Yes because angular momentum and linear momentum are conserved seperately.
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...
No!
I had it all wrong. So angular momentum and linear momentum are conserved seperately.
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...
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