I'm working through a problem and I don't understand how to approach it.

(Not a duplicate of this - asking about a different thing)

A square hoop with side length $l$ and vertices $PQRS$ (labelled anticlockwise) is horizontal and rotates with a constant angular velocity $\omega$ about the vertical axis going through $P$. (Gravity is ignored). At time $t=0$, a small bead is initially at rest at the midpoint of the side $QR$, and the bead is free to slide without friction on this side of the hoop. What is the magnitude of the force that the hoop exerts on the bead, viewed in the stationary frame of reference and the rotating frame of reference?

So far I have tried considering the equation for acceleration in a rotating frame of reference with constant $\omega$, where if $S$ is a stationary frame and $S'$ is a rotating frame, then we have $$\ddot{\mathbf{r}}_{S}=\ddot{\mathbf{r}}_{S'} + 2 \mathbf{\omega} \times \dot{\mathbf{r}}_{S'} + \mathbf{\omega} \times (\mathbf{\omega} \times \mathbf{r}_{S'})$$

from which we can deduce $\ddot{y}=\omega^2 y$ and hence $y=\frac{l}{2}\cosh{\omega t}$ (where $y$ is the distance of the bead to the vertex $Q$) which came up earlier in the problem (this comes from plugging into the equation above and taking the component along the side $QR$).

However, for finding the magnitude of the force exerted by the hoop on the bead, it seems like I need to consider centrifugal force and/or Coriolis force, but it doesn't seem obvious how exactly the motion of the hoop is affecting the bead.

Given the solution to the differential equation above, we can see that the bead moves with strictly increasing speed towards the vertex $Q$, reaches it at time $T=\omega^{-1} \cosh^{-1} 2$, and then stays there, in which case the force exerted by the hoop on the bead must simply be the force needed to keep it in a circular orbit of radius $l$, so for $t>T$ (by standard circular motion) the magnitude of the force would be $m \omega^2 l$ (viewed in the stationary frame). In the rotating frame, it would seem to be $0$.

However, it really doesn't seem obvious how to find the force exerted on the bead before it reaches the vertex $Q$, given that the bead is moving in its own way on the side $QR$ and the whole hoop is also rotating. How do you find this in each frame?


1 Answer 1


from the free body diagram you obtain

$$ \begin{bmatrix} 0 \\ \ddot y \end{bmatrix}=m\begin{bmatrix} \omega^2\,l+2\omega\dot y \\ \omega^2\,y \end{bmatrix} +\begin{bmatrix} N \\ 0 \end{bmatrix} $$

you already used the second equation, from the first equation you obtain the force $N$


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