Centripetal Motion problem -- A bead sliding around a horizontal loop

[Rubberduck2005]

Homework Statement

A bead sliding around a horizontal loop of radius R with a friction co efficient of U has a speed V0 at t=0, First find V(t) then the angle by which the bead has rotated by time t

Relevant Equations

.

To be honest I am a bit clueless first with how to interrupt this question I think the bead is going around a wall type thing where there is friction both in the up and X direction. Some hint to get some ideas running would be great

 

[jbriggs444]

Homework Statement:: A bead sliding around a horizontal loop of radius R with a friction co efficient of U has a speed V0 at t=0, First find V(t) then the angle by which the bead has rotated by time t
Relevant Equations:: .

To be honest I am a bit clueless first with how to interrupt this question I think the bead is going around a wall type thing where there is friction both in the up and X direction. Some hint to get some ideas running would be great

What is the net normal force of the wall+floor thing on the bead, given that the bead has velocity v?

 

[haruspex]

I do hope an integral expression is accepted as the answer for the time.

 

[Rubberduck2005]

jbriggs444 How would you determine net force using it's velcoity?

 

[jbriggs444]

jbriggs444 How would you determine net force using it's velcoity?

If you know mass and acceleration, you can deduce net force. That's Newton's second law.

Given a velocity and a circular trajectory there is a well known formula for acceleration.

 

[Fred Wright]

I think the easiest way to solve this problem is to employ Lagrangian dynamics. I'm not sure if the OP has reached that level of education but at any rate we consider a suspended horizontal loop with a bead sliding with a contact coefficient of friction $\mu$, initial velocity $v_0$ and mass $m$ . If we assume the frictional force is proportional to velocity we can write the Lagrangian,
$$
\mathcal L=e^{\mu t}[T-V]
$$
where $T = \frac{mv^2}{2}$ is kinetic energy, $V=0$ is potential energy (the loop is horizontal, so no potential energy). The position of the bead and its velocity is written,
$$
\begin{align*}
& x=R\cos(\theta)
& \dot x=-R\dot \theta \sin(\theta)\\
& y=R\sin(\theta)
& \dot y=R\dot \theta \cos(\theta)\\
&v^2=\dot{x}^2 + \dot{y}^2=R^2\dot {\theta}^2

\end{align*}
$$
Thus our Lagrangian is,
$$
\mathcal L =e^{\mu t}(\frac{mR^2\dot{\theta}^2}{2})
$$
and
$$
\frac{d}{dt}\frac{\partial{\mathcal L}}{\partial \dot \theta}=0
$$
I won't take it any further because I've been scolded for completely solving problems on this forum. However the OP can take the partial derivative of $\mathcal L$ w.r.t. $\dot \theta$ and then differentiate w.r.t. to time. Observe that $\ddot \theta = \frac{d \dot \theta}{dt}$ and remember $v=R\dot \theta$. Integrating once will give an expression for $\dot \theta$ in terms of $\dot \theta_0$ and $t$ and integrating again will give $\theta$ in terms of $t$, $v_0$, $R$, $m$, and $\mu$.

 

[jbriggs444]

If we assume the frictional force is proportional to velocity

Why would the frictional force be proportional to velocity?

 

[Fred Wright]

Why would the frictional force be proportional to velocity?

It's a phenomenological assumption people make in these types of problems when employing Lagrangian dynamics. Whether or not it's a valid assumption can only be born out by experiment.

 

[jbriggs444]

It's a phenomenological assumption people make in these types of problems when employing Lagrangian dynamics. Whether or not it's a valid assumption can only be born out by experiment.

We already have a good model for kinetic friction specified in the problem statement. We are given a coefficient of kinetic friction. Going against that model is improper.

a friction co efficient of U

It is a fairly straightforward to produce a formula for the force of kinetic friction as a function of velocity. However, since OP is hung up and not yet able to produce that formula, I will not explicitly give the formula other than to state that it does not take the form: $F=-kv$.

 

[Fred Wright]

We already have a good model for kinetic friction specified in the problem statement. We are given a coefficient of kinetic friction. Going against that model is improper.

Since we are given a coefficient of friction, I don't see how my argument is improper. Granted, the problem statement didn't say, "assume the frictional force is proportional to velocity" but also the problem statement didn't mention the mass. I don't see how to solve this problem without making some assumptions.

 

[jbriggs444]

Since we are given a coefficient of friction, I don't see how my argument is improper. Granted, the problem statement didn't say, "assume the frictional force is proportional to velocity" but also the problem statement didn't mention the mass. I don't see how to solve this problem without making some assumptions.

The problem statement indicates that the frictional force is most certainly not proportional to velocity.

Mass is obviously irrelevant. If you write down the equations of motion, mass will cancel.

 

[Fred Wright]

The problem statement indicates that the frictional force is most certainly not proportional to velocity.

Mass is obviously irrelevant. If you write down the equations of motion, mass will cancel.

I don't see where it says that but this certainly isn't worth a flame war. I'm probably wrong and I apologize for muddying the waters.

 

[jbriggs444]

I don't see where it says that but this certainly isn't worth a flame war. I'm probably wrong and I apologize for muddying the waters.

As I read the problem, what we would like to calculate is the radial+upward force between wire and bead. The tangential force that we really care about will be $U$ times this value. (I'd rather use the greek letter $\mu$, but we can use $U$)

In order to calculate the radial force, we can use the fact that the bead is traveling around a circular trajectory with radius $R$. (I'd rather use lower case $r$, but we can use $R$)

Here is where the formula for centripetal acceleration will come in handy. We know $v$ and $R$. So we can quickly find the resultant acceleration $a$. It will not be directly proportional to $v$. It will be directly proportional to $v^2$.

Newton's second says that $\sum F = ma$.

But we are not done yet. There is at least one more force acting on the bead that we have not yet quantified. Which is where I will leave it for OP.

Unfortunately, this is all the easy part. It gets harder when we have to solve the resulting differential equation. I've not tried that part yet.