Was staring at my paper for about an hour trying to figure out what to look for. Can’t believe my teacher expected me to make this kind of connection. I wouldn’t have figured this out in 100 years. Thank you very much!!!!
Great lecture. It helped my mood and morale to know that, as an student, you couldn't solve the problem and did not know where to begin. This was precisely my position until I watched your video. Sincere thanks from a chemist relearning mechanics.
This would be very weird - imagine that you were taking a walk and chilling out, and then all of a sudden some person came falling upwards through a hole in the ground, HAHA!
The rotation is just a mathematical analogy, but in reality the "angular" frequency does not require rotation in the real physical world. Simple harmonic motion with a mass oscillating on the end of a spring is such an example.
Aren't there two opposing forces acting on the falling vehicle in this frictionless tunnel ? One is the gravitational attraction force toward the center of the Earth exerted by the shrinking sphere and the other one also a gravitational attraction force exerted by the mass contained in the growing spherical cap above the vehicle . It seems that all models recognize only the first force and completely ignore the second. Why? I welcome your responses.
The matter outside the sphere defined by its position, does not apply a force on the vehicle, only the mass inside the shrinking sphere as it approaches the center of the Earth.
Michel: Please explain, why wouldn't it? If a small object with mass m was touching the center of a flat surface of a huge spherical cap with mass M floating somewhere in the empty space, would you argue that Newton's Law of Gravity (F =GmM/x^2) does not apply if, at the same time, this small object was attracted by a nearby sphere? I would argue that it does. The center of gravity for the cap would be the same as its geometrical center, i.e. z = (3/4)(2R-h)^2/(3R-h) where h = cap height and z = centerline distance from flat base. Note that in this case x = r + z where r is the distance of the small object from the sphere's center.
Think about it this way. If you are at the center of the Earth. All the mass around you pulls on you but all the forces will cancel and thus there is no force of gravity at the center. This same principle holds no matter where inside the sphere you are.
@@MichelvanBiezen Thank you for your explanation, however, why would it be wrong to assume that the gravity force balance acting on the vehicle falling to the center of the Earth is determined by an expanding spherical cap above the vehicle (F upward) and by a shrinking truncated sphere below the vehicle (F downward)? The motion would still be harmonic, although math-wise this model is complex and probably requires a numerical solution. Sorry, but I'm stubborn and have spare time while in "house arrest" during this pandemic.....
I like you tenacity. Typically I encourage my students not so spend too much time trying to figure what they did wrong, but rather learning how to do it correctly. As it turns out we have an entire playlist of videos that shows how to calculate the gravity caused by mass distributions which will show you why it works that way: PHYSICS 18.1 GRAVITY WITH MASS DISTRIBUTION
When we are dealing with a spring, the k represents the spring constant. k = (force on spring) / (elongation of the spring) Here we need to find the "equivalent" k for the Earth system.
Thanks for the post. Is there any reason why the orbital velocity of the object on Earth's surface through an arc length of 1/4 the Earth's circumference is the time it takes for half this trip straight through?
that's not quit right. you forgot for F^1 to gave the force of the mass of the earth already covered with the force of the remaining mass therefore, f^1=GM^1m/R^2 -G(M-M^1)m/(r-R^2/2)^2 as for F^1=F_1-F_2
Yeah@@MichelvanBiezenman.. Binary Beefburgers are the way forward when it comes to Space Travel.. B=MC Squared... Burgers = Mac Sauce and Chocolate Shake to the Crack Pipe Power of Two Baby.!! 0️⃣🤜🧢🍔💥😜1️⃣
Was staring at my paper for about an hour trying to figure out what to look for. Can’t believe my teacher expected me to make this kind of connection. I wouldn’t have figured this out in 100 years. Thank you very much!!!!
Wonderful!
Thank you so much for this. There's almost nothing about this type of problem on YT besides just qualitative descriptions of it.
Great lecture. It helped my mood and morale to know that, as an student, you couldn't solve the problem and did not know where to begin. This was precisely my position until I watched your video. Sincere thanks from a chemist relearning mechanics.
if i could i would subscribe 100 times !!!! YOU ARE THE BEST I THANK YOU.
thank you!! I study engineering and your videos have helped me so much
Absolutely remarkable, saved me on my assignment. Thank you!
Glad it helped!
This would be very weird - imagine that you were taking a walk and chilling out, and then all of a sudden some person came falling upwards through a hole in the ground, HAHA!
Sir if there is no rotation here so how can we take omega
The rotation is just a mathematical analogy, but in reality the "angular" frequency does not require rotation in the real physical world. Simple harmonic motion with a mass oscillating on the end of a spring is such an example.
Aren't there two opposing forces acting on the falling vehicle in this frictionless tunnel ? One is the gravitational attraction force toward the center of the Earth exerted by the shrinking sphere and the other one also a gravitational attraction force exerted by the mass contained in the growing spherical cap above the vehicle . It seems that all models recognize only the first force and completely ignore the second. Why? I welcome your responses.
The matter outside the sphere defined by its position, does not apply a force on the vehicle, only the mass inside the shrinking sphere as it approaches the center of the Earth.
Michel: Please explain, why wouldn't it? If a small object with mass m was touching the center of a flat surface of a huge spherical cap with mass M floating somewhere in the empty space, would you argue that Newton's Law of Gravity (F =GmM/x^2) does not apply if, at the same time, this small object was attracted by a nearby sphere? I would argue that it does. The center of gravity for the cap would be the same as its geometrical center, i.e. z = (3/4)(2R-h)^2/(3R-h) where h = cap height and z = centerline distance from flat base. Note that in this case x = r + z where r is the distance of the small object from the sphere's center.
Think about it this way. If you are at the center of the Earth. All the mass around you pulls on you but all the forces will cancel and thus there is no force of gravity at the center. This same principle holds no matter where inside the sphere you are.
@@MichelvanBiezen Thank you for your explanation, however, why would it be wrong to assume that the gravity force balance acting on the vehicle falling to the center of the Earth is determined by an expanding spherical cap above the vehicle (F upward) and by a shrinking truncated sphere below the vehicle (F downward)? The motion would still be harmonic, although math-wise this model is complex and probably requires a numerical solution. Sorry, but I'm stubborn and have spare time while in "house arrest" during this pandemic.....
I like you tenacity. Typically I encourage my students not so spend too much time trying to figure what they did wrong, but rather learning how to do it correctly. As it turns out we have an entire playlist of videos that shows how to calculate the gravity caused by mass distributions which will show you why it works that way: PHYSICS 18.1 GRAVITY WITH MASS DISTRIBUTION
Seems like something Jules Verne would be interested in.
This is on my test!!
Great job 😊
keep up the good work
Why would we assume it is a restoring force? How does that make intuitive sense?
A "restoring" force is defined as a force that will push the object to the equilibrium point, which in this case would be to the center of the Earth.
Were does gravity start?
If you chose not to assume a constant density, could you not assume a reasonable function for it and use it in this derivation?
Yes, you can use a function approximating the density, but then the problem would not be a pure oscillator
@@MichelvanBiezen 👍
Brilliant as always.
thanks a lot sir
👍big thumbs up
can i ask k in f equation stands for what
When we are dealing with a spring, the k represents the spring constant. k = (force on spring) / (elongation of the spring) Here we need to find the "equivalent" k for the Earth system.
Very good 🙏🙏🙏🙏🙏
yes, this is a very interesting problem.
Grateful to you sir !
Glad it was helpful. 🙂
@@MichelvanBiezen a lot..
Thanks for the post. Is there any reason why the orbital velocity of the object on Earth's surface through an arc length of 1/4 the Earth's circumference is the time it takes for half this trip straight through?
The time it takes to go through the Earth is the same as the time for a satellite to go around the Earth once. (A one to one relationship).
that's not quit right. you forgot for F^1 to gave the force of the mass of the earth already covered with the force of the remaining mass therefore, f^1=GM^1m/R^2 -G(M-M^1)m/(r-R^2/2)^2 as for F^1=F_1-F_2
Have a look at Gauss's law
great video!
adamsın hocammmm
great! great! thanks a lot! :-)
Tunnel should be frictionless sir and it can be done by applying principle of electromagnetism
can you be my professor instead of my bad one
why is this considered simple harmonic motion if there was no spring
As long as there is a force that acts like F = - kx there will be simple harmonic motion. That force does not need to come from a spring.
I love your bow ties
Sizzling Sublick's Superfast Superman Sausages Baby.!! 0️⃣🍔1️⃣
Interesting analogy! 🙂
Yeah@@MichelvanBiezenman.. Binary Beefburgers are the way forward when it comes to Space Travel.. B=MC Squared... Burgers = Mac Sauce and Chocolate Shake to the Crack Pipe Power of Two Baby.!! 0️⃣🤜🧢🍔💥😜1️⃣
keep up the good work