# Tag Info

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If you have a general oscillator with restoring forces and damping forces: $$F_{rest}=-kx,\text{ } F_{damp}=-b\frac{dx}{dt}$$ Such that the equation of motion would be: $$m \frac{d^2x}{dt^2}=-b\frac{dx}{dt}-kx$$ $$m \frac{d^2x}{dt^2}+b\frac{dx}{dt}+kx=0$$ This equation has solution of the form : $$Acos(\omega x + \theta)e^{t/\tau}$$ Where you can ...

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$Ma$ expresses a force. Friction is: $f = μN = μmg$ right? The car accelerates with a certain acceleration which results to a certain force. Once the magnitude of that force ($Ma$) becomes equal with the maximum magnitude of friction ($μmg$), the car skids. I hope this helped.

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There should be different between static frition coefficient $\mu_s$ and the kinetic friction ceofficient $\mu_k$, and $\mu_s$ must be greater than $\mu_k$. If $\mu_s \lt \mu_k$, there will induce non-physical phenomena. Another concept is that the kenetic friction (as long as two badies have relative motion) is a constant $f_k = \mu_k N$ within the simple ...

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Your first question is answered here: Torque - Why does a block topple and net torque when net force is non-zero?. I understand that static friction causes rolling without sliding, and comes into the picture to prevent sliding motion of the ball. As static friction between two surfaces takes a maximum value, is there some sort of a limit to this rolling ...

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Since friction occurs due to the irregularities on the surface of materials where between two bodies when these irregularities "bumps" meet, charged particles get very close to each other and exert electromagnetic force (attractive) note: $F_{EM} \sim 1/r^2$, magnitude of the EM force increases as the distance between charged particles get smaller. ...

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I know that the friction is μmg Not always true. $\mu mg$ corresponds to the maximum friction, $f_{\rm max}$. The static friction force $f$ can be anywhere between $0$ and $f_{\rm max}$: $$0\leq f\leq f_{\rm max} = \mu N$$ To solve the problem, first, write down the net force and torque equations $$ma=F-f \qquad \text{and}\qquad I\alpha = fR$$ Using $v=\... 2 In the rotation frame of the disc, the bug (mass$m$) position$\vec{r}'(t)$relative to the disc. Note that$|\vec{r}'| = | \vec{r} | = r$, the disc frame differs from the prime system only in the angular velocity. The bug moving outward suffers the following forces: The kinetic frictional force,$ -\mu_k m g \hat{v}'$; the centrifugal force,$m \omega^2 \...

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Yes, it is correct. The maximum static friction is $F_A^{max}=\mu_S Mg$, being $M$ the mass of the wheel, and its torque is $\tau_A^{max}=R\mu_S Mg$, so the maximum angular acceleration without slipping is $$\alpha^{max}=\frac{\tau_A^{max}}{I}=\frac{R\mu_S Mg}{I}$$ Let $F$ be the exerted force, which produces an acceleration $$a_{CM}=\frac{F}{M}$$ so $$\... 1 If there is sufficient friction, the solid sphere will reach first. In the case of pure rolling without sliding (slipping), yes. But if their is no friction, will both of them reach at the same time? In the case of pure sliding without rolling, yes. 0 The formulas for friction (static and kinetic) are based on empirical models for friction, that is, models based on observation and experimentation. So yes, they are approximations. For a more detailed discussion, see: http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kin Hope this helps. 0 You are correct. The f=\mu N formula is indeed an approximation. Every material will have imperfections. However, in many cases, the above formula offers a 'pretty good' description of what goes on. 0 Friction is a non conservative force. A conservative force can be written as the negative gradient of some potential, namely$$\vec F_{\rm conservative} = -\nabla U$$Since friction is non-conservative, therefore it cannot be described as a gradient of some potential. Gravity on the other hand is a conservative force. So V can indeed be a gravitational ... 0 If a round object (say ring or a sphere) is kept on a rough inclined plane of inclination \theta and coefficient of friction \mu is such that it exactly balances the component of weight of that object, that is, mg \sin \theta=\mu mg \cos \theta, then it's acceleration would be 0, right? One thing that was hinted at by the other answers but not ... 4 such that it exactly balances the component of weight of that object I'm not sure this assumption is correct. The (static) friction force is only holding the contact point stationary, not the entire object. So there doesn't seem to be an obvious reason to think that this friction must be balancing out the entire weight. then it's acceleration would be 0, ... 2 You have to consider where the force applies on an object. For a sphere or hoop it's going to effectively act on the object's center of mass. Assuming uniform shape and density that's the center. If you consider forces this way then there's counteractive forces from the weight of the object on the incline plane that cancel each other, the force of friction ... 3 This is the mistake you are making: mg \sin \theta=\mu mg \cos \theta Friction will never be equal to \mu mg \cos \theta. Why? Friction will be static and will only try to prevent the slipping between the point of contact and the ground. The net acceleration of point of contact of ground \ne g\sin\theta due to the torque produce by friction in ... 1 See my post here: Where does a torque-invoking force belong in work energy theorem? (instead of tension, think friction). Essentially, there is work done while translating and work done due to the rotation. These works cancel. 1 The problem is that the student forgot about Newton's third law: if a first object exerts a force on a second, the second must exert a force of equal magnitude and opposite direction on the first. Thus all force coming from interactions between A and B, which define our system, must come in pairs. This is where the student failed. For example, P is a force ... 0 \mu is the frictional coefficient for static friction since the tire is rolling without slipping. In the non-sliding case, \mu N give a maximum value for the frictional force, not necessarily the actual amount. These links may be helpful: https://www.sciencedirect.com/topics/engineering/rolling-tyre https://www.jstor.org/stable/44562957?seq=1 ... 3 I believe it was initially shown that the drag force is proportional (under certain conditions) to the dynamic pressure q and the frontal (projected) area A. The proportionality constant of this relationship is what we call a drag coefficient C_D.$$F_D\propto qA\quad\Leftrightarrow\quad F_D=C_DqA$$Nice and neat. It turns out that this paramter C_D ... 2 Yes it would slide down, if the objects that applies the vertical force also has a zero coefficient. This could be achieved by a pusher with a (frictionless) roller on the end, or the object that applies the force has the same attributes. 1 If you apply your force sideways (trying to push the block into the wall), and everything is frictionless and there are no deformations, then yes, the block would indeed slide down, as there are no forces to pull it up. 0 The definition of the static friction is a force that opposes relative motion between two surfaces. Thus, if you pull the top block with some force F, the static friction force will oppose sliding (such that the top and bottom blocks do not slide with respect to each other). This implies they must have the same acceleration (otherwise they'd slip). This is ... 1 The mistake is here the work done on a body equals the change in total kinetic energy ... Using value of Wf from argument 1 W_f is not the work done on the cylinder. See in your calculation that you used W=\oint_{\mathbb{R_{a}}}^{R_{b}} \mathbf{F} \cdot d \mathbf{R} where \mathbf{R} is the position of the center of mass. This is calculating the ... 1 I believe that the frictional force is forward. Why? This is only true v_{com}<\omega R. Note that work done is defined as : dW=F\cdot ds where ds= Point of application of force. It is not the displacement of centre of mass. In the case of skidding , friction will always try to oppose the relative motion of the point of contact and the ground . ... 0 I have deleted my previous answer as it clearly didn't get to the heart of the issue for you. I'll try again: Your first equation is the work done by friction during the non-slip portion of the motion, during which rotational kinetic energy is converted to translational. You are forgetting the skidding portion of the problem, during which time the work done ... 1 Why would you need to touch something for there to be friction or heat? Imagine moving an air-blower over a pile of balls. You are indirectly pushing the balls around, thus imparting kinetic energy to them. In much the same way, as your finger moves over the pen the electrons in your hand interact with and push around electrons in the material, warming it up.... 0 The practical answer, corresponding to real rigid blocks, is that there is no friction force at the second block. The reasons are: the macroscopic contact between blocks doesn't mean a full microscopic contact as happens inside each of the blocks. The original question assumes no deformation, that is: rigid bodies. It translates in real life for blocks ... 1 Sliding friction does work on the sliding body because there is relative displacement between the sliding body and the surface upon which it slides; the point if contact does move (slide) before rotating away from the surface. (Pure rolling friction does no work because there is no relative displacement.) For a rigid body there is no dissipation of energy ... 1 In the case of no-sliding, there is a one to one mapping between the points of the surface of the roll and the ground. After any length l of the circunference of the roll, corresponds the same length l. In this case of braking sliding, there is a difference \Delta l = l_g - l_r, that is, the length on the ground is greater that on the roll. The work is ... 1 After making the diagram , I realized there is no way to know whether friction will do work or not, unless more information is provided. The net velocity\hat i  is given by : v_0 - (v\cos\theta+ \omega r)  Friction won't do any work if the above sum is 0. Since you mentioned , that the body is not in pure roll, It should indicate that the the net sum is ... 0 Imagine that the blocks are not actually touching. They are as close as you may want (or very far away if you prefer, it does not make any difference). What happens when you apply that force in the first block? Nothing, it will not move, since the static friction is more than enough to overcome the force being applied. As @David White, the friction is equal ... 0 We see this by considering the relative motion between close points (see the standard book from Batchelor from which I took most of the following). Let us consider the velocity \vec{u} at position \vec{x} as well as the neighboring velocity \vec{u}+\delta \vec{u} at \vec{x}+\vec{r} at time t. Then to first order we can write$$ \delta u_i = r_j \...

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Yes, assuming the the surface on which the plane rests is frictionless (or has a small enough coefficient of friction such that the static friction between plane and earth is 'small enough' it will move in the opposite direction). Consider the following sketch: I drew the FBD for the block and the incline (or the plane, if you will). I assumed no friction ...

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Yes, the plane will move too by momentum conservation, but there is no contradiction here. Energy gets dissipated, momentum is not. In fact, it's possible for a system to lose virtually all kinetic energy while still having a large momentum. For example, if $mV$ is large but $V$ is much smaller than 1/large, then $mV^2$ is very close to zero.

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I try to interpret this weird description. The greatest-sloep line means the vertical line. And the pulling rope forms an angle $x$ with the vetical line: $\tan x = 3 / 4$. The vertical plane means the vertical line (greatest-slope) and rope both are passing through the body, define the vertical plane. In a comprehensive way of dscription: The body weight $... 0 static friction does not dissipate energy because there is no displacement at the point in which the force is being made, assuming the cylinder climbs without slipping. The static friction force will be vertical, because the cylinder's edge tends to move downward at the vertical surface, so the friction will be opposite to this potential motion and point ... 1 If a flow is consistent with a rigid rotation, then you can always "move" into a (non-inertial) reference frame where you observe the flow to have zero speed everywhere. As a result, even though such a rotating flow can induce changes in the stress state of the fluid, those changes must be consistent with a hydrostatic state of the fluid, which ... 0 I am not sure what exactly you want to do but you can study motion in terms of translation of center of mass and rotation about center of mass in cases of combined translation and rotation. 1 When you apply a force less than maximum value of static friction between blocks will accelerate the lower block and the system will move without slipping (between surfaces of block). When you draw free body diagram of lower block, you will see that it only has 1 horizontal force which is due to static friction between the blocks. Hence there is no way that ... 1 There is a condition for object to be rolling without slipping. As show above by BioPhysicst, the static firctional force$f$equals to (adopts the inertial moment for a cylinder of mass$M$and radius$R$:$ I = \frac{1}{2} M R^2$, $$f = \frac{1}{3} M g \sin \theta \tag{1}$$ and this frictional force is arised from the normal froce of$M$and the ... 1 The ball bearings in the arms are just to make it heavier. Why ball bearings? I don't really know, but I guess it's because they already have a supply of ball bearings and a machine that inserts ball bearings. Maybe they also like the way it looks. Maybe they like the way you can grab the inside of one of the ball bearings with your fingers and swing the ... 3 The issue is that for a no slip condition, the static friction cannot be equal to the component of the weight parallel to the incline. Setting up Newton's second law (for linear and rotational motion), we have$^*$$$mg\sin\theta-f=ma$$ $$fR=I\alpha$$ Imposing the no slip condition$a=R\alpha$, we can determine that $$f=\frac{mg\sin\theta}{1+mR^2/I}$$ So, the ... 2 Bobsleds go faster the harder (and cooler) the ice beneath it is. That is, the more slippery the surface, the higher the speed the bobsled can attain. And ice is slippery because there is a thin layer of water on it (the surface of the bobsled must also be very smooth). So what would happen if we used butter instead of ice? Well if you freeze butter, it will ... 0 Initially, you have some gravitational potential energy, namely$\mathrm{PE_{grav}}=mgh$. This is your initial potential energy. Then at the end, you transform some of that energy into elastic potential energy, namely$\mathrm{PE_{spring}}=\dfrac12 kx^2\$. This is your final potential energy. Since there is friction, then it will do work on your system, ...

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First, using conservation of energy, calculate the kinetic energy of the block at the bottom of the incline before it encounters the spring. Then, calculate the work (using force times distance) done by the forces on the block, friction and the spring, to slow the block to zero kinetic energy, and that will give you the distance traveled. I believe the ...

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