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1

Say you have a weight tied to each side a a rope which is strung over a pulley with friction. Here's a really easy way to see why the tensions on each side of the rope can't be equal. Imagine a really stiff pulley - in other words, ${\bf F}_{friction}$ is high. If that's the case, it'll be possible to balance unequal loads on this pulley system - i.e. a ...


4

You seem to be saying that friction couldn't speed it up, because nothing else is moving that fast. Well, how fast is it moving? We can imagine the gyroscope axis parallel to the z axis, and the casing to be aligned such that the x axis goes through it. If the casing is tipped slightly, the gyroscope resists that turning and one side of the shaft has firm ...


3

Actually, in your rotating torus your are mimicking gravity, which is point outward. I.e. the outside of the torus acts as a floor. The centrifugal acceleration would be $g\approx\omega^2 R$. This $g$ plays the same role as the gravitational acceleration on liquid or gas pressures under normal gravity conditions, so you can say $\Delta p = \rho g h$, or in ...


0

You can analyze this by imagining a tiny tiny gap between the two masses. Physically we have exactly that, as the electrons at the surface of the first block are certainly not in contact with the electrons at the surface of the second block. Then we have a series of two collisions: the first is the initial impulse, the second is the collision between the two ...


0

If you assume that the fluid is incompressible, I'm relatively sure that it could be shown that 100% efficiency is theoretically possible. You could use a variety of mental models to do this. I would prefer to think about conventional hydraulics lifting some weight. If you match the weight to the pressure, then you could raise an external object some (m g h) ...


0

Turbines (impellers) have fins that work more like aircraft wings and sails than the buckets of a old water wheel. They have have very high lift/drag ratios which translate the high efficiency. It is friction and heat loss that kill efficiency not the work of the turbine (how much it spins).


0

I'm going to just use $32^∘$ below; it doesn't make a difference. Your equation isn't correct. You should have $F_x−f_k = 0$ or $Fcos32∘−f_k = 0$. The x-component of F is $F_x = F.cos32^∘$. Writing $F_x.cos32^∘$ doesn't make logical sense. Why? Shouldn't I be able to use $∑F=0$ in this problem to find the answer? Yes, you can.


3

There are three possible forms of motion for a small object in a $\propto \frac{1}{|r|}$ potential: hyperbolic motion parabolic motion elliptic motion The first and second possibilities are unbound, so the object comes from infinity and will go back to infinity afterwards. The third one is bound, which means that the object cannot escape the ...


2

Objects in orbit come pretty close. If you don't mind venting the cabin or taking a walk outside, even air drag can be very nearly eliminated. All you have left are very small forces due to being in a non-inertial reference frame, and drag from the very, very thin atmosphere. Neither would be noticed without some very precise equipment. To reduce these ...


1

Have you heard of superfluidity? It happens when you cool liquid helium below about 2 Kelvin. The helium then will flow freely and without any friction. If you induce a current vortex in liquid helium, it will remain flowing until the end of time (however, you cannot draw energy from it, because the liquid is frictionless) or until it warms up again. So, in ...


0

If your friend's energy+ yours= F1, then you would see your own energy expenditure halved, which we know cannot be the case. If your friend helps you push the object, then you are no longer applying the same force, or (lazy answer) the force is no longer localised and motivates the part of the object most subject to friction. So, once the object is first ...


0

The glass bob will reach the ground earlier as acceleration due to gravity is independent of a falling body's mass. Being an insulator, no induced current is developed in it due to Earth's magnetic field.


2

There is, effectively, only gravitation and friction acting on the pack of gum. However, the friction is not that strong (it is mostly independent of the velocity of the book, and dynamic friction is weaker than static friction) and it doesn't have that much time to act. Hence it doesn't affect the momentum of the gum noticeably. This is very related to the ...


3

I'm sure everyone has had that concern when we encountered the definition for the first time, in school. There is a valid reason why this definition is still persisted with, despite the deficiency that you hit on. The most popular (and simple) forces in physics (also the ones with which we begin learning physics) are conservative forces, implying that the ...


31

If you're pushing a 10-ton truck and it's not moving, you are not doing any work on the truck because the distance $ds=0$ and the nonzero force $F$ isn't enough for the product $F\cdot ds$ to be nonzero. Your muscles may get tired so you feel that you're "doing something" and "spending energy" but it's not the work done on truck. You're just burning the ...


2

The proof is most easiest if we use the vector notation. We have $$\vec \tau = \int {d\vec \tau } = \int {(\vec r \times dm\vec g)} = \left( {\int {\vec r dm} } \right) \times \vec g$$ where I have used the assumption that near the earth $\vec g$ is constant. Now according to the definition of center of mass we have, $${{\vec r}_{cm}} = \frac{1}{M}\int ...


-1

Its a violation of Newton third law..because Newton third law only valid when 1)same kind of force ie when there is force of same nature So electromagnetic and block forces are not of same nature so Newton third will not hold


6

Backspin! Those shots in which the cue ball "draws" backwards after hitting the target ball involve backspin. Without backspin, the cue ball cannot reverse direction. Consider what happens when the cue ball is not spinning at all when it hits the target ball. The cue ball will come to a dead stop if it hits the target ball straight on. Think of Newton's ...


1

The direction the ball will take depends on the angular momentum. The velocity with which the ball moves or bounces backwards but the chief determinant is the spinning effect of the incoming ball.


5

First of all, if the collision is elastic, the distribution of momentum in between the components is completely determined by momentum and energy conservation! This statement is most obvious in the center-of-mass frame where the total momentum is zero and the two objects are moving in opposite directions. The momentum conservation (the total momentum is ...


2

This is low-Reynolds number particle sedimentation. It turns out that the problem is strikingly difficult, despite the simplicity of the setup and even of the equations (Stokes plus dynamics of pointwise solid particles). Check the webpage of E Guazzelli who's been working a lot on this. However, I believe you can get a fair rendering with simply a ...


1

It would crash into the earth because the Earth's gravitational field is not uniform and, even if said ring were to be perfectly positioned, ignoring the effects of wind, strikes from cosmic debris (not a lot that low in the atmosphere), change in mass of the ring (e.g. corrosion), change in shape of the ring (due to e.g. gravitational forces, heat ...


1

The angular momentum of the ring would stay constant. So if the ring were built on the earth and then lifted up to 100 feet and fastened together, it would would initially rotate at the same rate at the Earth's surface. The prevailing winds would probably disturb its orbit and I suspect it would eventually crash into the Earth. I think it would take extra ...


0

I'm guessing that you understand clearly the effect of precession here. The reason why the wheel starts to fall down is that when we explain the change in angular momentum of the wheel, we say that the angular momentum vector only changes in direction- right? But the angular momentum vector of the wheel doesn't only point outwards; it also points upwards (it ...


0

From a mathematical standpoint, any collision in which no mass is lost is described by two equations: Conservation of energy: $ m_1v_1^2 + m_2v_2^2 = m_1v_1'^2 + m_2v_2'^2 + E $ Conservation of momentum: $ m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' $ You know the masses and initial velocities, so this reduces to a system of two equations in three unknowns ($ ...


1

To parameterize the degree of inelasticity you use the "coefficient of restitution" which is 1 for elastic processes and 0 for inelastic processes. This is described by $$ \text{coef. of restitution} = c_R = \frac{\text{final relative speed}}{\text{initial relative speed}} = \frac{v_2 - v_1}{u_1 - u_2} \,. \tag{*} $$ This also tells you how to compute the ...


1

Perfectly elastic and perfectly inelastic collisions are just limiting cases on a scale of how much kinetic energy is retained. As noted in @Nathan's answer, if you work in the center-of-mass frame, a perfectly inelastic collision results in 0% of the kinetic energy retained, while perfectly elastic collisions have 100% of kinetic energy retained. So, you ...


3

There is not such thing as a "partially elastic" collision. Classical collisions between particles can be separated into two categories: elastic and inelastic. Elastic collisions are defined as collisions in which no energy leaves the system (i.e. $E_i = E_f$). All other collisions are inelastic, as some energy is lost ($E_i > E_f$). A perfectly inelastic ...


0

Neither, however... If you are standing on such a board, there is a simple way to propel yourself, assuming you can change direction (i.e. steer) the board. You cannot propel yourself forward, but you can propel yourself sideways, by pushing the board to one side. This gives you some sideways velocity. Then (before you fall over) turn the board so it is ...


0

You can't actually propel yourself forwards or backwards in this way (unless you are taking advantage of significant friction in the bearings). Moving your body forward or backward would cause the platform to move in the opposite direction but only so much as to leave the person+platform system's center of mass unchanged. Another consideration. Where would ...


0

Your derivation of the additional force on the scale for the falling rope is wrong, both cases yield the same results. If I understand correctly you are comparing the effect of a rope falling on a scale, to the similar fall of a liquid, something like this schematically: The problem of your reasoning is related to a misconception of the stagnation ...


3

Many students confuse the term work in physics with the conventional term of work. Your body wastes energy when you push something, and when that something doesn't move... 100% is wasted in the biological efficiency. 1st step: forget the concept of how hard it would be for you to do it. How much work is a table doing by holding up a 1kg weight? zero. It ...


1

It'll be exactly the acceleration due to gravity, because no other force is acting on it. Edit answering comment: You're thinking about velocity. Acceleration isn't something an object has. Objects accelerate because there's an external force acting on them. When B is towing A, it exerts a force on A which causes A to accelerate. When B lets go, that ...


3

In a sense your question is entirely apropos. The critical distinction to be made, however, is that the two forces can only be summed to zero if we are talking about the two-body system. Each body, considered in isolation, experiences an unbalanced force and thus experiences accelerated motion. The system, however, only experiences mutually-canceling ...


0

To get the initial conditions you guess and look at whether the guesses fit the physical situation. For example, suppose that you have a mass on a spring and you are holding mass so that the spring is slightly stretched and then release it. The mass is not moving at t=0 so the initial condition is that the velocity is zero at t=0. But you could imagine other ...


0

Does matter really exist and cease to exist in a subatomic field or in some other place according to the quantum theory? or is it talking about how some particles have a superposition? Your question uses a lot of jargon and you seem to be rather unclear about what that jargon denotes, so I'm going to explain what the jargon denotes. To measure ...


0

This may be cheating, but I think the problem is easier if you use conservation of energy. If you set the gravitational potential energy reference to the height of m1, then initially you have $$ U_i = k\frac{q^2}{d}. $$ The final energy will have a gravitational potential energy and an electrical potential energy: $$ U_f = mgh + k\frac{q^2}{r} $$ A bit of ...


0

I worked the problem out a ways and it involves quite a bit of tedious algebra. The technique I used was to include the electric force in the Fx and Fy equations by looking at the angle that $ \hat{r}$ (the vector between the two charges) makes between the charges. For example, the equation i came up for Fx is $T_x - \frac{Kq^2}{r^2} \cos{\theta} =0$ where ...


1

Assuming the train doesn't accelerate during the ball's fall, it will land in the spot you aimed at. Think about it this way. Before you drop the ball, it is moving along with the train (i.e. it has some horizontal speed). When you drop it, the ball still has this speed, and since an object in motion tends to stay in motion unless you exert a force on it ...


0

This is the classic horse and buggy brain teaser. If the horse pulls on the buggy, the buggy must pull back on the horse with an equal force so who moves anywhere? The answer lies in the fact that the buggy pulls back on the horse because it is being ACCELERATED. That is where the force comes from. In the case of the two boxes, the second box (one at the ...


0

(Not an answer as such, more an extended comment and suggestion). You wrote "On the other hand, the 1-D elastic collision problem can't be solved with Conservation of Momentum alone - Conservation of Energy is also required. " Actually the 1D elastic collision problem can be solved with (i) Conservation of Momentum (COM) (ii) application of symmetry (iii) ...


1

Let me make you precise about conventions, because your notations are unconventional. Suffix in a vector quantity is always represent the component along specific direction, component of velocity $v$ along x-axis is denoted by $ v_x.$ However, your question is precise irrespective of your conventions used. Insight into your problem You have made precise ...


1

Assuming that the train is an inertial frame of reference (non-accelerating) and if we neglect both air friction and the effects of gravity, then the ball will move in a straight line away from you at the same speed which you threw the ball at until some other force acts on the ball. This situation would look the same as one where you threw the ball while ...


13

Let me first go through this without friction or air drag. You say $v_y$ along the $x$-axis and the train moves with $v_x$ along the $z$-axis. This is a little inconsistent. I will use the velocities, but not your description of the axes. So the train moves in the $x$-direction, the ball is thrown into the $y$-direction and it the $z$-direction is up-down. ...


-1

The bullets time in the air, having been fired horizontally, depends on it's velocity. In general that varies from about 900 fps to near 3000 fps for a rifle, so obviously the time it takes to hit the ground varies as well.


1

You are very close. Just to review what is going on, the period is given by \begin{equation} T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{I}{k_{eff}}} \end{equation} where $I$ is the moment of inertia of the system and the torque is proportional to the angle by which the pendulum has been displaced with a coefficient that I'm calling $k_{eff}$ in analogy to ...


1

A better electrical analogy to Newton's second law might be inductance: $$ V = L \frac{\mathrm d i}{\mathrm d t} $$ The only reason this looks different is that physics has a name and conventional symbol for the derivative of speed, but electronics does not have a name for the derivative of current. So let's just pretend that the word acceleration does not ...


0

I'm not sure if I'm just misinterpreting your question, but we can most certainly influence the acceleration of an object directly. To do this we must take into consideration what is actually happening as we accelerate; Acceleration, in physics, is the rate at which the velocity of an object changes over time. While it is a sum of the net forces over the ...


0

A South Easterly wind comes from the South East your diagram seems to show a North Easterly wind. As the ship is steaming into the wind the apparent velocity should be greater than the true velocity.


1

We should think a bit more carefully what force, mass and acceleration really are: For simplicity, we consider a classical point particle. The force $\vec F$ is something externally applied to the particle, it is a property of its environment. Most often, it is the gradient of a potential, $\vec F = - \nabla V$, but it need not be. In general, it is some ...



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