MrJBphysics

[MrJBphysics @MrJBphysics]

@elqam Yes you’re absolutely correct. I noticed that a few weeks ago and smacked myself for that 🤦🏻‍♂️ Fortunately the primary lesson is about experimental design, and correct rotational inertia’s will be given as needed in unit 7 where it matters most

[MrJBphysics @MrJBphysics]

@AndrewSenseney ...manually calculate the value. If it's a common shape like a hoop, disc, sphere, etc. then you can use the known pre-determined M & R relationships. If it's an 'unusual' shape, then it becomes far more complex to calculate, so finding it experimentally (above) is the way to go.

[MrJBphysics @MrJBphysics]

@AndrewSenseney For angular momentum, you need that angular speed AND the rotational inertia of the object. For "I" there isn't a way to directly measure it... you would have to indirectly find it, by applying known torques and measuring the resulting angular accelerations...OR...

[MrJBphysics @MrJBphysics]

@mantej05 You’re welcome!

[MrJBphysics @MrJBphysics]

@mantej05 I’m pretty sure I went through the answers (with brief explanations) on the next video. It may be after the next lesson though (at the end of the video)...I can’t remember.

[MrJBphysics @MrJBphysics]

@fancyview And the condition is no longer met. Like in this case. I hope this helps 👍🏻

[MrJBphysics @MrJBphysics]

@fancyview In conclusion. That derived equation is a specific condition to relate v and R for an object IN circular orbit. A new orbit requires a new speed, but that doesn’t mean just because you change one the other must adjust appropriately. Instead, adjusting one just ruins the circ orb

[MrJBphysics @MrJBphysics]

@fancyview ....that doesn’t mean just because the speed decreases, it can move itself to a higher orbit. A higher orbit requires MORE energy... and if the satellite slows down, it has less energy, and it will definitely “fall”...

[MrJBphysics @MrJBphysics]

@mantej05 And at r2, the speed there is actually greater than the v that your equation would give you. So v doesn’t equal sqrt(GM/r)...at either location...if it did...then it would be in circular orbit and wouldn’t be changing R values

[MrJBphysics @MrJBphysics]

@mantej05 Unfortunately no... the equation you derived is for the speed required for a uniform circular orbit. Because this is in an elliptical orbit it means that that condition is not met. In fact, at R, v is less than what that equation would give you (which is why it doesn’t stay at R)

[MrJBphysics @MrJBphysics]

@mantej05 So the tangential speed on the outer edge is completely independent of the linear speed of the disc across the floor. So now let’s address the only way the linear speed of the object can change: with the application of a net force. Horizontal F’s are balanced so no linear acc

[MrJBphysics @MrJBphysics]

@mantej05 V=rw compares the tangential speed (relative to the CoM) to the angular speed. That doesn’t mean that has to be the linear speed of the CoM relative to the ground though... those 2 are only the same when something is rolling w/out slipping (not what we have here)

[MrJBphysics @MrJBphysics]

@RileyJeffries1 @JacobsPhysics Woo!! Great news

[MrJBphysics @MrJBphysics]

@AlinaSterenfeld @GregJacobs That’s great news!

[MrJBphysics @MrJBphysics]

@DanielZou4 @JacobsPhysics Good luck Daniel!