Just found out for the disk with axis at rim, it is (3/2)(M)(R^2) so I=(1/2)(9m)(2r)^2+(3/2)(m)(r)^2 and with simplification I=19.5mr^2 so for initial L=Iw so L=(18.5mr^2)(21.4) and L=395.9mr^2 so if Li=Lf then 395.9mr^2=(19.5mr^2)w and w=20.3025641
so the KE=(1/2)(19.5mr^2)(20.3025641) and...
Ok, so if I used I=(x)MR^2 the only difference would be the x in front of MR^2, so instead of (1/2) it would be 1/some number. So After consulting my book and a brief web search, I cannot find a rotational inertia for a disk with the axis on the rim. So either a) how do i find this, or b) what...
Alright so since it is a disk, then I=(1/2)MR^2 so since there are two disks, should it be I=(1/2)(M)(R^2)+(1/2)(m)(r^2) where M is the mass of the large disk and m is the mass of the other disk, and R is radius of large disk and r is radius of small disk? So if that is the case then it is...
Ok so for the initial KE, I did KE=(1/2)(I)(w^2) so KE=(1/2)(10m)(21.4rad/s) and
KE=107 m x rad/s
Now I re-read the problem and I realized I am unsure how to calculate the Angular velocity after the slide. Would I have to use L=Iw to find initial angular momentum and set the initial L= to...
Ok, I think I know how to solve for the angular velocity and KE of the system before sliding, but How do I do it after the other disk moved to the edge. Maybe we haven't gone over this in class yet, but I have another similar problem and I do not know how to do it, so I will attempt it, and wait...
Homework Statement
A uniform disk of mass 9.00m and radius 2.00r can rotate freely about its fixed center like a merry-go-round. A smaller uniform disk of mass m and radius r lies on top of the larger disk, concentric with it. Initially the two disks rotate together with an angular velocity...
Thank you all, all I had to do was multiply the area I calculated by the mass, now that I think about it it was weird how I didn't use the mass at first to calculate work. and for the last three all I had to do was what I said in the original question sqrt((2W)/3)! Again thanks for the help!
Is the area the force it takes to move the particle? If it is then since W=f*d I would multiply the area by the length of the segment? for (a) it woul be 28*4?
Homework Statement
Figure 7-39 (attached) gives the acceleration of a 3.0 kg particle as an applied force moves it from rest along an x-axis from x = 0 to x = 9.0 m. The scale of the figure's vertical axis is set by as = 8.0 m/s2. How much work has the force done on the particle when the...
Well, it is online homework so I put zero in and it is right, so, but thanks. I would just like an explanation for parts b and c because my professor does not do examples in class.