Rotational motion problems

Dear /sci/, I cannot for the life of me figure out how to do these two problems. If any kind person can provide a solution, please for the love of god make it as simple as possible since I'm retarded.

Love, Anonymous.

A smooth, thin rod, mass M=4.1 kg, and length L = 1.3 m, is spinning in a horizontal plane about a fixed vertical axis passing through one of its ends at an angular speed of w = 1.2 rad/s. A small ring of mass m = 0.010 kg, initially placed on the rod very close to the axis of rotation starts slowly sliding down the rod towards the other end of the rod. Calculate the speed of the ring when it reaches the other end of the rod

Anwser: 1.6

A homogeneous disk of radius R= 0.2m rotating at constant angular speed 21 rad/s about its central axis is carefully placed so that it simultaneously touches, both, a horizontal floor and a vertical wall. The coefficients of friction between the wall and the disk and the floor and the disk are the same mu = 0.05 . Calculate the angle, in degrees, the disk covers before stopping.

Answer: 1.6

For the first problem, set up a differential equation for the acceleration along the rod: r'' = w^2*r (where r(t) is the distance from the axis of rotation at time t). Solve it, get rid of the arbitrary constants by setting known values (t=0->v=0, and t=0->r=r_0, where r_0 is small but nonzero). Then find t for r(t)=1.3, and then put that t into r'. As r_0 approaches 0, you'll get 1.56. Which turns out to be L*w, although I can't immediately see why...

I can't seem to visualize the second problem. What force is pressing the disk against the wall/floor?

>>2
Firstly for the second problem I just copy and plasted what was shown to me.. when I tried to solve it I couldn't figure out what force was pressing the disk against the wall but I thought that gravity was holding the disk on the ground. Perhaps since the disk was on the ground it would also roll toward the wall and thus keep it there. As for how to calculate that.. well thats why I'm here.

Secondly,
I see where you're going with the first problem.. but as I've said, I'm kinda retarded so a few things aren't making sense to me in your explaination. So a few things to clear up:

1) r(t) = distance = v * t  ...right?
2) I intergrate r'' = w^2*r (not sure where you got this from but I'll go with it) to get r' = (w^2 * r^2)/2 ; then r = (w^2 * r^3)/6
3) Given that r(t) = v*t, I set t = 0 to get that r' has  constant of zero, and r also has a constant of zero (probably totally wrong here...)
4) Find r(t) = 1.3 ...uh how?

That ends my train of thought. So.. I would much appreciate it if you could dumb down your explaination even more.

>>3
r(t) = distance = v * t  ...right?
Only if v is constant for all t, which is not the case. Instead, r = ∫v dt (aka r' = v).
r'' = w^2*r (not sure where you got this from but I'll go with it)
r'=v, and equivalently r''=a, which is the acceleration along the rod, which in this case is the same as the centrifugal acceleration. The formula for centrifugal acceleration is w^2*r.
There's no need to integrate it, you can just look up the solution somewhere, eg. the first one on this page: http://eqworld.ipmnet.ru/en/solutions/ode/ode-toc2.htm
For a=-w^2, x=t, y=r, you get:
r(t) = c_1 sinh(tw) + c_2 cosh(tw)
Now to figure out c_1 and c_2, use known values. You have r(0)=r_0 (you can't set this to zero because if it was right at the centre it wouldn't feel any outward force), and r'(0)=0 (no speed at t=0).
Then do the r(t)=1.3 thing, etc.

Ok, let's see how far this gets you.

Ok so r(t) = c_1 sinh(tw) + c_2 cosh(tw)
When r(0) = c_1 sinh(0 * w) + c_2 cosh(0 * w)

Since r(0) = r_0, then:
r_0 = c_1 sinh(0 * w) + c_2 cosh(0 * w)
sinh(0 * w) = 0; cosh(0 * w) = 1
r_0 = c_2 ...??

r'(t) = w(c_1 sinh(tw) + r_0 cosh(tw))
since r'(0) = 0, then:
sinh(0 * w) = 0; cosh(0 * w) = 1
thus: r'(0) = w(0 + r_0); so w = 0... which makes sense cause theres no speed at time zero.. but i dont get what c_1..

Yeah.. I have no idea wtf I'm doing.

First part is correct, but you have the wrong formula for r'(t). (Note that w=1.2, a constant, so it can never suddenly be zero.)

The first problem, I see two possible solutions.

Soln a: v = angular velocity * L = 1.2 * 1.3 = 1.56

Soln b: This is more complicated. It uses two concepts, conservation of angular momentum and moment of inertia.
The moment of intertia (I) for the thin rod rotating on one end is I = 1/3 M L^2. The ring is simplified to be a point object, so I_ring = m r^2
Angular momentum (P) is moment of intertia times angular velocity. P = I w
conservation of angular momentum means intial angular moentum is going to be the same as the final.
For I_ring = m r^2, the r is 0 for the ring initially, it is L for the ring finally.

I w1 = Iw2 + I_ring w2
=> w2 = [M/(M+m)] w1

v = w2 L = 1.556

PS. you teacher is gay for making the answer with only 2 significant digits.

>>7
=> w2 = [M/(M+3m)] w1

fixed. though i am not rly shure the actualy soln is right.