Hi. I think you did a fine job solving this problem. I also enjoyed watching your next video, of the simulation. I want to thank you for all the work you've done on the exercises that are posted on my website. You've drawn attention to the fact that the exercises section of the website, which is 17 years old now, was originally generated using a (now defunct) code generator, before I knew anything about coding for the web - it's not very pretty, out of date (using deprecated tags, for example), and badly needs attention. I am working with one of our volunteers now, to rewrite at least the exercises part of the site (and maybe the rest of the old FrontPage-generated code) so the content can be presented in a more user-friendly way, particularly for mobile users, and to make it easier to maintain and expand in the future.
+15:45 I am confused! At 15:45, the distance the ball moves is shown as 12/49 v0^2/ug. When I did my own calculation, the ball moves this distance due to slipping. IN this respect, the derivation is 100% correct. However, as soon as it starts to slip, the ball also starts to rotate. Initially slowly but faster later. Eventually, the ball rolls fast enough and will not slip any more. I believe that the center of the ball moves due to the rotation while it is slipping. According to my derivation, this term becomes 5/49 v0^2/ug. The reason why I am confused is if tthe correct answer is 12/49 v0^2/ug or if it is the sum of the two motions to give 17/49 v0^2/ug (from 12/49+5/49). What do you think?
Yes, you want to use calculus in this case as well as in a more general case. In this case, the accelerations in the linear and rotation are a constant. The author used the well know equations, v=vinit+(acceleration)*t and d=1/2(acceleration)*t^2 + vinit*t. I have checked the result shown in +15:45 using calculus and found that the derivation was correct. Please, refer to my comment posted.
Hmm, I solved this problem in a slightly different way and got the answer (5v^2)/(18mu*g), which is the same except for the numerical coefficient. Idk what happened, but something about these weird fractions is off putting Edit: I used the wrong moment of inertia, which explains the difference in numerical coefficients
![Felix "Unfair" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/Oswujxm2Ag0/mqdefault.jpg)
Hi. I think you did a fine job solving this problem. I also enjoyed watching your next video, of the simulation. I want to thank you for all the work you've done on the exercises that are posted on my website. You've drawn attention to the fact that the exercises section of the website, which is 17 years old now, was originally generated using a (now defunct) code generator, before I knew anything about coding for the web - it's not very pretty, out of date (using deprecated tags, for example), and badly needs attention. I am working with one of our volunteers now, to rewrite at least the exercises part of the site (and maybe the rest of the old FrontPage-generated code) so the content can be presented in a more user-friendly way, particularly for mobile users, and to make it easier to maintain and expand in the future.
+15:45
I am confused! At 15:45, the distance the ball moves is shown as 12/49 v0^2/ug. When I did my own calculation, the ball moves this distance due to slipping. IN this respect, the derivation is 100% correct. However, as soon as it starts to slip, the ball also starts to rotate. Initially slowly but faster later. Eventually, the ball rolls fast enough and will not slip any more. I believe that the center of the ball moves due to the rotation while it is slipping. According to my derivation, this term becomes 5/49 v0^2/ug. The reason why I am confused is if tthe correct answer is 12/49 v0^2/ug or if it is the sum of the two motions to give 17/49 v0^2/ug (from 12/49+5/49). What do you think?
Thanks! Really appreciate it helped me understand the problem a lot!
Isn't the Ffk clockwise, making it negative? Thanks for the video!
keep the playlist title for the sake of organization please!
Thank you a ton, this really helped!
Glad it helped!
No Calculus?
Yes, you want to use calculus in this case as well as in a more general case. In this case, the accelerations in the linear and rotation are a constant. The author used the well know equations, v=vinit+(acceleration)*t and d=1/2(acceleration)*t^2 + vinit*t. I have checked the result shown in +15:45 using calculus and found that the derivation was correct. Please, refer to my comment posted.
Hmm, I solved this problem in a slightly different way and got the answer (5v^2)/(18mu*g), which is the same except for the numerical coefficient. Idk what happened, but something about these weird fractions is off putting
Edit: I used the wrong moment of inertia, which explains the difference in numerical coefficients
glad you figured it out.