You filmed this video when I was just a primary school children in 2009, and I become a university student and watch your video in 2017. It definitely OLD But GOLD!! That feeling is amazing!!
Wow. These are by far the clearest and most consistant videos I've found. For someone taking Physics C independent studies, they're a godsend, especially for things like this that weren't covered in B. Thank you so much. I might actually do well on the exam in May now.
First of all, your videos are great! I just wanted to point out one thing you didn't mention. For friction you used the formula for max static friction so the sphere must be on the verge of slipping. This problem can can be solved without this assumption but with a little more algebra.
At 1:13, to clarify, since the ball is not rotating there is zero static friction, but isn't it so that there still could be some dynamic/kinetic friction?
@Synonym4GOAT cos(theta) is always equal to the length of the adjacent side divided by the length of the hypotenuse. Sin(theta) is always equal to the opposite side's length divided by the length of the hypotenuse. In this video the force of gravity is the hypotenuse.
aren't 1) a=alpha*R and the 2)a in translational motion different ? Isn't 2) just the acceleration of the centre of mass and 1) can define acceleration at any point on a body ?
First, talking speed, v=(omega)R is true for a point at the rim (relative to the axis) and for the center of mass (relative to the earth). The same is true for "a". Assuming no slipping. For a point at R/2, the a =(alpha)R/2 and v=(omega)R/2 relative to the axis.
hello! your videos have been very helpful. however, i believe you made a slight error. Force of the static friction cannot be assumed to equal mu times normal force, only that the force of static friction is less than or equal to mu times normal force
You filmed this video when I was just a primary school children in 2009, and I become a university student and watch your video in 2017. It definitely OLD But GOLD!! That feeling is amazing!!
Wow. These are by far the clearest and most consistant videos I've found. For someone taking Physics C independent studies, they're a godsend, especially for things like this that weren't covered in B. Thank you so much. I might actually do well on the exam in May now.
This is probably the best channel for physics on UA-cam, but I wish there were more videos though.
Man I miss 2009, life was so much easier
Great video! What program did you use for the simulation at the beginning? I thought it was very revealing of the concepts.
First of all, your videos are great! I just wanted to point out one thing you didn't mention. For friction you used the formula for max static friction so the sphere must be on the verge of slipping. This problem can can be solved without this assumption but with a little more algebra.
At 1:13, to clarify, since the ball is not rotating there is zero static friction, but isn't it so that there still could be some dynamic/kinetic friction?
i thought no friction means no friction at all like no rotational, static or kinetic
@Synonym4GOAT cos(theta) is always equal to the length of the adjacent side divided by the length of the hypotenuse. Sin(theta) is always equal to the opposite side's length divided by the length of the hypotenuse. In this video the force of gravity is the hypotenuse.
What program is that?
All the examples are really clear! Thnx!
@lasseviren
what about mgsintheta ,does it not provide any torque?
bcoz its perpendicular to the radius.
What software are you using for the simulations?
sooo much more helpful than my teacher! thank you!!!
hello.u r video is very useful ...its very interesting nd understandable..thanks fr the video!
aren't 1) a=alpha*R and the 2)a in translational motion different ? Isn't 2) just the acceleration of the centre of mass and 1) can define acceleration at any point on a body ?
First, talking speed, v=(omega)R is true for a point at the rim (relative to the axis) and for the center of mass (relative to the earth). The same is true for "a". Assuming no slipping. For a point at R/2, the a =(alpha)R/2 and v=(omega)R/2 relative to the axis.
at 4:00 wouldn't a component of the weight supply some torque?
Whats the program called that you were using in the begging of the video?
a(lineer) is not equal to a(tangential) in this problem, isn't it? If it is, then mu=(1/3)tanx. But, it's not true in every case. Am I wrong?
hello! your videos have been very helpful. however, i believe you made a slight error. Force of the static friction cannot be assumed to equal mu times normal force, only that the force of static friction is less than or equal to mu times normal force
I think you forgot to bring the R on the bottom over when you found your final a{t}
2:25 "If it's salad... a salad sphere" is what I heard. LOL
I love this
@UchihaKat
I'm taking this course independently too!
and I couldn't agree with you more (:
Im pretty sure he has his x and y componets switched...
it should be for example a=(mgcos(o)-Ff)/m
I believe it its called Interactive Physics
nvm... btw these videos are very helpful great job
this was filmed when I was 9
Nope, Vince. He's correct, it's a= (mgsin(theta) -Ff)/m