Cross Product Torque (with a Cross Product Review)

[@rahuldubey7490]

Sir dont stop making videos we all are in support of you I am from india and i hve shared this channel in my scl grp all students of my class says that they had not learnt physics in this interesting way before

[@FlippingPhysics]

Thanks for the love my friend!

[@thetientran]

Wow I never knew we needed to know some linear algebra for Physics C!

[@julielangenbrunner9212]

Looks great! (I like how Billy is muttering while Bobby is listing off the matrix elements in the first example.)

[@FlippingPhysics]

Thanks. I am experimenting with having BBB say little things while the others are talking.

[@rudyberkvens-be]

I do not understand how the direction of the torque can be in the direction of the rotation axis. Then how can this axis turn around? It hàs to work perpendicular to it.

[@kartik7633]

Sir when we rotate keychain in our finger then why it moves upwards?? I hope you understood my question

[@carultch]

I did not understand your question. Perhaps you could be more specific in your description of the situation.

[@technotux7835]

Thanks man keep up the good work

[@FlippingPhysics]

Thanks, will do!

[@shyamdas6231]

Sir, I was looking for your videos on electromagnetism. I couldn't find them.

[@FlippingPhysics]

I do not have any ... yet

[@TheAnimammal]

If T is defined as T = r x F, and we change r, what happens to T does it stay the same and F changes? How can you claim that the definition of L = r x p works differently? What evidence do you need to be able to see the truth?

[@carultch]

If you change r while holding F constant, torque will also change. Suppose we leave the directions of r and F alone. torque will remain in the same direction and will change magnitude in proportion to the change in r. What is the claim that L = r cross p works differently for angular momentum? Because the mathematics of that formula doesn't work any differently. You may be confused because r also may impact other factors that come in to play in a problem you have in mind. If you look at an object's angular momentum from a reference point that is half the distance from the object as you previously examined, the angular momentum will be half as much. As an example, consider a rotating target assembly on a vertical axis, which has a target that is 1 meter from the axis, and another target that is 2 meters from the axis, which has a moment of inertia of 20 kg-m^2. The assembly is initially at rest, and is free to rotate after the collision. We fire a bullet perpendicular at each target, and the bullet has an inelastic collision with the target in both trials. The bullet is 10 grams and is fired with a speed of 100 m/s in both trials. Between trials, we bring the assembly to rest and remove the bullet. When you apply conservation of angular momentum to solve this problem, you get 0.05 rad/s for the trial that strikes the inner target, and 0.1 rad/s for the trial that strikes the outer target. About twice as much for the outer target, because the bullet's initial angular momentum was twice as much. The first trial has an angular momentum of 1 kg-m^2/s, and the second trial has an angular momentum of 2 kg-m^2/s. I did account for the bullet remaining part of the rotating assembly after the collision, and it ended up being in the rounding error with the data given.

[@gravysnake78]

ahh so the C in Physics C also means Linear algebra, good to know as a Physics 1 student

[@Johan-yy9pk]

thanks

[@J.R.Krishna]

❤❤

[@rudyberkvens-be]

Funny humor