0.1 kg-m^2, calculated via m*L^2. If we account for the apple's size, and treat it as a uniform solid sphere, we will calculate moment of inertia about the top of the string through 2/5*m*r^2 + m*L^2, and get the moment of inertia to be 0.100036 kg-m^2, which rounds to the same answer we had before. The apple might as well be treated as a point mass, since its size is insignificant compared to the 1 meter string length.
Now, let's shorten the string to so that the center of the apple is 10 cm below the top of the string. Treating the apple as a point mass, gets us a moment of inertia of 0.001 kg-m^2. Treating the apple as an extended object, accounting for its shape as a uniform solid sphere, we will instead get 0.001036 kg-m^2, which we'd need to round to 0.00104 kg-m^2 for 3 significant figures. You now see that the 3 cm radius of the apple makes a difference, compared to what it would be if treated it as a point mass.
I think aarti Sangwan from india and are you from flipins
Mr clutch
i hate physics that is all
U r welcome
Thanks sir
You are welcome.