A bead-mass oscillatory system problem

[Zayan]

Homework Statement

A bead of mass m can slide on a smooth straight horizontal wire and a particle of mass 2m is attached to the bead by a light string of length l. Initially, the particle is held in contact with the wire and the string is just taut. Then, the particle is released to fall under the gravity.
Find the tension when the string makes 37 degrees with the horizontal.

Relevant Equations

W=∆K.E
m1v1=m2v2
a= mv²/r

I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame?

I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work energy theorem, the second would be by momentum conservation in horizontal direction but for that I require the ground frame velocity. After finding the velocity I think I am supposed to write the equation of motion of particle wrt bead as it is performing circular motion wrt to it.

 

[wrobel]

Introduce an angle $\alpha$ between the string and the wire and a coordinate x that determines the position of the bead.
Express in these terms the equation of energy conservation and the equation of momentum conservation in the horizontal direction. Find the constants of these first integrals by using the initial conditions.
Differentiating these equations in t you can obtain the equations of system's motion.
Then use 2nd Newton to find string's tension.

 

[Zayan]

Introduce an angle $\alpha$ between the string and the wire and a coordinate x

With respect to what do I write the coordinates? The bead is moving itself and if I wanted to write the coordinates wrt let's say, the left end, I would require the velocity of bead at all time.

 

[wrobel]

The equations mentioned above contain the terms $\alpha,\dot\alpha,\dot x$

 

[Zayan]

The equations mentioned above contain the terms $\alpha,\dot\alpha,\dot x$

How do I conserve the energy. There is also vertical velocity so shouldn't dy/dt also be taken into account

 

[wrobel]

$$\boldsymbol v_B=\dot x\boldsymbol e_x+\dot\alpha r \boldsymbol e_\alpha,\quad r=|AB|;$$
$$\boldsymbol a_B=\ddot x\boldsymbol e_x+\ddot \alpha r\boldsymbol e_\alpha-\dot\alpha^2 r\boldsymbol e_r.$$
$$\boldsymbol e_x=\cos\alpha\boldsymbol e_r-\sin\alpha\boldsymbol e_\alpha$$

$$(m_A\boldsymbol v_A+m_B\boldsymbol v_B,\boldsymbol e_x)=c,\quad \boldsymbol v_A=\dot x\boldsymbol e_x;$$
$$\frac{m_A}{2}|\boldsymbol v_A|^2+\frac{m_B}{2}|\boldsymbol v_B|^2-m_Bgr\sin\alpha=h,$$
$$m_B \boldsymbol a_B=m_B\boldsymbol g+\boldsymbol T,\quad \boldsymbol T=T\boldsymbol e_r.$$

I know that it is not a tradition of PF to present a detailed solution but I believe that in this particular case it is acceptable since the problem looks too hard for OP

 

[Zayan]

$$\boldsymbol v_B=\dot x\boldsymbol e_x+\dot\alpha r \boldsymbol e_\alpha,\quad r=|AB|;$$
$$\boldsymbol a_B=\ddot x\boldsymbol e_x+\ddot \alpha r\boldsymbol e_\alpha-\dot\alpha^2 r\boldsymbol e_r.$$
$$\boldsymbol e_x=\cos\alpha\boldsymbol e_r-\sin\alpha\boldsymbol e_\alpha$$




View attachment 366396

$$(m_A\boldsymbol v_A+m_B\boldsymbol v_B,\boldsymbol e_x)=c,\quad \boldsymbol v_A=\dot x\boldsymbol e_x;$$
$$\frac{m_A}{2}|\boldsymbol v_A|^2+\frac{m_B}{2}|\boldsymbol v_B|^2-m_Bgr\sin\alpha=h,$$
$$m_B \boldsymbol a_B=m_B\boldsymbol g+\boldsymbol T,\quad \boldsymbol T=T\boldsymbol e_r.$$

I know that it is not a tradition of PF to present a detailed solution but I believe that in this particular case it is acceptable since the problem looks too hard for OP

Brother your notation is killing me even more. Can't I do something simpler like finding the components of the velocity to get the horizontal momentum conservation equation. Also, I am sorta confused about what direction the particle would be having its velocity from ground frame. Would it be perpendicular to the rope? Because I don't think so. I believe the velocity of particle woth respect to bead would be perpendicular to rope instead.

 

[wrobel]

ok I quit:)

 

[Zayan]

I lowkey got the answer after trying 10 times. T= (690/361)mg. Can someone confirm

 

[Steve4Physics]

@Zayan, a few thoughts...

This looks a relatively advanced problem – halfway between a simple pendulum and a double pendulum. It would typically be solved using the Lagrangian; are you familiar with this?

The angle of 37 degrees is suspicious – it’s not a convenient fraction of 180 degrees and it’s a bit too big for using the small-angle approximation. But, interestingly, it’s a commonly used approximation for the smallest angle in a 3-4-5 triangle.

Your given answer of (690/361)mg seems unlikely as it’s an ‘exact value’, which doesn’t sound appropriate.

 

[Zayan]

@Zayan, a few thoughts...

This looks a relatively advanced problem – halfway between a simple pendulum and a double pendulum. It would typically be solved using the Lagrangian; are you familiar with this?

The angle of 37 degrees is suspicious – it’s not a convenient fraction of 180 degrees and it’s a bit too big for using the small-angle approximation. But, interestingly, it’s a commonly used approximation for the smallest angle in a 3-4-5 triangle.

Your given answer of (690/361)mg seems unlikely as it’s an ‘exact value’, which doesn’t sound appropriate.

I used the approximation from the 3,4,5 triangle. Also, no I am not familiar with lagrangian mechanics I'm still in HS.

Here's the value in trig ratios:
[2mgSin𝜃(7+2Cos²𝜃)]/(1+2Cos²𝜃)²

 

[TSny]

I used the approximation from the 3,4,5 triangle. Also, no I am not familiar with lagrangian mechanics I'm still in HS.

Here's the value in trig ratios:
[2mgSin𝜃(7+2Cos²𝜃)]/(1+2Cos²𝜃)²

I get the same result. I found the calculation to be fairly tedious. Easy to slip up.

 

[Zayan]

I get the same result. I found the calculation to be fairly tedious. Easy to slip up.

Thanks bruh

 

[ruben96]

For future reference, I write here the main ideas behind the solution. One has to clever distinguish between the lab frame, which is inertial, and the comoving frame of $A$, which is non-inertial. Let $\textbf{r}$ be the position vector of $B$ in the lab frame, $\textbf{x}_A$ the position of $A$ in the lab frame, and $\textbf{r}'$ the position of $B$ in the frame of $A$. Then we have
$$ \textbf{x}_A + \textbf{r}' = \textbf{r} $$
and for the accelerations
$$ \textbf{a}_A + \textbf{a}' = \textbf{a}.$$
Let us study the motion of $B$ in the frame of $A$. One can either use cartesian coordinates, with basis vectors $\{\hat{u}_x,\hat{u}_y\}$ or polar coordinates with basis vectors $\{\hat{u}_{r'},\hat{u}_t\}$, referring to the radial (along the wire) and tangential direction, respectively.
They are related by the transformation
$$ \begin{pmatrix}\hat{u}_x \\ \hat{u}_y \end{pmatrix}=\begin{pmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{pmatrix} \begin{pmatrix}\hat{u}_{r'} \\ \hat{u}_{t} \end{pmatrix}.$$
From
$$\textbf{r}'=l\cos\theta\hat{u}_x -l\sin\theta \hat{u}_y=l\hat{u}_{r'}$$
we obtain the velocity
$$\dot{\textbf{r}}'=-l\dot{\theta}\hat{u}_t=v'\hat{u}_t$$
and the acceleration
$$\ddot{\textbf{r}}'=\textbf{a}'=a'_t\hat{u}_t+a'_{r'}\hat{u}_{r'}$$
where $a'_t=-l\ddot{\theta}$ and $a'_{r'}=-l\dot{\theta}^2=-\frac{v'^2}{l}$.
The position of $A$ in the lab frame is
$$\textbf{x}_A=x_A\hat{u}_x = x_A\cos\theta\hat{u}_{r'}+x_A\sin\theta\hat{u}_t$$
and its acceleration is
$$\ddot{\textbf{x}}_A=\textbf{a}_A=\ddot{x}_A\cos\theta\hat{u}_{r'}+\ddot{x}_A\sin\theta\hat{u}_t.$$
The radial component of the equation of motion of $B$ in the lab frame is
$$ 2ma_{r'}=-T+2mg\sin\theta$$
If we insert $a_{r'}=a_{A,r}+a'_{r'}$, then we actually get the equation in the moving frame. Notice the appearence of a fictitious force, as it is the case in non-inertial frames:
$$-T+2mg\sin\theta=2m\left(-\frac{v'^2}{l}+\ddot{x}_A\cos\theta\right).$$
The relationship between the absolute values of the velocities $v_B=||\dot{\textbf{r}}||$, $v_A=||\dot{\textbf{x}}_A||$, $v'=||\dot{\textbf{r}}'||$ is easily found to be
$$v_B^2=v'^2+v_A^2-2\sin\theta v_A v'.$$
We need to eliminate $v_A$ and $v_B$ and express $v'$ only in terms of the angles. To that end, we can use the conservation of momentum along the $x$ direction, in the lab frame:
$$mv_A+2m(-\sin\theta v' +v_A)=0$$
from which $v_A=\frac{2}{3}\sin\theta v'$ and $v_B^2=v'^2\left(1-\frac{8}{9}\sin^2\theta\right).$
Finally, we use the conservation of energy in the lab frame:
$$\frac{1}{2}mv_A^2+\frac{1}{2}(2m)v_B^2-2mgl\sin\theta=0$$
from which $v'^2=\frac{2gl\sin\theta}{1-\frac{2}{3}\sin^2\theta}$. We plug this into the radial equation of motion (taking the absolute values on both sides):
$$\lvert -T+2mg\sin\theta\rvert=\bigg\lvert -\frac{2m}{l}\left(\frac{2gl\sin\theta}{1-\frac{2}{3}\sin^2\theta}\right)+2T\cos^2\theta\bigg\rvert$$
from which we get
$$T(\theta)=2mg\sin\theta\frac{7+2\cos^2\theta}{\left(1+2\cos^2\theta\right)^2}.$$