Röd (Swedish for Red) is the eighth studio album by Swedish alternative rock band Kent. It was released as digital download exclusively through the band's website on 5 November 2009 and physically on 6 November 2009. The first single from the album, "Töntarna", was released as digital download on 5 October 2009.
Röd is available in a standard edition and a deluxe edition box. The deluxe edition box version features the 11-track CD, a USB flash drive with high quality MP3 files as well as AIFF files, three 10" records which between them contain the whole album, and a 118-page book containing lyrics, abstract pictures and photographs. Due to distribution difficulties the deluxe edition was delayed until 11 November 2009.
My line of thinking is as follows:
\omega_{PQ} = \frac{v_{\perp}}{\ell} = \frac v\ell \frac{\sqrt3}{2}
Similarly for rod ##QR##
\omega_{QR} = \frac{v_{\perp}}{\ell} = \frac v\ell \frac{\sqrt3}{2}
Is my reasoning correct?
This is the initial setup of the problem:
The electric field due to the ring is:
$$E = \int\frac{k(dq)}{(\sqrt{R^2 + x^2})^2}\frac{x}{\sqrt{R^2 + x^2}} = \frac{kqx}{(R^2 + x^2)^{3/2}}$$
the force on the rod due to this Electric field produced by the ring is:
Consider a differential element...
This was the question
(The line below is probably some translation of upper line in different language)
For disc it was ma^2/2
For ring it was ma^2
For square lamina it was 2ma^2/3
For rods
It was different
Please explain
Thank You🙏
I(i)w(i)= I(f)w(f)
I(i)= 1.08 x 10-3 kg·m2
w(i)= 0.221 rad/s
I(f)= mr^2 + I(i) = (5 x 10^-3)(.138)^2 + (1.08 x 10^-3)
(1.08 x 10-3)(.221) = ((1.08 x 10^-3)+9.22 x 10^-5))w(f)
w(f) = (2.3868 x 10^-4)/(0.00117522)
w(f)= 0.203094 rad/s
This is my attempt; however, I cannot seem to get it...
Ok, so basically: There is a frase with a particle moving horizontally with velocity v and a rod parallel to the ground, with rest length L, falling with speed u vertical constant. The rod bounces off on the ground. Describe what happens in the particle frame and find the angle the rod makes...
Assuming no friction anywhere, no drag and perfect inelastic collision
Using conservation of mechanical energy i can determine the rotational speed of the rod right before collision occurs.
mgh=1/2*i*w^2
center of mass falls 1/2*L so we have:
M*g*1/2*L = 1/2*(1/3*M*L^2)*w^2
Solving for w...
Hi everyone :)!
I resolve this problem with components method and trigonometry method.
My results with components method its okay, but i can´t obtain the correct VE velocity.
Im sure that the problem its in the angles, but i don't know how to fix it.
The correct answers:
-Angular velocity...
Hi , I've been trying to manage a solution in my head and i think I'm on the right path , i just need some approval and maybe some tips.
So it's obvious I can't solve this without integration because law's only apply to point charges , and i can't shrink this object to a point as i could do with...
(A) and (B) are obviously wrong but I think both (C) and (D) are correct.
At the top, the forces acting on the mass are tension and weight, both directed downwards so the equation of motion will be:
$$\text{Tension}+\text{Weight}=m.a$$
$$\text{Tension}=m.a-\text{Weight}$$
Based on that...
The first part of the problem seems easy enough, the free electrons in the wire would move in a circle owing to an electric field that would be induced in the rod which would provide the centripetal force for the same (Please correct me if I am wrong). So we have $$eE=mω^2x$$, where e is the...
I am not sure if the explanation below is enough. This is a high school level question.
When rubbing occurs between glass and silk, then heat energy is produced which provides the energy needed to free up electrons in outermost orbits of atoms in silk or glass. But silk has very tightly bound...
Hello,
Given the statement a described above. To find the forces at point D I drawn a kinematic scheme and FBD of rod CD. But why am I allowed to ignore the mass of 50 kg, the forces at point B and point A? I know the are some rules about this, but I just can't remember them anymore.. The...
(A) incorrect, because opposite signs attract, and the sphere would've been drawn to the charged rod.
(B) correct, according to the answer key, but if the charge of the sphere and the charge of the rod are the same, then wouldn't they repel each other? I'm confused as to why this is the correct...
Hello,
It might sound silly, but when I try to calculate the kinetic energy of a rotating rod to form the Langrangian (and in general), why it has both translational and rotational kinetic energy?
Is it because when I consider the moment of Inertia about the centre I need to include the...
I have a dilemma.
If I look at the diagram and say the sum of the forces in x and y direction has to be zero, then I will simply conclude that the force of a wall on a revolvable rod is the blue N as drawn.
But what if the force is actually the green N? To me, it makes more sense because if...
Hello All
I pivot a long rigid rod at one quarter its length and gradually accelerate the tip of the short end to close to the speed of light.
An observer is standing some distance away from the mechanism, so that he/she can see the whole length of the rod.
What would the shape of the rod...
Hi guys,
I don't really know how to solve this problem.
The point is finding ##\omega## when ##m_2## passes from ##m_1##'s original position.
Ideally, I'm thinking about some conservation of energy/momentum to apply here, but I'm quite confused.
Any hint?
I am having trouble solving part 2, for
$ W_{\frac{n(n+1)}{2}} \leq 2^{n} (n-1) + 1 , n \geq 0 $
I know that $W_{m} \leq 2*W_{m-k} + 2^{k} – 1, 0 \leq k \leq m$
Let $m = \frac{n(n+1)}{2}$
So now $W_{\frac{n(n+1)}{2}} \leq 2*W_{\frac{n(n+1)}{2} - k} + 2^{k} - 1, 0 \leq k \leq...
The rod itself should have 3 translational+2 rotational DOF.
The particle on top of the rod has one additional DOF.
So total should be 6. But answer given is 4.
What I'm thinking wrong?
Charge QQ is uniformly distributed along a thin, flexible rod of length LL. The rod is then bent into the semicircle shown in the figure (Figure 1).Find an expression for the electric field E⃗ E→ at the center of the semicircle.
Hint: A small piece of arc length ΔsΔs spans a small angle...
I have some difficulties in solving this problem. This is what I did.
I wrote down the equation of motion for the masses. For the first point
\begin{equation}
m\ddot{\textbf{r}}_1=\textbf{F}_1=q\dot{\bar{\textbf{r}}}_1\times...
1) Applying conservation of linear momentum:
$$m.u = M.V + m.v$$
where ##V## is final linear speed of the rod
$$V=\frac{m.u-m.v}{M}$$2) Applying formula of circular motion:
$$V=\omega . r$$
$$\omega = \frac{\left(\frac{mu-mv}{M} \right)}{\frac{1}{2}L-x}$$
Is this correct?And can this be...
Hi,
I started with calculating the moment of inetria of the rod:
I = ⅓ML^2 + M(3/2 * L)^2 = 31/12 ML^2
and I thought that the reaction force in the first case will be equal to centrifugal force:
F1 = Mω^2*(3/2)L
Angular velocity is calculated from the conservation of energy:
Mg3/2*L=1/2 * Iω^2...
Say I have a rod with a notch at the very centre and another ro with a notch at the very end. Both rods are identical in length radius materials etc. Who would be impacted by the notch more? And if then both were to support a load at the end of the rod (both in tension), what effects would the...
I realized that the tension in the rod is not uniform and found it to be ##T = Wx/L## I found this by splitting the rod into two sections one of length ##x## and the other of length ##L-X## where x is the length from the base of the hanging rod
To find the total elongation in the rod I...
I want to ask the direction of normal force acting on the rod by the rim of the bowl. Is the direction perpendicular to the rod or will it be directed horizontally to the left?
My guess would be horizontally to the left because the normal force would be perpendicular to the "plane / surface"...
Question diagram, attempt at solution below
I need to cancel some of the terms in the moment equation but a not sure which ones to start with. I don’t know μ so can not calculate FA, so should probably substitute FA = RB2.
A metal rod held in hand and rubbed with wool will not show any sign of of being charged. However, if a metal rod with a wooden or plastic handle is rubbed without touching its metal part, it shows signs of charging. Why??
This problem honestly got me in big confusion.
I managed to find the angle ##\theta## at which the rod rests by equalling the components of weight and Lorentz's force... but from this point on I really don't know how to manage the harmonic oscillation part.
Note, I put this question here because I imagined it fit the best here, but mods feel free to move it if there's a better place.
So my girlfriend asked me a question today I couldn't answer. Imagine you have a crystal rod (for simplicity), and you push it a bit at one end, in the axial...
Okay so I actually have the answer because my teacher basically just gave it to me, but I would really like to know why I was even wrong in the first place. Here's my steps:
1. Knowing the momentum transfer per unit area is described by: 1/A dp/dt = S/c. I can begin by relating some known...
Since there is no friction, there is no radial force acting on Ball B after the pin is remove. Therefore the radial acceleration of Ball B is zero. I don't know how to determine the transverse components of the acceleration of Ball B. I looked at the textbook solution. It takes moment about the...
The soultion used polar corrdinates. Acceleration in polar corrdinates have radial and transeverse components.When calculating the acceleration of collar respect to the rod, the solution only calculates the radial component of acceleration. Is it because the collar is on the rod, so the...
There used to be sold a style bicycle handlebar bags that used what I think is a formed spring steel rod that fit over the handlebars and looped under the handlebar stem that supported a handlebar bag. For whatever reason this style does not appear to be available any more. I think it is a...
Method 1: Simply conserving angular momentum about the the fixed vertical axis and conserving energy gives ##v=3##, which is correct according to my book.
Method 2: Conserving angular momentum when the two rings reach distance ##x## from the centre gives
##(0.01+2x^2) \omega =0.9##
Also in the...
I'm not sure I understand why I need to use ##d##.. Maybe they want me to have the potential be zero at ##A##?
In any case, I have found$$V(B)=\alpha k\int_0^L\frac{x}{\sqrt{b^2+\left(x-\frac{L}{2}\right)^2}}dx+C=\frac{\alpha...
Summary:: Classical problem about a pendulum!
The problem itself:
My FBD:
I want to solve the problem with vectors, I think that you can use energy principle somehow. If we define the vector ##\vec{O}_B=\begin{bmatrix}0\\ -1\end{bmatrix}## and define a rotational matrix where...
Okay, i know that as a ball collides with a pivoting rod on an axis, the ball has angular momentum. Therefore after the collision, the ball is stopped or slowed, and the rod swings.
The ball provides a force and torque to the rod. But if I isolate the ball, isn't the only thing acting on the...
Ateempt of solution:
There are two key coordinates in this scenario, the leftmost tip of the rod, which in ##S'## is ##C_{0} = (t', 0, ut',0)## and the opposite tip
##C_{1} = (t', L,ut',0)##
An angle ##\phi## could be found through a relationship such as ##tan(\phi) = \frac{ \Delta x}{ \Delta...
$$\tau = I\alpha$$
$$FL/2 = I\omega^2L/2$$
$$T = 1/\theta \sqrt{F/I}$$
would this be correct?
I came up with this more basic question to solve a slightly harder question so I do not know the answer to the above-stated problem.
I have seen the solution and understand it. The solution defines θ to be the angle between the falling rod and the table. It then equates initial PE to the final (PE+KE) where the final PE=0 and the final KE to be (1/2) Iω2+ (1/2) ma2ω2 to finally obtain ω = √(3g/2a)
But i would like to know...
I have some threaded rods on my 3D printer that I want to straighten. After searching Youtube for a quick easy method that doesn't involve a million "guess and check" steps, I found this:
This guy seems to have concocted some method using Fourier series to straighten his rods. Not sure...
After solving using energy conservation, I found the angular velocity at 37° to be omega=2.97/(L)^½
Tension and the weight (dm)g are the two forces acting on the tip dm
To find the resultant force, I resolved the centripetal force and tangential force to find the centripetal force as
F=...
I'm struggling doing point 5, i have no idea how to solve that question. In point 1 i obtained the following result:
## I=\frac{ML^2}{2}## calculating the integral of dI, the infinitesimal moment of inertia of a small section of the rod of length dr.
2) Through the conservation of angular...