A rotation is a circular movement of an object around a center (or point) of rotation. The geometric plane along which the rotation occurs is called the rotation plane, and the imaginary line extending from the center and perpendicular to the rotation plane is called the rotation axis ( AK-seez). A three-dimensional object can always be rotated about an infinite number of rotation axes.
If the rotation axis passes internally through the body's own center of mass, then the body is said to be autorotating or spinning, and the surface intersection of the axis can be called a pole. A rotation around a completely external axis, e.g. the planet Earth around the Sun, is called revolving or orbiting, typically when it is produced by gravity, and the ends of the rotation axis can be called the orbital poles.
1) To be in equilibrium, it must be $$\begin{cases}F_{centr}-T=0\\ T-mg=0\end{cases}\Rightarrow F_{centr}=T=mg\Rightarrow m\omega^2 R_0=mg\Rightarrow R_0=\frac{g}{\omega^2}$$
2) It is intuitive that this equilibrium is unstable but I don't know how to formally prove this.
3) In ##R_0## the...
The figure is shown; the measurements were taken on two consecutive observing nights. The Ordinate is the flux normalized to continuum and the abscissa is the wavelength scale. You can see the "bumps" indicated by the arrows referring to some Starspot as the spot moves on the profile; assuming a...
This was the answer key provided:
My questions are the following:
if the force required for rotational equilibrium is more than the limiting static friction, then the body will rotate aka slip over the surface. When it slips, the frictional force will be kinetic and not static, right?
If I...
The conservation of energy equation is basically GPE is converted to KE of block and KE of cylinder.
To get the correct answer, the KE of the cylinder is 1/2mv^2, where m is its mass and v is the velocity of its COM (which is the centre of cylinder).
However, I viewed the cylinder as rotating...
The solution is simple by noting that the total angular momentum of the system is constant. (Though I overlooked this)
Instead, I went ahead analyzing the individual angular momentum of both drums.
Let ##L_a## and ##L_b## be the angular momentum respectively. ##M_a##, ##M_b## be the...
I am trying to derive the tangential acceleration of a particle. We have tangential velocity, radius and angular velocity. $$v_{tangential}= \omega r$$ then by multiplication rule, $$\dot v_{tangential} = a_{tangential} = \dot \omega r + \omega \dot r$$ and $$a_{tangential} = \ddot \theta r +...
Hello!
I was reading two things:
1) tidal locking (as explained in the Wikipedia article:https://en.wikipedia.org/wiki/Tidal_locking
where it is stated that, because of internal friction caused by the body of water being attracted to the moon and deforming, the kinetic energy of the system...
This comes from a list of exercises, and setting ##m_1 = 5.4kg##, ##m_2 = 9.3kg## and ##F=5N##, the answer should yield ##2.19m/s^2## (of course, supposing the answer is right).
If I knew the radius ##R## of the cylinder, I could find its momentum and use it to find the linear acceleration...
Hello there, I have a question regarding this problem. I have no problem with part A. However, in part B, my solution manual states that the hollow cylinder will reach the bottom last. Why is it? I mean shouldn't the solid cylinder and the hollow one reach the bottom at the same time? you know...
Hi, I have the following problem:
A homogeneous disc with M = 1.78 kg and R = 0.547 m is lying down at rest on a perfectly polished surface. The disc is kept in place by an axis O although it can turn freely around it.
A particle with m = 0.311 kg and v = 103 m/s, normal to the disc's surface at...
Let's begin with the first point.
a.I) Apply a generic boost in the y-z plane (take advantage of the arbitrariness in deciding the alignment of the y and z axes).
\begin{equation*}
B_{yz} =
\begin{pmatrix}
\gamma & 0 & -\gamma v_y & -\gamma v_z \\
0 & 1 & 0 & 0 \\
-\gamma v_y & 0 &...
Hi,
I take a big number of disks to composed a circle of a radius of 1 m, the blue curved line is in fact several very small disks:
I take a big number of disks to simplify the calculations, and I take the size of the disks very small in comparison of the radius of the circle. The center A1...
So when the rotation starts some water will move upwards and in the vertical part of tube.
I know hat centripetal force will be given by
F=mv²/r
Now I though of taking r as centre of mass of the water system but I don't know what to take the value of m as?
Should I only consider the water...
From a freebody analysis I got,
$$ \vec{r} \times \vec{F} = |r| |F| \sin( 90 - \theta) = (R-r) mg \cos \theta$$
and, this is equal to $$ I \alpha_1$$ where the alpha_1 is the angular acceleration of center of mass of small circle around big one,
$$ I \alpha = (R-r) mg \cos \theta$$
Now...
I watched a video that showed how to calculate the center of gravity of a horizontal bar suspended from two wires, one attached to each end. Each wire was then attached to a vertical wall. The angle each wire made with the wall it was attached to was given. They treated it as an a example of...
When the lamina rotates about A, FA must act on B (because it is the farthest away) perpendicular to AB (so that all of FA contributes to rotation).
Same argument is valid for rotation of lamina about B as well.
Having noted that, I tried two approaches:
Approach 1-
If I assume that the...
I was solving a problem for my quantum mechanics homework, and was therefore browsing in the internet for further information. Then I stumbled upon this here:
R is the rotation operator, δφ an infinitesimal angle and Ψ is the wave function.
I know that it is able to rotate a curve, vector...
Below is the attempted solution of a tutor. However, I do question his solution method. Therefore, I would sincerely appreciate it if anyone could tell me what is going on with the below solution.
First off, the rotation of the matrix could be expressed as below:
$$G = \begin{pmatrix} AB & -||A...
It's often said that you don't feel earth rotation because the gravity acts against the centrifugal force.
Of course this is true but also your body is turned around once each 24 hours.
So I wonder on a planet which is rotating once each 3 seconds and has same g=9,81:
Would you feel the rotation?
I was talking to someone about the equilibrium of fluids and we reached at some stage where we had to prove that in an external field the translational forces add to zero along with moments (torques) should also add to zero. The first one was quite easy but during the discussion of second...
Hello, I am a computer science major and Ex-Biology grad student, my knowledge in physics is humble, but I got a little curious when my professor derived the expressions of moment of inertia for different objects.
The moment of Inertia of a thin disk is 1/2MR2, but it is the same as the moment...
Summary:: Calculating the inclination angle
A stick is on two springs with spring constants D1=500N/m and D2=300N/m. Consider the stick is without mass and can rotate around the point E, which is distant from spring 1 with 0,1m and from spring 2 with 0,8m. A force F=100N pulls the stick up...
I have tried doing the obvious thing and multiplied the vectors and matrices, but I don't see a way to rearrange my result to resemble the initial state again:
##(\mathcal{D_{1y}(\alpha)} \otimes \mathcal{D_{2y}(\alpha)} )|\text{singlet}\rangle = \frac{1}{\sqrt{2}}\left[
\begin{pmatrix}...
For a cylinder rolling down an inclined plane, does the tangential velocity of a point a distance R from the axis of rotation equal the velocity of the center of mass?
Answer choices: N2L for Translation, N2L for Rotation, Both, Either
1. You are asked to find the angular acceleration of a low-friction pulley with a given force exerted on it.
My solution = N2L for rotation
2. You are asked to find the angular acceleration of a low-friction pulley due to...
A uniform rod AB of length ℓ is free to rotate about a horizontal axis passing through A. The rod is released from rest from the horizontal position. If the rod gets broken at midpoint C when it becomes vertical, then just after breaking of the rod. Choose multiple answeres from the below...
(The answer given in the text says ##\boxed{T_1\; >\; T_2}## but, as I show below, I think it's just the opposite).
I begin by putting an image relevant to the problem above. Taking a small particle each of the same mass ##m## at the two positions, the centripetal forces are ##T_1 =...