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Draw a FBD for a small angular section of the rope (subtending an angle $\delta\theta$), showing the normal reaction from the pulley and the friction force when the system is about to slip and then let $\delta\theta\to0$. You should get: $$\frac{\mathrm{d}\,T}{\mathrm{d}\,\theta} = -\mu\,T$$ subject to $T(0) = 10^6{\rm N}$ where $\theta$ is the angular ...


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Here is a method using energy. Assume the mass starts a distance $A$ from its equilibrium point and it moves past the equilibrium point a distance $B$ before turning around. The initial spring energy is equal to the final spring energy plus the energy lost due to the work by friction: \begin{align} \frac12 kA^2 &= \frac12 kB^2 + FA + FB \\ 0 &= ...


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You've basically got it; you can often analyze constraint forces like this by asking "what would happen if my constraint force disappeared?" For example, if you have a door which is constrained to be on two hinges, the bottom hinge must actually push the door outward, away from the door frame. This is easily seen if you imagine what would happen if that ...


1

"It will soon fall off" - indeed. Hence the part of the question that says "at what value of $\theta$ does the particle leave the surface" By first seeing what force normal to the surface is needed to keep the particle "on the surface" and following the curvature, and then finding when this force is less than gravity, you break the problem into two ...


3

Torque is a mathematical object called a bivector, produced by taking the wedge product of $\mathbf{r}$ and $\mathbf{F}$. Bivectors can be thought of as area elements of planes; the magnitude is equal to $rF \sin \theta$ and the plane itself contains $\mathbf{r}$ and $\mathbf{F}$. By a great coincidence, in three dimensions, there are three distinct planes ...


2

Torque is a vector whose direction is always out of plane. The same with angular velocity $$ \vec{\tau} = \vec{r} \times \vec{F}$$ $$ (0,0,\tau) = (x,y,0) \times (F_x,F_y,0) = (0,0,x F_y - y F_x)$$ $$ \vec{v} = \vec{\omega} \times \vec{r}$$ $$ (v_x,v_y,0) = (0,0,\omega) \times (x,y,0) = (-y\, \omega,x \,\omega,0)$$ I always think of planar quantities as a ...


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Deceleration for the situation you describe is acceleration in the opposite direction of the initial velocity vector. Let's say it takes 2 seconds for the boat to stop, and let's ignore the upward movement of the boat when it hits the tires and attempts to ride over them. Say the boat's initial velocity when it hit the tires was 4 meters/sec. Suppose it ...


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Sorry, but the currently accepted answer is not precise enough. In the situation you discussed, there is no applied force acting on the object, but there is still the normal force $N$, which prevents the object from "falling through" the surface due to gravity. The most common model of kinetic friction is $F_k = \mu N$, where $\mu$ is the coefficient of ...


1

Friction comes into play whenever there is relative motion between the surfaces in contact or a tendency of motion between the same.There need not necessarily be an externally applied force on either of the bodies,that is,there need not necessarily be a relative acceleration initially,merely relative motion or a tendency for the same. It is the frictional ...


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Solving this particular set of equations would yield two values for $u1$ and as such two values for $u2$ - 10 and 5 ,that is, if the value of $u1$ were 5 the value of $u2$ would be 10 and vice-versa.You may arrive at this by solving a quadratic equation in $u1$ (or for that matter,$u2$) Now,you may have, while solving quadratics seen we,sometimes acquire ...


1

No. Even if you include some additional things for angular momentum there are still many things not in Newton's laws of motion. For instance, the fact that mass doesn't change from one value on Tuesday to a different value on Wednesday. Newton talks about mass in the Principia but it isn't about motion or forces so it isn't part of his Laws of Motion. ...


1

Since drag is the resistance force of an object moving through a fluid (a fluid friction term), then you can make the physical argument for the value of $b$. This frictional force exists only at the boundary between the object and the fluid itself (a surface force), this means that the drag force must be independent of the mass of the object, thus $b=0$. ...


1

The way I like to phrase pretty much all of dimensional analysis is, "you can only take an arbitrary mathematical function of dimensionless parameters: mathematics doesn't directly deal in any other sorts of functions." When you see $[[R]] = \text m, ~~[[M]] = \text {kg},~~[[v]] = \text{m/s},~~[[\rho]] = \text{kg}/{\text m^3}$ your first question needs to ...


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You'll find a one parameter family of solutions, because you have 4 independent quantities while in this problem you have the 3 independent units for mass, length and time. In your solution, you can see that the freedom to choose the parameter b comes from the fact that the density of air divided by the density if the object is dimensionless. To fix b ...


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Well, it's not energy, its power $P$. $~~~~~~~~~~~~P = \int F \cdot dv$ And since power is the derivative of energy $P = \dot E$, your world makes sense again ;). Regarding your problem I agree with Yanping Cai, the kinetic energy of the car $E_{kin}$ must be converted into potential energy of the spring $E_{spring}$. $~~~~~~~~~~~~\frac{1}{2} m_{max} ...


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This answer may not look like a typical answer, but I am attempting to instill a key concept, so please bear with me. For this type of problem, where you are investigating a possible solution, UNITS are EXTREMELY important. What are the units of momentum? What are the units of force? Note that if units do not match across an equal sign, the answer is ...


1

I'm sure you can consider the force and integrate it over displacement to calculate the total work has been done. But why don't you step back and consider the conservation of energy? In short, $$\dfrac{1}{2}mv^{2} =\dfrac{1}{2}kx^{2}$$ $k$ is the minimum spring constant it requires.


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Problems that depict situations where the tensions are same on ropes on both sides of the pulley are ideal situations.It is stated so in order to minimize any complexities that may arise if the pulley was to rotate.Now, if the tensions were not equal on both sides, the pulley would experience a net non-zero torque and hence a net angular acceleration and ...


1

Because the pulley possesses mass, you need to apply a non-zero net torque to it to increase its angular acceleration (assuming that is the goal here). If the tensions were the same on both sides of the contact point between the string and the pulley, there would be no angular acceleration.


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Suppose someone suggests that following a perfectly elastic collision, two billiard balls are each traveling twice as fast as they were before (and opposite to their original directions). You can't prove him wrong using conservation of momentum, but you can prove him wrong using conservation of energy. Therefore conservation of energy has implications that ...


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The answer is that there is no simple answer. The way that energy and momentum get split up in the aftermath of a collision depends on the details of the collision itself, and there is nothing in the conservation laws themselves that influences this. The simplest case is in one-dimensional collisions, where both objects are constrained to move along the ...


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Isaac Newton observed the actions and reactions of objects in motion and recorded his observations as three famous laws. These laws (plus the conservation of momentum and of energy) can be used to explain how momentum and velocity are distributed among the objects coming out of a collision: (1) An object in motion will remain in motion with the same speed ...


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Moment of inertia depends on how far away mass is from the axis. In a ring of radius $R$, all the mass is $R$ from the axis. For a single particle $R$ away from the axis... well, all the mass is $R$ from the axis.


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This text is excerpted from Richard Fitzpatrick's Newtonian Motion: "It should be noted that Newton's third law implies action at a distance. In other words, if the force that object $i$ exerts on object $j$ suddenly changes then Newton's third law demands that there must be an immediate change in the force that object $j$ exerts on object $i$. Moreover, ...


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Moments of inertia are additive. Suppose you have particle $A$ with a moment of inertia $I_A$ and particle $B$ with a moment of inertia $I_B$. Then the total moment of inertia of both particles is just $I_A + I_B$. You can imagine a ring as being made up from lots of point particles, all at a distance $R$ from the central axis. In that case the moment of ...


2

Firstly, you must qualify moment of inertia by the axis it is taken about. If you translate this point, the inertia also changes as described by the parallel axis theorem. So you question relates to the moment of inertia of a ring about the ring's axis of symmetry normal to the ring's plane, and to a point on this plane at the same distance from this ...


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If you imagine drawing the potential as a function of position and pick a point and zoom in on it, then if you zoom in enough (and the potential is differentiable) then it will look like a line. But the line might not be horizontal, when you assume the potential is spatially constant to first order you are assuming that straight line it looks like happens ...


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Yes, there is no reason for the supersonic aircrafts making a significantly larger amplitude of sound as compared to the ones that go at the speed of sound. (But yes, the aircrafts going with a speed power than that of the sound create the sound waves in such a way that they do not interfere and the enhanced amplitude is not generated whereas when the ...


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The sonic boom is not just the combination of the engine noise and ordinary sounds of the plane, smashed together because of the speed. It is the propagating effect of the air being smashed into by the plane, faster than the air can get out of the way. The amount it compresses the air depends on how fast the plane is going, and the loudness of the sound you ...


1

Your $E$ is potential energy in the rubber, which transforms to kinetic $K$. So your starting velocity $v$ will be: $E=1/2 mv^2$ From conservation of energy: $0-1/2 mv^2=0-mgh$ $h=v^2/2g$ and $v^2=2E/m$ Confirmed Interestingly enough the rubber does not obey Hook's law, and you need a lot more work if you want to find out what really ...


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A better question in this case is to ask if we drop a feather from a height, we see that the feather settles on the ground quietly. I you can under this which is a lot easier to explain and much mor intuitive. So as any object falls in a fluid (in this case air) it firstly accelerates and finally reaches constant velocity termed as terminal velocity, at ...


1

Just before the ball reaches the ground, all of its molecules are coming down with almost an equal speed that is the speed of the ball.(Although, due to the non-zero temperature of the ball, the molecules are also vibrating about their mean position wrt COM frame of the ball).And thus the ball possesses a systematic macroscopic kinetic energy. Now when ...


1

It has been pointed out that this cannot simply be done by examining the mass distribution (first and second moment of mass). But there is a way to "look inside" most common objects: Take a CT scan. Not sure if you consider that "typical" lab equipment - but it's equipment I have in my lab... Of course depending on the size of the object and the material ...


0

If you're dropping a projectile into sand, the potential energy that you began with ends up being converted into kinetic energy (from the sand thrown out from the collision), sound energy, and thermal energy. Ultimately, the thermal energy is the only surviving energy after any appreciable time though.


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The contact with the rail creates a kinematic center of rotation where the reaction forces meet. The rail car will tend to rotate about this center as a result of side loads. If the center is above the center of mass, the rail car acts like a hanging pendulum. A small deflection will cause a restoring torque opposing the swing. If the center is below the ...


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Previous answers are great and explain the dynamics very well. I'd like to point out that this can be explained just as easily in a static situation. Imagine the weight the shaft has to carry. You don't even have to imagine a curve to note that the weight will automagically center (and lower the center of gravity of) the train in the first image. In ...


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In both diagrams in the question, the left wheel has a smaller radius at the contact point than the right wheel. Because they're fixed to a common axle, in any given amount of time, the right wheel will travel a greater distance than the left, so the axle as a whole will rotate anti-clockwise (when viewed from above) about a vertical axis. As it does this, ...


0

If the mass was simply attached at the current position and let free, your energy method is correct. If the mass was attached by hand and slowly allowed to stretch the spring until it remains fixed at a stable equilibrium point, the force method would have been correct. However, for your given situation I believe the question is asking you for the first ...


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The two approaches do apply to different situations. Let's first look at the "energy method". Here you describe what happens to the different energy components as the mass starts to fall from $x=h$. The initial energy is indeed $mgh$. Now, after that the mass has fallen a distance $x$, the energy is composed of three components: the residual potential energy ...


0

Your force considerations are wrong. Note that there is only force on the block (i.e. $kx$); but there are actually two forces which are extending the spring. One is being applied by the block hanging(which you have considered). But you have not considered the force on the spring due to it's attachment to the ceiling. There will be an additional equal ...


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To see why the first configuration is used rather than the second, perform the following experiment: Hold a bowl in your hand and place a small ball inside. Move the bowl in circles at various speeds and observe the behavior of the ball. Now turn the bowl over and balance the ball on top. Again, move the bowl around and observe the ball. Which is more ...


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Shift the upper configuration to the left a short distance at equilibrium. Result: the left wheel goes a little up, the right goes a little down, the train tilts clockwise, the center of mass is to the right of the centerline between the wheels, and therefore the center of mass provides a restorative force to push the train back to the right. Shift the ...


-2

I can see a difference by thinking about the flanges... Given that 1) the flanges are on the inside of the wheels 2) the right handside of the diagrams is the outer part of the track where the flange will press against the rail... compare and where the red lines indicate the plane of the flange.... in the upper case the flange neatly pushes ...


-1

The answer is no. The net force at the maximum elongation points has the same magnitude. This is because the rest point of the spring is modified by the gravitational weight of the mass. The mass oscillates around this new rest point, and at the points of maximum amplitude the net force is the same. One can tell that the gravitational force can be ...


2

In accordance with Hooke's law the force is linear with distance. Incorporating gravity only means that the equillibirum position of the spring has changed, the "zero" around which it oscillates. The gravitational pull is already compensated by the spring. Thus the magnitude of the force is euqal at $-A$ and $+A$. Edit: When the gravitational pull on the ...


5

The equilibrium position in this case is not where the spring is not stretched, it is actually stretched by a $\Delta x$ amount with $F_{spring}(0) = k\Delta x$. So the spring force on point A is a little smaller than in point -A, since $ F_{spring}(A) = -k(A-\Delta x)$ and $ F_{spring}(-A) = k(A+\Delta x)$ so it compensates the "extra" force. You have ...


0

If the person is moving in a straight line and then takes a right turn then the answer simply is because of inertia. The person initially moving in a st line would continue to move in that direcn and hence when he takes a right turn it would appear that he turns left but actually he is simply trying to maintain his earlier motion.. If he is moving,in a ...


0

As in the comments, on a flat surface, one needs to raise a square's center of mass for it to roll: its center's distance from the ground varies cyclically between $r$ and $\sqrt{2}\,r$ as it rolls where $r$ is the square's halfwidth. So there is a potential barrier to the motion: one needs to input energy to raise the wheel's potential energy - and that of ...


0

A slab can slide, but it will not move unless the static friction force (which is proportional to the weight $Mg$ of the slab by the static friction coefficient $\mu_s$) is exceeded by the force $F$ pulling on the slab parallel to the surface on which it is supposed to slide. So, there will be no movement unless $$ F>\mu_s Mg $$ Alternatively, if you want ...


2

The basic physics in laymen's terms Okay, so the basic idea is: an object in motion tends to stay moving at the speed that it's moving. When we apply this to rotational dynamics we have an interesting effect: an objects speed goes linearly with the radius it is from the center it rotates around. So if something is rotating with a period T, it must go a ...



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