# Homework Help: 2 interesting problems

1. Apr 14, 2004

an interesting problem

I came along this problem, it is quite interesting. However, I have worked with the problem to long and than you can't see any new ways. So I thought, maybe you guys could try it with a fresh mind. Good luck, it is a tough one (or I fail to see something obvious)
imagine a jojo on a string which can be represented by three clinders, the middle one(with the wire around it) with mass m1 and diameter/2(I don't know what that is in english, in dutch it is straal)R1, and the two cilinders on the sides of it both have mass m2 and R2.
first give the rotational inertia on a horizontal axis through the center of mass point.
that was easy, 1/2m1*r1^2+m2*r2^2.
the second question was: calculate the accelaration when the jojo falls down( it rolls down the wire and the axis all stay in the same alignment).
here I got stuck. I thought, you can analyse the movement as a pure rotation around the point where the wire is connected to the middle cilinder. then the angular momentum becomes 1/2m1*r1^2+m2*r2^2+(m1+2m2)*r1^2 (paralel axes law.) speed is then expressable as (m1+2m2)hg=1/2(Ip)W^2. However, if you fill this in and try to express the speed, you get a gigantic formula that can't be right. I must be taking a wrong route, after all, I need the acceleration, not the speed, but I don't know how to come there.
Help would be greatly appreciated.(I am always a bit messy in my work, so it is possible that I made a mistake in the start of the answer)
sorry for my poor english, especially on physics therms.
faithfully,

Last edited: Apr 15, 2004
2. Apr 14, 2004

### NateTG

You might be better off posting each question in its own thread. That said:

Question 1:
How about considering this as an energy problem? You should be able to figure out the torque and angular distance the that yoyo goes through.

3. Apr 14, 2004

### Staff: Mentor

YoYo problem

I assume the question was: calculate the rotational inertia about the center of mass? If so, that's fine. (What you are calling "angular momentum" is actually the rotational inertia.)
Again, what you are calling "angular momentum" seems to be the rotational inertia. I believe that you made an error in applying the parallel axis theorem. Redo that calculation as you will need it.

In any case, there is nothing wrong with treating the motion as instantaneously a pure rotation. I wouldn't bother calculating the speed. Instead, calculate the acceleration directly using Newton's 2nd law for rotation. You know the torque about that point and the rotational inertia (recalculate it).

PS: I agree with NateG. Take your second problem and create a new thread. (Then delete that post in this thread.)

4. Apr 15, 2004

Of course, that's much more logical. I wanted to take the differential formula of the speed, but that's the hard way I guess.
so the T=FR. F is g(m1+2m2), R is R1. T is R1*g(m1+2m2). T=Ia. the mistake I made in using the paralel axes theorem was that I didn't use all the mass, isn't it? the therm added should be (m1+2m2)R1^2, so I is 1/2m1*R1^2+m2*R2^2+(m1+2m2)R1^2. a= T/I so a is (R1*g(m1+2m2))/(1/2m1*R1^2+m2*R2^2+(m1+2m2)R1^2). Then, At(acceleration of the center point) is a*R1. so At(what I need)=(R1^2*g(m1+2m2))/(m2*R2^2+(1.5m1+2m2)R1^2).
Somehow that doesn't seem right.
The problem is that it isn't possible to simplify the formula since i can't turn the divider into one therm.
Is this right then?
thanks

Last edited: Apr 15, 2004
5. Apr 15, 2004

### Staff: Mentor

Looks right to me.

6. Apr 16, 2004