# Tag Info

28

What keeps a bicycle up is a variety of things, but it all comes down to the front wheel, which can move left/right. The bike is always out of balance, and if it starts to fall to the left you unconsciously turn to the left, which moves the point of support (the wheel on the surface) to the left, which arrests the fall and may start the bike falling to the ...

19

By the principle of relativity, you will not fall over – assuming that you know how to use the bike and you won't be deliberately "confused". The principle says that the laws of physics have the same form in all inertial frames that are moving by a constant velocity relatively to each other. The reference frame associated with the moving sidewalk is as good ...

7

The energy of an element of a traveling wave is not constant. Halliday-Resnick-Krane is right. For a string of density $\mu$ and tension $T$ the kinetic energy of an element $dx$ is $$dK=\frac 12\mu dx\left(\frac{\partial \xi}{\partial t}\right)^2.$$ For the potential energy we have $$dU=Tdl,$$ where $dl$ is the stretched amount of the string. A small ...

4

The second derivation is correct, as explained by Diracology. However, the first derivation is 'sort of' correct, in the sense that the location of potential energy can be ambiguous. For example, consider the three following systems. A mass on a stretched spring. A mass sitting on a table. A charged mass next to another charge. These three systems have ...

4

Yes: I've done that. I used to have a device for the purpose, commonly called "rollers". It's like a treadmill for bicycles.

4

That formula holds for a simple pendulum of length $L$ in a gravitational field $g$ and released from rest with an amplitude $\theta_0$. Since this system is conservative its mechanical energy is constant and equals the gravitational potential energy when it is released. Setting the zero of potential energy at the fixed point of the pendulum, the mechanical ...

4

Multiply both sides of the equation by $\Delta l$ to obtain: $$k\Delta l=EA\frac{\Delta l}{l_0}$$ The relationship on the left is the tensile force F. The quantity $\frac{\Delta l}{l_0}$ is the tensile strain. The Young's modulus E times the tensile strain is equal to the tensile stress $\sigma$:$$\sigma=E\frac{\Delta l}{l_0}$$ The tensile stress times ...

3

Tl;DR: It is mainly do to the skier's positioning on the ski and how being thrown off-balance by the avalanche affects it. Concerning the first part of the first question: 1) If all free falling objects accelerate at the same rate (this was on a fairly steep mountain section), why did we get "trapped" into the avalanche, when our acceleration already ...

2

But my problem is that I don't understand how a force can be a vector, in my head I see it as a direction vector and some power number Right. If the direction is a "unit" vector, then you can compare the magnitudes of different forces to compare the strengths. But you can multiply the magnitude and the direction to get a new vector that contains ...

2

Using Torricelli's law (https://en.wikipedia.org/wiki/Torricelli%27s_law) you get $v=\sqrt{2gh}$ irregardless of how big the opening is. Now you can calculate how much mass is leaving the tank at any time by multiplying the volume that is leaving the tank by its density: $$\Delta m=A\Delta s\cdot\rho$$ with $\Delta m$ the mass leaving the tank at any second, ...

2

(Source of image: Mohsin Khan, http://cslearners.blogspot.com/2009/08/equation-of-motion.html) Here they are! All the formulas. Sorry to say, you cannot find anything if you have only acceleration and distance. Think like this, Say you have an object that has an acceleration of 2 m/s^2. and if i say that if travels a distance of 2 meters. It can travel ...

2

If you draw a velocity-time graph you will see that you do not have enough information to find the initial and final velocity. The gradient of the graph is fixed because it is the acceleration. The distance is the area under the velocity-time graph. As you will see from the graph you can draw an infinite number of trapeziums (or triangles) which satisfy ...

2

Newton's third law is a statement that momentum is conserved, so it is equivalent to the law of conservation of momentum. Conservation of momentum follows from a fundamental symmetry (of the action) called space shift symmetry and as far as we know this applies to all our physical theories. So Newton's law is still valid but has to be treated with some care ...

2

But when in space, there is no such thing called direction. In space, you have to first choose a frame of reference in order to measure or calculate vector quantities. The choice is arbitrary but whatever choice you make, you will end up with the same answer. Some choices make the calculations easier. For example, you could identify a nearby small ...

2

Finding the missing equations Coordinate transforms just complicate the issue. The heart of the matter is that in n dimensions you have n degrees of freedom for the velocities of the COM of each sphere, and you only have n momentum conservation equations plus one energy conservation equation. That means you need an additional n-1 equations to solve the ...

1

Force has both a magnitude and direction, which are the properties of a vector. The magnitude can be given by $\vec{F} = G\frac{M_1M_2}{\vec{r^2}}$ where $G$ is the gravitational constant and $M_1,M_2$ are the masses. the distance from each other is represented by the vector $\vec{r}$ which will be the displacement from the origin.

1

Half of your questions are concerning Newton's law: Why is it non-uniform? Because the object would have different densities in different parts, so weight would be greater in some parts? Yes. Think of a car. It is in contact with the ground in four places and pushes down causing four normal forces. If the car is heavily loaded with bagage in the ...

1

We have to talk about Physical Strength last, for two reasons: (1) we have to clearly define the other physics terms first, and (2) Physical Strength gets us into biophysics so we'll have to talk about how muscles generate force, etc. "Force" is a push or a pull or a twist (if a twist, then it is also called "torque"). Force is what is required to ...

1

If the object does not slide, the lowest point on the base will form a pivot about which the object can turn. If the vertical line through the CG is inside the base, then the weight forms a torque about this pivot which turns the object back onto its base. If this line falls outside the base, the torque turns the object away from its base.

1

For the first part of your question, you have to realize that your net velocity (the one that you plug into the expression for centripetal force) is the vector sum of the surface velocity and your velocity relative to the surface. If you were running West as fast as the earth turns East, you would "stay in place" and the sun would appear to stop moving in ...

1

A force applied uniformly to a uniformly dense object (or with a uniform-force-per-unit-mass even if the object is not uniformly dense) is equivalent to the same total force applied at the center of mass of the object. The equivalence means the same total force and the same total torque. In particular, if the center of mass is chosen as the origin of the ...

1

The cheap and easy answer to this is that the double pendulum is considered chaotic because it is very sensitive to small perturbations in initial conditions (amongst other things). Showing this mathematically may be difficult (see the Lagrangian formulation for the dynamics), but if one looks at the animations on the Wikipedia page showing the trajectory ...

1

Your formula's wrong. You've got $v=\frac12 at^2$, whereas that's the formula for $y$=height. Velocity's actually $v=at$ (with $a=9.8\mbox{m/sec}^2$).

1

The equation of motion for the ball from the time it bounces till the time it hits the ground again is $$y = v_0t - \frac{1}{2}at^2$$ where ground level is $y=0$, and $v_0$ is the velocity going up after adjusting for the coefficient of restitution, and $t$ is the time since the bounce. This equation will take the ball through its peak and back to the ...

1

Gravity is an attractive force, and classical Newtonian theory is adequate to answer in the affirmative: starting from two objects in space at rest, the attractive potential is 1/r , r their distance, and they will fall on each other. The reason planets and satellites have a stable orbit is due to the solutions of the kinematic equations when there exists ...

1

Assuming the spring is "ideal" (massless) you actually have 2 masses. You can describe your problem as the motion of the center of mass, and either of the masses. And if no external force is exerted on your system, you are only left with the motion of 1 mass relative to the center of the mass of the system. Let's say your masses are m1, m2, and the spring ...

1

Yes. All objects are gravitationally attracted to one another. Even people. Therefore, the earth will draw you to itself no matter how far out you are.

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To do the equations of motion you need positions from an inertial reference frame. Say a wall far away. Call positions of the two masses $x_1$ and $x_2$. The spring force (tension is positive) is $$F = k (x_2-x_1 )$$ and the two equations of motion \begin{align} F & = m_1 \ddot{x}_1 \\ -F & = m_2 \ddot{x}_2 \end{align} All this is trivial. ...

1

Now the real problem I'm having is trying to decide what the forces are acting on the system in order to come up with my differential equation? As there are no external forces the linear momentum of the system will be constant of motion and the constant can be taken as zero as well. If m1 and m2 are the masses at any time described by position ...

1

Here's the illustration that I would use1: The blue spheres represent blobs of material, where a blob could be any number of atoms or molecules. The red lines represent the bonding force between the blobs. Young's modulus $E$ represents the strength of the bonding force, i.e. the strength of exactly one bond. When you stretch the material, all of the ...

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