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7

No, you don't need to account for the mass of humans. We're a part of the total mass. Children who grow and gain weight are merely transferring mass from other parts of the Earth to them. The mass of humanity versus the mass of everything else on the Earth is a zero sum game. To see whether the Earth is changing mass, you need to look outside the box (or in ...


7

What acts here is called impulse (of a Force) Suppose balls A, B are made of stainless steel and (m = 0.1 Kg r = 0.03 m) they collide. B is at rest (v = 0): Ball A will exert on b the Impulse of a Force $J$ and its velocity, momentum and KE will increase: $$J = [F . t] = \Delta p$$ If you know exactly of what steel the balls are made you can calculate the ...


6

Your position is well-taken. But note that the discussion does not mention any rotation of the turntable... Since the turntable is frictionless, there is no way that the rotary motion, if any, of the turntable can influence the block. The only force between the turntable and the block is the normal force balancing gravity. Any motion of the block, caused ...


6

Yes you need to include the mass of the 7 billion or so humans in the total mass of the Earth, though since their total mass is something like $5 \times 10^{11}$kg and the mass of the Earth is around $6 \times 10^{24}$kg humans make up only about 0.00000000001% of the total mass. We don't make any great contribution to the escape velocity. The point of your ...


5

There is approximately 7 billion people on earth. If we take some average of about 75 kg each, that means $$ m_{\rm population}=7\cdot10^9\times75\,{\rm kg}=5.25\cdot10^{11}\,{\rm kg} $$ The mass of the earth is roughly $$ m_{\rm earth}=5.97\cdot10^{24}\,{\rm kg} $$ which is about 13 orders of magnitude bigger than the total population of the world. So no, ...


5

Does it have something to do with the curvature of the Earth which is assumed to be spherical You'll probably groan when you read this answer since it isn't nearly as complicated as you might think. Essentially, there is factor of $\pi$ since the angular frequency $\omega = 2\pi f = \frac{2\pi}{T}$ A well know result from the linearized pendulum ...


4

The formula for the time period of a pendulum (for small angles of displacement from mean position) is $$T = 2\pi\sqrt{\frac{L}{g}}$$ Now, $L$ here is the length from the point of suspension to the center of mass of the bob. For illustration, assuming the bob is spherical, as the water leaks out, the center of mass will shift downwards, increasing $L$ and ...


4

The concept of inertia is indeed useful in two ways. I think your notion of it as a technical promotion of the everyday word "sloth" (without the baggage given it by the Roman Catholic translation of the "deadly sin" Ἀκηδία) as extremely close to the mark. In physics the notion of "inertia" has two, very alike uses: The first is practical, through a weak ...


4

First notice that simply by considering the dimension of the parameters involved, one can deduce that the time period of oscillations should go like $$T\propto\sqrt{\frac{\ell}{g}}. $$ This is because $g$ is acceleration hence has the dimensions of Length over Time squared and so the only way the quotient can have the dimension of time is to have the ...


4

The question of mass has arguably been one of the two most important issues in physics (the other being the electromagnetism). Physics has tried to uncover the true nature of mass for hundreds of years, to no avail so far. Not surprisingly, its description is somewhat circular: “In physics, mass is a property of a physical body which determines the body's ...


3

This can be answered in two different frames of reference. If we look at this problem from the perspective of an outside observer, then as the wheel moves forward, the bottom of the wheel doesn't move at all and the top of the wheel moves at twice the speed of the wheel itself. In this frame, as the wheel moves forward, the centripetal force provided by the ...


3

If you have another model, then use it to generate a set of predictions, and let's compare those to the predictions we get in the world we see. If you think there's some ether filling the universe, then write down some properties it has, and make some predictions with those properties. Talking in vague generalities will only lead you down a rabbit hole of ...


2

In this type of collision where you have what amounts to a very quick change in velocity, the force is called an impulse force and it is best to think of the equation a little differently. For example, instead of: $$ \sum F = \frac{\Delta mv}{\Delta t} $$ Think of $\int F \mathrm{d}t$ being equal to the change in momentum, that is: $$ \Delta mv = \int ...


2

Since the box is attached to the center of the table by a string, if you set it in motion with a tangential velocity, it will keep moving in a circle around the center, with the string supplying the required centripetal force. The movement of the box does not need to match how the table itself rotates -- as User58220 explains the turning of the table cannot ...


2

Your maths is almost correct. I think you should treat the lever as two separate rods which each have a pivot at one end. If the mass per unit length is $\sigma = M/L$, then the following applies to the two rods of length $r_1$ and $r_2$ (with $r_2=L-r_1$), where I split the rod up into infinitesimal chunks of length $dr$, mass $dm=\sigma dr$ and moment of ...


2

Let's make a concrete example with numbers: Suppose that $v_a = 6m/s$ and $v_b = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6, v_{cm} = p/M = 2$ . According to the conservation of energy and momentum: Kinetic energy and momentum are conserved only in a perfect elastic collision, if the bodies stick together the collision is inelastic an ...


2

There are two issues at play here. If you lift something upward at constant speed, then the acceleration $\vec a$ is zero. This means that the net force $\vec F_\text{net}$ is zero by Newton's second law ($\vec{F}_\text{net}=m\vec{a}$). As long as something moves with constant velocity, all of the forces add up to zero (i.e., they cancel out). Yes, that is ...


2

The equation for period of a pendulum is; $$T=2π\sqrt{\frac{L}{g}}$$ This equation holds for constant lengths, constant gravitation and small angles(such that the string is not horizontal and there is tension in the sting). Assuming these are true, you are not doing anything wrong concerning the experiment. Note that there is no mention of mass in the ...


2

The first one is correct. The problem is that in general the force acting on the particle will no longer be conservative in the moving reference frame and we can no longer associate the potential energy $U$ to it. To see why this is so, realize that a force $\vec{F}$ is conservative if and only if: $$\oint_C\vec{F}\cdot d\vec{r}=0$$ to any closed curve $C$, ...


2

The bottom of the wheel is a different part of it at every moment. If you follow a particular point on the wheel, you'll see it moves down and slows in forward motion until it touches the surface at zero speed and immediately starts to move up and accelerate forward again. Up to twice as fast at the top to catch up and get on the forward side again and then ...


2

It comes from the work made by the gravitational force of the big object that attracts the Earth. I updated my answer to a previous question from you. But remember that the description in terms of conservation of energy and the use of potential energies is an alternative description (usually simpler) than that using all the forces involved (specially ...


2

I'll try to explain it like I would to my kid, as soon as he gets there. If you make the same pendulum swing in a horizontal circle, and look at it from the side, you see the same harmonic motion. The only difference is, it stays at the same height all the time, but since we are dealing only with small angles, that's not much of a difference. Now comes, ...


1

The earth also 'gains' the positive work the same way the rough floor does, only it's not apparent at first. To understand how, let's go to the microscopic level to see what actually happens when you rub an object against a rough surface. You have the molecules and atoms of the object above, and those of the surface below as in the image shown. The ...


1

The comments seem to provide the answer. $$ L = p \times r $$ $$ L = m ( v \times r ) $$ Thus if r (the distance to the focus point of the orbit) changes, the velocity can change without the angular momentum changing. This is not possible for circular orbits where v is always perpendicular to r and the magnitude of r is constant. However, in elliptical ...


1

Suppose a body(very big but not bigger than Earth) moves against gravitational force of Earth. Let us take a real situation: a large rocket is fired straight up. It takes kinetic energy from the chemical explosions; from momentum conservation part of the available energy from the explosion moves the earth in the opposite direction. The gravitational ...


1

As you can see, $\rho$ only depends on a single variable: $r$. Thus, it should be intuitive that one can do this problem by integrating only over the variable $r$. To see what you are supposed to do, consider what happens if you fix $r$: You obtain a spherical shell (as was pointed out in the comments). The moment of inertia of a spherical shell is quite ...


1

The problem is that "the bottom of the wheel" is not a specific physical part of the wheel. It is a role or description that applies to each part of the wheel as it moves around the axle. You could just as well wonder how the top of the wheel can travel twice as fast as the wheel, and still stay connected. The answer is that the double speed "top of the ...


1

This is a side-effect of treating angles as dimension-less. For translational systems, we have \begin{align*} [\text{linear momentum}] &= [\text{action}][\text{length}]^{-1} \\ [\text{force}] &= [\text{linear momentum}][\text{time}]^{-1} \\&= [\text{energy}][\text{length}]^{-1} \end{align*} Correspondingly, for rotational systems, we have ...


1

This might be one of those problems that looks really simple on the surface but it's surprisingly complicated to prove. There might be a simpler proof, but I couldn't come up with one. Let's stick to the "ideally rigid" scenario. At an instinctive level, I think everybody will agree that the friction "ought to be zero". It took me several rounds to find a ...


1

So my question is: is this formula, that I frequently encounter in syllabi and books, correct? Where lies my mistake. The procedure $$\mathbf F = \frac{d\mathbf p}{dt} = \frac{d}{dt}(m\mathbf v) = \frac{dm}{dt}\mathbf v + m\mathbf a,$$ is based on the erroneous idea that the equation $$ \mathbf F = \frac{d\mathbf p}{dt} $$ is valid for systems with ...



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