Recent content by Justforthisquestion1

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Does gravity produce torque around the cm? It should not?

Yes because angular momentum and linear momentum are conserved separately. Now: i have a time T in which the rod rises up and falls down again, and does exactly one revolution. So omega * T = 2 * pi and 0.5 * m * v^2 = m * g * h and also v * t - 0.5 * g *T^2 = 2* h This is because it needs to...

Gravity is to be neglected during the initial acceleration For the rest please kindly look to my entry before this one

No! I had it all wrong. So angular momentum and linear momentum are conserved separately. So for v and the conservation of linear momentum i can say p = m * v --> F * delta t / m = v But for omega and the conservation of angular momentum i can say p * x = I * omega omega = p * x / I Edit: Okay...

Thats kind of a tricky question. i am using the cm as axis of rotation? x * p = angular momentum? --> x * p = l * omega with p = F * delta t an I = 1/12 * m * l^2 we get: omega = 12 * x * F * delta t / (m * l^2) and then i would get the speed of cm by multiplying omega with l? Or do i use F *...

Thank you for both your answers! I think what a dont understand is, is whether the point of force application has to be included when calculating v or omega. Also the body will translate and rotate which confuses me even more. So lets say: F*delta t = m * v --> v = F * delta t / m and since i...

Honestly i have very little idea. F * delta t = p F * delta t /m = v So i know the speed of the rod And i know that however high the rod is supposed to go, when its back down it should have done excactly one revolution. I have the feeling that I should So probably i have to use something like...

Yes that is correct. Okay i will do it for ccw and then forget about it :D

Got it thank you so much Now i can forget about yoyos forever!!!

In my mind, the solution for ##theta = 0## should be ##Fmax= (mu * m * g * 3*R) / (2 * b +R)

Almost! The -2b are the problem! it should be +2b! Thank you! i didnt think of this easy yet effective method to test this solution Do you have any idea where the mistake could come from? If not then i will write my solution on a paper and scan it!

First of all i used ##a## = ##alpha## * ##R## So i switched the ##a## in m * ##a## in Newtons second law for ##alpha## * ##R## so now i have ##m * alpha * R## = F * cos(theta) - F(s) then i simply adjust that to ##alpha## = (F * cos(theta) - F(s)) / (m * R) Next i insert this ##alpha## in...

I chose my coordinate system with i_hat to the right j_hat down. Doesnt that make cw positive? Otherwise i would also have to correct the other lines? So i would now have F(s) = mu *(- mg + F*sin(theta))

So i finshed. Could you please tell me, if i have it correct? F(max) = (3*mu*m*g*R) / (R*cos(theta)+3*mu*R*sin(theta)-2*b) Please let that be correct :D