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6

The issue is that the formula that connects force and potential gets an extra term when the force depends on velocity ${\bf v}$. The formula reads (see e.g. Ref. 1) $$\tag{1} {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}},$$ rather than just $$\tag{2} {\bf F}~=~ - \frac{\partial U}{\partial {\bf r}}. ... 4 Friction causes the chalk to stay on the chalkboard. While the chalkboard appears smooth, under a microscope its surface is rough. Chalk is a much weaker material than the chalk board. When it is forced across the chalk board, small parts of chalk ('dust') are broken and remain trapped by friction in the surface asperities of the chalk board. The rougher ... 3 According to conservation of momentum, the center of mass of a system cannot accelerate without external forces. In other words, if the center of mass starts out at rest (which is generally a good procedure in simulations), then it should always stay at rest. It is normal for numerical errors to introduce deviations, but the motion you are seeing looks ... 2 When talking theoretically, an ideal flywheel rotating clockwise or anticlockwise would face the same magnitude of frictional forces in opposite directions. But if the flywheel is made such so as to rotate in one direction only or has worn out and/or has been damaged might produce different magnitudes of frictional forces. The surfaces which is rotating on ... 2 From another comment: I wonder why the object with greater mass does not accelerate slower Maybe it will help, to think of it like this: If you (theoretically) split up the object in many, many (let's say infinitely many) pieces of equal mass, then gravity can pull equally in each of those pieces. They will all fall with equal acceleration g. If ... 2 Forces are only affecting acceleration. Not any other parts of motion. Think of Newton's 2nd law:$$\sum F=ma$$If the sum of forces is zero as in your case, nothing accelerates. If it was standing still, then it continues standing still. If it was moving then it continues moving! It only takes a force to change a motion. Not to keep it up. 2 Your equation for the damped solution is wrong. In order to match the boundary conditions (initial velocity = 0) you have to add either a phase, or a \sin term. I prefer the phase. If initial velocity is zero, the derivative must be zero:$$A_t = A_0 e^{-t/\tau} \cos(\omega t + \phi)\\ v_t = A_0\left(-\frac{e^{-t/\tau}}{\tau}\cos(\omega t + \phi) - ...

2

Consider the diagram below of a ball on a horizontal surface: Newton's Laws tell us that if no net force acts on the ball it will remain in a constant state of motion ($v=0$ or $v=\text{constant}$). Consequently, if no net torque acts on the ball its state of rotation will also remain constant ($\omega=0$ or $\omega=\text{constant}$). Where friction does ...

2

I gather that the large source of error you are worried about is the ability of the experimenter to accurately hit the start/stop button on the stopwatch at the start/stop of the ball's journey down the ramp. What is the approximate magnitude of error we'd expect? Before I directly answer your question, let's estimate how bad the experimental error will be ...

1

To see that your integral expression does not make any sense, imagine that $\vec{r}(t)=( x(t),y(t))$ describes a circle. Then the line integral of the force around the loop gives the change in potential energy, which should of course be zero, $$\oint \vec{\nabla} \phi \cdot \vec{dl} = \Delta \phi =0.$$ But if you insert the actual values from your ...

1

Let's look at this problem from the point of view of equations of motion, see diagram below: Firstly let's make a few assumptions. Ball and cube are of same weight ($mg$) and same size. Simple friction model $F_f=\mu F_n$ holds and $\mu$ is independent of speed. Both objects are completely stationary (no sliding, rolling or tumbling) at $t=0$, at which ...

1

The system is subject to a non-zero net force in the horizontal direction and no friction, so it will experience constant acceleration (of the center of mass). Superimposed on that motion with be the anti-symmetric oscillation of the two masses on the spring. If the masses are both $m$ and the spring is characterized by constant $k$ the angular frequency of ...

1

It depends on the surfaces how high the friction becomes. If they are smooth and clean enough how do they "know" that they are separate surfaces? It is possible for them to actually weld into a a single part. Although this is difficult to achieve in practice - it is annoying when you don't want it to happen.

1

If you are sliding across the surface, then "static friction" is not applicable. Consider first your motion on a merry go round without sliding. At any instant, your tangential velocity is the same as the tangential velocity of the surface under your feet. Since the two velocities are the same, no instantaneous frictional force is required to keep you moving ...

1

Just because an object is in motion, that doesn't mean that there is a force acting on it. Newton's 1st law states that an object in motion will stay in motion unless acted on by an outside force. So the car's momentum is what keep's it going as it coasts. The forces acting on the car would be gravity, the normal force (the ground pushing up on it), and ...

1

The idea is that, roughly speaking, the block moves down the gradient of the slope. Because the slope changes its direction of motion, it pushes the block left and right in roughly equal measure, and because it has a short period, the block never moves with significant horizontal velocity relative to the fixed axes. Thus, the horizontal speed of the block ...

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I assume that friction is an external velocity dependent force in your simulation code. Since you have such external forces, your total energy, total angular momentum, total momentum are likely not to be conserved. In your case, the friction is a phenomenological external force, but similar behavior could also be simulated with a large particle, moving in a ...

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Because the work done by friction is converted into rotational kinetic energy of the cylinder, since friction provides the torque to roll down the cylinder.

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Answering your Title question. NO, the objects do not change their state of motion if all the forces cancel each other. I will assume that the friction on the object is maximum. That means $200N$ is the maximum friction on the surface. Now the force needed is $200N$ to overcome the static frictional force. So if initially you apply $200N$ force, there will ...

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