第四届广州户外运动节3月25日在白云山开幕
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百度 中国钢铁制造商表示，真正令人担忧的是这些关税带来的潜在连锁效应，因为中国生产的钢铁正在流入全球市场其他地区，而在其他国家竞争在加剧。
Applies to questions of primarily educational value - not only questions that arise from actual homework assignments, but any question where it is preferable to guide the asker to the answer rather than giving it away outright. Please READ THE GUIDANCE IN META before asking homework-like questions.
0
votes
How do you determine the velocity and acceleration components of the sleeve?
starting with the position vector $~\vec R_{op}$
$$\vec R_{op}=\begin{bmatrix}
x\\
y\\
\end{bmatrix}=r\,\begin{bmatrix}
\cos(\omega\,t) \\
\sin(\omega\,t)\\
\end{bmatrix}$$
form here the ve …
- 14.3k
3
votes
How to derive the relation between Euler angles and angular velocity and get the same form a...
How to derive the relation between euler angles and angular velocity
\begin{align*}
&\text{The equations to calculate the angular velocity $\vec{\omega}$ for a given transformation matrix $S$ are: …
- 14.3k
1
vote
Pauli vector in spherical coordinates
the Cartesian sphere components are:
$$\vec{R}=\begin{bmatrix}
x \\
y \\
z \\
\end{bmatrix}= r\,\left[ \begin {array}{c} \sin \left( \vartheta \right) \cos \left(
\varphi \right) \\ \sin \l …
- 14.3k
2
votes
How to prove Lorentz invariance for rotations?
Why a Rotation matrix $R$ don't "destroy" the Lorentz invariant
Prove what Mr. Avantgarde wrote
\begin{align*}
& \text{General Lorentz Transformation matrix}\\\\
\Lambda&=\begin{bmatrix}
…
- 14.3k
0
votes
Constrained motion of the vertices of a quadrilateral
Perhaps this help you to solve the problem
thus $\begin{aligned}2u=\dfrac {\Delta S_{1}}{\Delta t}\\
2v=\dfrac {\Delta S_{2}}{\Delta t}\end{aligned}$
$\begin{aligned}\Rightarrow \\
v=\dfrac {\Delt …
- 14.3k
0
votes
Velocity of the touching point between 2 rotating circles
The velocity at the contact point I is:
$$v_I=\omega_1\,a_1-\omega_2\,a_2$$
thus the components of the velocity in $R_0$ coordinate system are:
$$\vec{v}_0=v_I\,\begin{bmatrix}
\cos(\varphi)\\
\ …
- 14.3k
0
votes
Do the distances from directrix and focus in a parabola shaped trajectory have a physical me...
The projectile parabola is:
$$y(x)=\tan(\varphi)\,x-\frac{1}{2}\frac{g}{v_0^2\,\cos^2(\varphi)}\,x^2$$
where $$ 0 \le x \le \frac{v_0^2\sin(2\,\varphi)}{g}$$
the focus of this parabola is :
$$p_f=\fr …
- 14.3k
1
vote
The Components of two Forces Along the Axis are Equal
The components of $F_1$ and $F_2$ along u axis are not equal because the angle $\alpha$ is not equal to the angle $\beta$, the question supposed to be find the components of $F_2$ along u ?
$$F_1\cos …
- 14.3k
2
votes
Approximating the angle between the trajectory
The green line is the line that the "zig-zag" runner is running.
Thus:
$$s\cos(\alpha)=a\tag 1$$
So if he is doing this n times during the half-marathon you obtain
that:
$$n\,(s-a)=\Delta L\tag 2$$
W …
- 14.3k
0
votes
How to calculate the right path for this spaceship?
The Equation of motions are:
$$m\,\ddot r-m\,\dot\theta^2\,r+F_r=0\\
r^2\,\ddot\theta+r\,2\dot r\dot\theta=0\quad\Rightarrow\\
\dot\theta=\frac{h}{r^2}$$
where $~F_r=\frac{m\,\gamma}{r^3}$
form here y …
- 14.3k
0
votes
Understanding of a pendulum with (perpendicular) moving masses
This is not the answer of your quation ,but the solution of the your problem.
to solve the problem the position of $~m_2~$ must be constant
starting with the components of the position vectors give …
- 14.3k
0
votes
Accepted
Find how the center of mass changing
you want to obtain the center of mass positions as a function of the geometry $~L,a~,b~$ and the masses $~m_i~$ . the equations are
$$\sum_{\rm torque (A)}=m_3\,d_3-m_1\,d_1-m_2\,d_2=0\\
\frac L2-(d_ …
- 14.3k
0
votes
Calculating Reaction Force on Circular Section
$$ R_x=R\sin(\alpha)\quad, R_y=R\cos(\alpha)$$
you know the normal vector $~\vec n=[n_x~,n_y]^T~$
from here the tangent vector $~\vec t=[n_y~,-n_x]^T~$
thus the angle $~\beta= -\arctan\left(\frac{-nx …
- 14.3k
1
vote
Accepted
Analyzing limiting case in a statics problem
Take the sum of the torques abbaut point A , you obtain
$$ M\,g\,\cos(\theta)\frac L2-F\,x=0$$
thus
$$ F=\frac {M\,g\,\cos(\theta)\frac L2}{x}\tag 1$$
with $~x=\frac {l}{\tan(\theta)}~$ you obtain
$$ …
- 14.3k
1
vote
Accepted
Using Lagrange's Equations
start with the position vector to the masses (inertial components)
$$\vec R_{P1}=\begin{bmatrix}
x \\
0 \\
\end{bmatrix}$$
$$\vec R_{P2}=\vec R_{P1}+a\,\begin{bmatrix}
\sin(\theta) \\
-\cos(\ …
- 14.3k