# Tag Info

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It seems to depend mainly on the spin. See the comsol.com simulation at https://www.comsol.com/blogs/physics-behind-baseball-pitches/ For the slider, in essence a lateral motion, Just spin it laterally, right to left if you are rightly, and it'll curve to the left. The axis of rotation would be vertical, if righty the direction of the angular momentum ...

1

In short, you are to think of the direction of the torque as pointing along the axis of the rotation it would induce in a rigid body initially at rest. But if the conception of torque as a vector out of the page seems artificial, that's because it is. Torque is not fundamentally a quantity that is a vector but a directed plane or directed area. Such an ...

1

There are probably lots of duplicates, so my apologies, but for clarity I will try a short answer, as the graphic from Wikipedia is particularly illustrative. The torque is perpendicular, ( orthogonal) to the other two vectors, so it could be the line where the hinges are located, depending on the direction of the other two forces. From Wikipedia ...

3

No you cannot assume that. The initial rotation is about the major axis, and it will continue to be so (in the absence of torque, and since you were already rotating about the major axis). Instead, since $\omega_2=\omega_3=0$, your equations for the evolution of the angular momentum don't require the moments of inertia to be the same.

4

If I understand correctly, you are asking if a meteor impact could (i) slow the Earth's rotation on its axis or revolution around the Sun enough to account for the 8 to 12-fold decrease in longevity of human-kind measured in Earth days/years; and (ii) cause 40 days of torrential rain, resulting in sufficient inland flooding to float a large wooden boat. An ...

2

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 ...

0

Drop a piece of paper and it glides sideways as well as flips. So aerodynamics (and hence the shape) affect the way things fall. Specifically aerodynamic forces have a center of pressure, which when ahead of the center of mass the body would rotate and flip, but if behind it will swing and stabilize at this orientation. This is the reason arrows, darts and ...

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

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, ...

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 ...

0

The thing is, the book claimed the catapult had no recoil during launch thanks to its fixed counterweight. Is this physically correct? Of course not. The projectile has been given some momentum, so the launcher must recoil by conservation of momentum. You can't dodge that. In this case the problem with the proposed mechanism is that after the (initially ...

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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 ...

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 $O$ ...

0

Somewhere in between $T_1$ and $T_2$ -- but exactly what depends on details about how the friction between the string and the pulley varies, how the string stretches under tension and the pulley deforms while being accelerated by the string ... All of these are things that cannot be deduced from an idealized picture such as this. In one extreme, if the ...

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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....

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 ...

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 ...

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Thank you for the update, WJ47. The slope of the blue tube looks very steep. Both the ruler/tube and the white cylinder (cellotape holder) look quite smooth, so I think there will be little friction, resulting in a mixture of rolling and sliding here. It is very difficult to predict how much of each. This is a rather 'messy' experiment, IMHO, difficult ...

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 ...

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, ...

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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 ...

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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: $P_z = p -... 10 A long rod, especially with additional masses at the ends, has a large moment of inertia and therefore can change its angular velocity only slowly. This means that if the walker gets off-balance, there is more time available to correct before he falls. 2 Displacing some mass closer to the axis of rotation reduces the moment of inertia$I$. Considering Earth as an isolated system (which is not), its angular momentum$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$Iwhile the tree ... 2 I would think that the whole atmosphere surrounding the earth is far heavier than the trees that were cut. The atmosphere turns with the earth and the changed position of trees would not even be noticed. 2 Flexibility plays no role, because the rod can be naturally bent. If the walker has no rod and his CoM is not aligned to the rope he hasn't a big chance of adjusting it. A rod contributes to the mass of the system (and the position of CoM) and distribute mass on a greater extension, and the man can shift it sideways or rotate it. Angular momentum mvr = L ... 2 You might hear the story about figure skating. When a rotating person expands his/her arm, he/she can slow down rotation. Same thing can happen with earth. Assuming the tree is trillion tons and you cut it and lift it up, you can slow down the earth. 25 Model the tree as a point massm$located some height$habove 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{... 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$is the total external torque. So if you cut the trees ... 1 To formalize @knzhou's comment: The answer, in a nutshell, is no. Your basic assumption that cutting trees reduces the Earth's mass is wrong, because trees don't leave the Earth when they are cut! Even if all trees left the Earth when cut, a lot of tree cutters plant trees to replace what they cut, and each tree is such a tiny amount of mass ... 1 If the rod were rotating around a point other than the center of mass, then the center of mass would be rotating around this point. That is, the center of mass will be accelerated to perform circular motion, but we are assuming here that there are no external forces acting on the system (other than the initial push), so the center of mass cannot accelerate.... 0 Apart from friction in the load (which cannot be predicted), the flywheel does not make any difference to the maximum rotation speed reached by the motor. If there is no friction, then given sufficient time the motor can accelerate the flywheel up to its own maximum unloaded speed, however small the torque supplied by the motor. An even higher speed can be ... 6 Here is a video that shows the ball behaviour you're describing. The phenomenon is explained by the CoandÄƒ effect: the tendency of a fluid jet to stay attached to a convex surface. Note that the ball does actually move in a kind of oscillatory motion. This is probably due to the water jet in the video not being highly stable. When more liquid is running ... 0 Yes I agree, the solution in the manual misses out some terms in the equation of motion, as well as the factor$\frac12$in front of$k\$. Probably some approximations have been made without explanation. Your equation seems correct to me. When the ball arm is vertical, and there is no rotational motion in the vertical plane, then balancing moments about O ...

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