# Tag Info

## Hot answers tagged rotational-dynamics

72

Your wire is not quite round (almost no wire is), and consequently it has a different vibration frequency along its principal axes1. You are exciting a mixture of the two modes of oscillation by displacing the wire along an axis that is not aligned with either of the principal axes. The subsequent motion, when analyzed along the axis of initial excitation, ...

25

Model the tree as a point mass $m$ located some height $h$ above the ground --- that is, forget the mass of the trunk and assume all the mass of the tree is in the branches and leaves above the ground. Then the moments of inertia of the tree before and after felling are \begin{align} I_\text{tree,up} &= m \left( (R+h)\cos\theta \right)^2 \\ I_\text{...

12

UPDATE : After looking again at the video, I agree that Floris' explanation seems to be correct and my explanation below is wrong. Slightly different frequencies of vibration in two perpendicular planes accounts more simply for a rotation which reverses one way then the other. Kinetic energy seems to decay constantly; it does not seem to be stored in an ...

10

A long rod, especially with additional masses at the ends, has a large moment of inertia and therefore can change its angular velocity only slowly. This means that if the walker gets off-balance, there is more time available to correct before he falls.

6

Here is a video that shows the ball behaviour you're describing. The phenomenon is explained by the Coandă effect: the tendency of a fluid jet to stay attached to a convex surface. Note that the ball does actually move in a kind of oscillatory motion. This is probably due to the water jet in the video not being highly stable. When more liquid is running ...

4

Cutting the trees and leaving them flat instead of vertical will diminish the moment of inertia of earth. The angular momentum of course will not change, but the speed of rotation will increase. However, I do not believe that the change is measurable with current instruments.

4

I know the example given sounds crazy but the physics behind it might be useful for someone learning rotational dynamics. The angular momentum of a system does not change if there are no external torques, since $$\frac{d\vec L}{dt}=\vec \tau,$$ where $\vec L$ is the total angular momentum and $\vec \tau$ is the total external torque. So if you cut the trees ...

3

I just add to above points which are mostly correct my two pennies. 1- the wire is loaded with a history of residue strains from the manufacturing and handling so the stiffness and elasticity of it is not homogenous lengthwise or even along cross-section. if you'd write the whiplash DE's in a finite element software it is not a linear differential equation ...

3

$$F_1=mg\cos\theta$$ $$F_3=mg\sin\theta$$ With no radial notion and Newton's 2nd: $$F_2=F_1$$ Using the simple model of friction, with $\mu$ the static coefficient of friction (no slipping is assumed): $$F_f=\mu F_2=\mu mg \cos\theta$$ Call $R$ is the radius of the sphere and $r$ the radius of the ball. The torque $\tau_1$ caused by $F_3$ about $O$ ...

2

If you accept that no external work was done, then if there is a change in the state of a system through which the kinetic energy changed, there must be a corresponding change in potential energy. The key to understanding the (rather poorly narrated) video is that the lecturer implies (at T=2:30) that $\Delta E=0$ from which it follows that \Delta KE= - \... 2 Flexibility plays no role, because the rod can be naturally bent. If the walker has no rod and his CoM is not aligned to the rope he hasn't a big chance of adjusting it. A rod contributes to the mass of the system (and the position of CoM) and distribute mass on a greater extension, and the man can shift it sideways or rotate it. Angular momentum mvr = L ... 2 Displacing some mass closer to the axis of rotation reduces the moment of inertiaI$. Considering Earth as an isolated system (which is not), its angular momentum$L$must be conserved: $$L=I\omega = \text{const}$$ Therefore if$I$goes down, the rotation frequency$\omega$must increase. But if we should also consider the reduction of$I$while the tree ... 2 I would think that the whole atmosphere surrounding the earth is far heavier than the trees that were cut. The atmosphere turns with the earth and the changed position of trees would not even be noticed. 2 You might hear the story about figure skating. When a rotating person expands his/her arm, he/she can slow down rotation. Same thing can happen with earth. Assuming the tree is trillion tons and you cut it and lift it up, you can slow down the earth. 2 For a good reference, try Landau and Lifshitz book on Elasticity. Suppose you have a wheel (approximated by an infinite cylinder) that is coming into contact with the flat ground. By symmetry, the area of contact will be a rectangle that will be very thin in one direction and very long along the length of the cylinder. Call the long length$a$and the small ... 2 If I read the description correctly, your experiment is set up to impart an initial velocity to the coin before your timing starts - as you drop it from slightly above the edge of the table, but don't start timing until you hit the edge (hear the impact). At the moment of impact, the center of mass of the coin will slow down a little bit - but it will still ... 2 Switching the electromagnet on and off at appropriate times will make the arrangement into a (not particularly efficient) electric motor. In order to use it to spin up the disks -- or just overcome the friction in the bearings for an extended time -- you will find that the electromagnet requires slightly more electric energy deposited into it in order to ... 2 In answer to your 1st question, whether the magnets attract or repel depends on how their poles are oriented with respect to each other. However, as they pass the confluence point C (ie position in 2nd image) the interaction will overall have no effect on the speed of the discs. For example, N & S poles approaching each other will attract and accelerate ... 2 Hold it in the center, make it vertical and either throw it, or give yourself some spin and just release it. If you hold it in the center and keep it horizontal, it will be like a spear, but if you hold it in the center vertically, it will also go without rotation. Hold with both the hands and throw like a a two hand hi-five. Throw like a javelin or spear, ... 1 As drawn the disc's angular speed$\omega$is too fast as related to the velocity of the centre of mass of the disc$v$for the no slipping condition ($v = R \omega$, with$R$the radius of the disc) to be satisfied. You can think if the frictional force as trying to accelerate the centre of mass whilst at the same time the frictional force applies a torque ... 1 It is not necessary to consider the torque on the ball. It only makes the situation more difficult to think about. It is best to stick to energy considerations. Because the ball does not slip, the friction force does no work. However, contrary to your assumption, this does not mean that rotational KE is constant. This is where your reasoning is going ... 1 No. As you realise, this will increase the moment of inertia, which will reduce angular acceleration since the torque which the motor can supply is limited. However, the maximum speed which the motor can reach is not affected, since this depends (mainly) on the aerodynamic force, which is the same - it depends on the shape and size of fan, but not its mass.... 1 There are mathematical ways of showing all this, but it is important to have intuition before starting to calculate. Here is my intuitive picture: The kinetic energy of the rolling cylinder has two components: The kinetic energy of its motion along the plane. The kinetic energy of its rotation about its own axis. The only source of kinetic energy is the ... 1 There is nothing wrong with your proposed approach, but I think maybe you may have misunderstood your physics teacher; most likely he/ she would like you to be completely comfortable and proficient with the analysis of dynamical systems from an inertial frame (i.e. unaccelerated frame) before shifting on to analysis from accelerated frames, which involve ... 1 Proceed by choosing a point O on the symmetry axis Z. Choose a differential fluid element having dimensions$\mathrm{d} r$,$\mathrm{d}x$,$\mathrm{d}y$where x is the polar angle in your horizontal plane of observation and y is the azimuthal angle perpendicular to this plane. An F.B.D of this element should easily give the following 2 equations:$P_z = p -...

1

To formalize @knzhou's comment: The answer, in a nutshell, is no. Your basic assumption that cutting trees reduces the Earth's mass is wrong, because trees don't leave the Earth when they are cut! Even if all trees left the Earth when cut, a lot of tree cutters plant trees to replace what they cut, and each tree is such a tiny amount of mass ...

1

If the rod were rotating around a point other than the center of mass, then the center of mass would be rotating around this point. That is, the center of mass will be accelerated to perform circular motion, but we are assuming here that there are no external forces acting on the system (other than the initial push), so the center of mass cannot accelerate....

1

For a pulley that has mass and moment of inertia, there must be a net torque on the pulley for the pulley to demonstrate an angular acceleration. Assuming that mass "M" is greater than mass "m", a net torque necessarily requires that the counterclockwise torque from mass "M", given by the equation Torque1 = T1(R), is larger than the clockwise torque from ...

1

Yes. Tension can vary if external forces are acting between the ends of the string - such as gravity (if the string has mass) and friction where the string makes contact with other objects (such as the pulley). For example, suppose you attach one end A of a uniform massless string to a support and the other end C to a vertically hanging mass M. This ...

1

The tension in a string is uniform Not always. The tension of the string is uniform in some cases: If String is mass-less and its particles don't move with respect to each other (i.e. string is inextensible or if it is extensible, it reach to its final tension). If String is mass-less and there is no friction between string and pulley. For string with ...

Only top voted, non community-wiki answers of a minimum length are eligible