Hot answers tagged rotational-dynamics
15
Applying the brakes makes the wheel stop turning in relation to the bicycle's frame but not in relation to the road. The bike's center of mass (especially with a rider pressing against the handle bars) is higher than the hub of the front wheel.
When the brakes are applied that mass has momentum toward the front of the bike that exerts a force on the front ...
7
First I wanted to turn to textbooks to solve that, but felt this way is going to be boring (especially if we are talking Mass Effect). So I decided to derive equations of motion for our system from the first principles.
Further I need a lot of trigonometry, so I'll use short-hands for cosine and sine: $$c_x=\cos x,\;s_x=\sin x$$
Let us start by introducing ...
7
First break the problem up using two free body diagrams.
Then figure out the kinematics at point A
$$ \vec{r}_A = \begin{pmatrix} x \\ 0 \\ 0 \end{pmatrix} $$
$$ \vec{v}_A = \begin{pmatrix}\dot x \\ 0 \\ 0 \end{pmatrix} $$
$$ \vec{a}_A = \begin{pmatrix}\ddot x \\ 0 \\ 0 \end{pmatrix} $$
and point B
$$ \vec{r}_B = \vec{r}_A + \begin{bmatrix} ...
6
I like your description of this cool bit of unintuitive physics. I find the best balance of $a$ to $b$ to $c$ to cost of the object involved is best for a (boxed) pack of playing cards.
The mathematical explanation for this (see also Wikipedia) is that when considered in the principal axis frame (i.e. the frame of reference that rotates with the body and ...
6
The center of mass of the bike+rider wants to keep moving forward (Newton's first law)
Since the center of mass is higher on the bike - when it goes forward the bike pivots around the point where the front wheel sticks to the road.
Imagine attaching a string to the middle of the wooden block and pulling it forward, while having the front wheel stuck to ...
6
There is simplified "cheat" method for thinking about this.
The hard way is to draw a free-body diagram of the bicycle, in which all the forces acting on it are made plain: the force of gravity acting through its center of mass, the force of friction from braking, and so on.
The cheat method is to visualize the bicycle in an accelerating frame of ...
5
No.
A top can offer stability like in a gyroscope, but it does not in any way reduce the mass of an object.
Some applications of a gyroscope are useful for transport, however. The popular Segway scooters use them to measure how far/fast it is tilting forward or backward and drives the wheels to compensate.. Similarly, motorcycles take advantage of the ...
5
You can always decompose a motion like this into two parts: (1) rolling without slipping and (2) slipping without rolling.
What is slipping without rolling? It means the object moves uniformly in one direction along the surface, with no angular velocity about the object's own center of mass. For instance, a box that is pushed along the ground can easily ...
4
First the answer: The motion is not quite stable, but only because of two subtle thing that your brain probably can intuitively feel:
the rate at which the outer ring (the one that is rotating around the vertical axis) rotates has to speed up and slow down as the inner ring becomes vertical and horizontal respectively. Once you slow down and speed up the ...
4
Start with the unperturbed gravitational potential for a uniform sphere of mass M and radius R, interior and exterior:
$$ \phi^0_\mathrm{in} = {-3M \over 2R} + {M\over 2R^3} (x^2 + y^2 + z^2) $$
$$ \phi^0_\mathrm{out} = {- M\over r} $$
Add a quadrupole perturbation, you get
$$ \phi_\mathrm{in} = \phi^0_\mathrm{in} + {\epsilon M\over R^3} D $$
$$ ...
4
I'm not sure if this is the dominant factor, but...
Once the coin begins to tip at all, there is torque due to gravity. If you work it out with your hands you'll see that this torque acts perpendicular to the angular momentum from the rolling of the coin and in the plane it rolls on. Thus, it acts to direct the coin in a circular path.
A quick experiment ...
4
Imagine the string runs over a pulley and is connected to a weight, like this:
The weight must fall, which means the red dot must move to the right. To make that happen, the yo-yo must roll to the right.
4
1) The first thing I notice is that you have stated that the velocity at the end of the ramp is $2\textrm{ m/s}$. Remember that the can is accelerating as it rolls down the ramp, so the equation $v=\textrm{d}s/\textrm{d}t$ is not applicable here for finding the instantaneous velocity at the bottom. The can does indeed average $2\textrm{ m/s}$ during its ...
3
It is the external energy that the hoop needs to spin. The Hamiltonian is a conserved quantity since it does not depend on time explicitly, but the mechanical energy (kinetic plus potential) is not conserved.
Note that:
$$E=K_1 + K_2 + U$$
where $K_2$ is the kinetic term which does not depend on velocities $\dot \phi$, then
$$L=K_1 + K_2 - U$$
and
...
3
There are lots of different examples of oscillatory systems that have essentially the same mathematical form. Let's start by just looking at one type of differential equation:
$a = \frac{d^2 x}{dt^2} = -\omega^2 x$
This equation has a general solution (you can check this)
$x(t) = A \sin (\omega t + \phi)$
which oscillates with a period of ...
3
The issue appears to be rather complex, so I do not aim at providing an exhaustive answer.
At a toy model level it is reasonable to model the eye as a "camera". Specifically, let us assume that a human eye "samples" at a maximum frequency of $\nu$, so that we may make use of the Nyquist-Shannon sampling theorem. Basically, given an instantaneous angular ...
3
In this answer, I will present a framework to use, and then I will frame the prior answers within that framework. Let me sum up the values we have here. I'll use the same notation (as best as possible) as everyone else and Wikipedia for an oblate spheroid where $a$ is the large, equatorial, radius.
Mark1, method in the question, $2 (a-b) = 21.6 km$
...
3
I look at two models of a "fat earth":
a spherically symmetric interior with an aspherical surface layer in hydrostatic equilibrium. This analysis generalizes from the constant density assumed in other answers and thereby exhibits the sensitivity of the flattening to the surface density. I compare the result to those of various other answers.
To estimate ...
3
Assuming your rotating object (e.g. the Earth) is rotating at a steady speed the only way to change it's apparent speed of rotation is if you're rotating around it.
You give the example of a geostationary satellite. This rotates around the Earth at the same angular velocity as the Earth rotates, so the Earth appears to be stationary (hence the name ...
3
You can think of ${\vec F}_{\rm net}$ as consisting of all of the external forces acting on a system of particles, irrespective of where those forces are applied. The only place where the center of mass comes into play is on the other side of Newton's second law: the ${\vec a}$ in $\sum {\vec F}_{i} = m{\vec a}$ is most assuredly the acceleration of the ...
3
$T$ must be accounted for when you sum the forces. The application of $T$ away from the center of mass is equivalent to moving $T$ to the center of mass and adding an appropriately sized couple.
3
First, efficiency of an electric motor is just output power divided by input power. Input power is your electrical input power, which is V*I. Output power is your mechanical output power, which is speed*torque.
Given that, we can see that efficiency for every motor is going to be 0% at no load (i.e., maximum speed at 0 torque). Efficiency will then ...
3
It does have a force. It is the centripetal force. Usually, the spinning body is a circular object, so all centripetal forces are balanced out to 0. For each point mass, there is another point mass on the opposite side. It is the symmetry of the system that keeps everything in the same place spinning.
2
It is simple. $\vec{\omega}_{[e]}$ are not the components of angular velocity seen in the reference frame attached to the rigid body itself. As you point out, that angular velocity is zero.
It is the result of mathematical manipulation. You have a set of relations between the basis vectors of the inertial frame and the rotating frame, and you use that to ...
2
Short answer: It's the geometry.
Consider a bucket which is quite stable having and has $r_{1} < r_{2}$. Note circumference $2\pi r_{1}$ is less than outer circumference $2\pi r_{2}$.Now if bucket is set into motion, the point which lies on inner circumference has to cover less distance compared to point on outer circumference. Since whole body is ...
2
I don't know if this is what OP is asking(v1), but the rotational temperature $T_{\rm rot}$ times Boltzmann's constant $k_B$ denotes a typical energy $k_BT_{\rm rot}$ involved in the difference $\Delta E$ between two neighboring rotational energy levels of, say, a molecule.
This fact follows from the classical formula $\frac{{\bf L}^2}{2I}$ for rotational ...
2
To a certain extent yes, if you take aerodynamics into account. Specially made top can deflect air to provide a lift force. Exaggerated example is a helicopter. A better example though is Frisbee.
But it's not because the mass reduced or something is done to gravity, it's just a force exerted by surrounding air.
By the way, George, provide a reference to ...
2
This puzzle confused me too a long time ago, and the answer you get is absolutely right. If you apply the same impulse on the side, the disk will still translate with the same linear momentum, but it will also rotate around the center. The rotation is given by the angular momentum impulse, which is the torque $Fr$ times the increment of time.
The reason ...
2
As explained on Wikipedia, the nice tensor form of the equations is
$$ \mathbf{I} \cdot \dot{\boldsymbol\omega} + \boldsymbol\omega \times \left( \mathbf{I} \cdot \boldsymbol\omega \right) = \mathbf{M} $$
This reduces to your equations if one diagonalizes the tensor of the moment of inertia $I$ and labels the diagonal entries etc.
The three components are ...
2
You can always repalce an off center force by the same force centered plus a torque $r\times F$.
So the trajectory and acceleration of the COM are the same you would get with the same force centered, so you can solve them independently of the rotational dynamics. The torque mentioned above is what drives the rotational dynamics of your rigid body.
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