# Rotating frame of reference

Can you help me to do this:

Two frames of references $S$ and $S'$ have a common origin $O$ and $S'$ rotates with constant angular velocity $\omega$ with respect to $S$.
A square hoop $ABCD$ is made of fine smooth wire and has side length $2a$. The hoop is horizontal and rotating with constant angular speed $\omega$ about a vertical axis through $A$. A small bead which can slide on the wire is initially at rest at the midpoint of the side $BC$. Choose axes relative to the hoop and let $y$ be the distance of the bead from the vertex $B$ on the side $BC$. Write down the position vector of the bead in your rotating frame. Show that
$\ddot y-\omega^2 y=0$ using the expression for the acceleration. Hence find the time which the bead takes to reach a vertex $C$.

I showed that $\frac{d^2\vec r}{dt^2}=(\frac{d^2\vec r}{dt^2})'+2\vec\omega\times(\frac{d\vec r}{dt})'+\vec\omega\times(\vec\omega\times\vec r)$ where $'$ indicates that it's done in rotating frame. $\vec r$ is position vector of a point $P$ measured from the origin.

I got that
$\vec r=r\cos\theta\vec i+y\vec j$
$\vec r'=(\dot rcos\theta-r\dot\theta \sin\theta)\vec i+\dot y\vec j$
$\vec r''=(\ddot rcos\theta-\dot r\dot \theta sin\theta-\dot r\dot\theta\sin\theta-r\ddot\theta\sin\theta-r\dot\theta^2cos\theta)\vec i+\ddot y\vec j$
$\omega\times\vec r'=-\omega\dot y\vec i+(\omega\dot r\cos\theta-\omega r\dot\theta\sin\theta)\vec j$
$\vec\omega\times (\vec\omega\times\vec r)=-\omega^2 r\cos\theta\vec i-\omega^2 y\vec j$

I suppose I have to write Newton's second law now, but I don't know which forces do I have in this motion.

• Your equations are pretty hard to follow. What is $\theta$ in equation one? And how did you write $\vec r=r\cos\theta\vec i+y\vec j$? Note that $y$ is the distance of the bead from $B$ and not from the x-axis, so I think this equation is incorrect. Maybe your approach to the problem isn't correct. I know another way to solve this, If you want I'll post it. – udiboy1209 Jul 28 '13 at 11:24
• $\theta$ is an angle between position vector and x-axis in rotating frame. I should choose axes $x$ and $y$ so that they lay along sides of the square. I chose that $x$ lays along $AB$ and $y$ lays along $AC$. Yes, please, write your solution. – gov Jul 28 '13 at 11:32

I see the problem with your equation now.

When differentiating $\vec r=r\cos\theta\vec i+y\vec j$, you have considered $r$ to be constant, which is wrong.
$r$ is given by $$r=\sqrt{l^2+y^2}$$ where $l$ is the side-length of the square. So $r$ will change with $y$, and you'll have to differentiate $r$ too.
This is where the math gets pretty ugly and dissuading!

To avoid that, what we can do is we can observe that in the rotation frame, the bead will experience an outward centrifugal force. This force will have a component along $BC$. That component can be written as(I'll be borrowing your variables) $$F_{BC}=m\omega^2r\sin\theta$$ $$F_{BC}=m\omega^2 \sqrt{l^2+y^2} \frac{y}{\sqrt{l^2+y^2}}$$

Thus by dividing by $m$ on both sides you get

$$\ddot y=\omega^2y$$

Note that this is the same as applying $\frac{d^2\vec r}{dt^2}=(\frac{d^2\vec r}{dt^2})'+2\vec\omega\times(\frac{d\vec r}{dt})'+\vec\omega\times(\vec\omega\times\vec r)$. It is just that this approach is more problem specific(and a lot easier too!).

• The problem is that I HAVE TO use this formula for acceleration. Ok, I will differentiate it again having on mind that $r=r(t)$ and I will see what will I get. Just tell me this: Which forces are acting on a bead? Gravitational and this one that you wrote? Are they both perpendicular to vertical axis? If they are, I can multiply expression for Newton's second law with Is $\frac{d\vec r}{d\theta}$ to get $0$ at the right side. – gov Jul 28 '13 at 12:06
• Newton's law wouldn't be of much help here. This is because the loop is constrainted by some external agent to maintain a constant angular velocity. Now this force will keep varying according to the changing moment of inertia of the system, and it will vary such that the loop maintains a constant angular velocity. The equations of motion you have will suffice to solve this problem. – udiboy1209 Jul 28 '13 at 12:11
• If you are still not satisfied, you can think of it this way. All the newton's second law equation will give you is a description of the beads motion, which you already have with you. Just differentiate the equation taking $r$ as variable and you should get your answer. – udiboy1209 Jul 28 '13 at 12:12
• I corrected expressions in question. Why there's no gravitational force in your answer? Why do you observe just centrifugal force? – gov Jul 28 '13 at 12:34
• I've got it. Gravitational force doesn't have component along $y$-axis. – gov Jul 28 '13 at 12:36