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9

Yes, it is possible to rise theoretically w.r.t a ground frame. But keep in mind that the rope-man system's center of mass must keep moving downwards because of the only external force acting on the system(gravity). The lighter M gets, the harder it will be for the man to rise, and it will become impossible in the limit the rope becomes massless. This is ...


7

Kepler's third law is irrelevant here. It applies to many (small) planets orbiting a (large) central star, not to a binary star system. If the stars have masses $m_1$ and $m_2$ and the radii from the COM are $r_1$ and $r_2$, then from the definition of the COM, $m_1r_1 = m_2r_2$. The (gravitational) central force $F$ acting on each star is the same, but ...


6

Here's some tennis racket physics from Rod Cross, including links to several Am. J. Phys articles (the physics educators' journal, thus excellent for learning from) and this excellent diagram: There are at least three "sweet spots": The node, at the center of the strings, is a point where the natural standing waves in a vibrating racket don't have any ...


5

Recall the difference between the weak and strong Newton's third law, cf. e.g. this Phys.SE post. If the internal forces satisfy the weak Newton's third law (but not the strong Newton's third law, i.e. without the collinarity assumption), then it is not guaranteed that the internal forces do no work, cf. e.g. Fig. 1. ^ F | | 2 x-----------...


5

Will a tennis ball go further if i hit it with the side of the racket? No. You want the racket to deform, not the ball. This means using the strings to elastically store energy and return it to the ball. The Ball The ball's deformation upon impact is undesirable because "a tennis ball is required by the rules of tennis to dissipate a fraction of ...


3

All you need to do is calculate the perigee distance $r_p$ that is the distance of closest approach. Then if $r_p < R_A$ your satellite will crash and burn. Once again we start from the vis-viva equation: $$ v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right) \tag{1} $$ The parameter $a$ is the semi-major axis of the ellipse, and it is related to the ...


3

Centripetal force is the name of the force that points towards the centre. This is in the radial direction. Tangential velocity is, as the name suggests, a velocity direction tangent to the circle. The radial and tangential directions are by definition always perpendicular - in the same way that the x and y axes are. You are of course right that if any ...


3

The rope pulls just enough that the pendulum doesn't fall to the ground, but follows an arc. The following picture shows you how work out the force for the static case (no motion of the pendulum): However, you need to take account of the fact that the pendulum is moving in an arc. When something moves in an arc, you need an additional force $F=\frac{mv^...


2

The motion (perpendicular to the wind) is called Ekman Transport. It results from a combination of the wind and the Coriolis Force (due to the rotation of the Earth), which together produce an underwater motion called the Ekman Spiral. Icebergs, which extend large distances below the surface, experience forces from the whole depth of the spiral. The ...


2

You are combinig two question, I am combining two answers. The system you describe consists of two points with masses. We know that: Every two points $A$ and $B$ lay on a single line; When line is rotated around axis intersecting it, all points of the line have same angular velocity $\omega$, except for the intersection with $\omega_0=0$; Centre of mass $...


2

I think the OP has made a mistake in applying second law of Newton. The law (about a particle) says: $$\Sigma \vec F=m\vec a$$ As it is seen, this is a vector equation. This means that corresponding components of both side of the equation must be equal. Although it is not said in the law's body, but it is obvious that we must write the equation above with ...


2

Which way does the y axis point? If the y axis is chosen to point up, (having the positive direction upwards) then you are right, normal force should be positive (it points upwards as well) and weight negative. Is it chosen to point down, then normal force is negative (points in the negative direction along the axis) and weight positive. Remember that ...


2

The centripetal acceleration always points toward the center of the circle. In this case, the center of the circle is below the car, so the centripetal acceleration points downward. Now, you'll notice that, in the given solution, the centripetal acceleration term is positive. That means the writer has chosen a coordinate system where positive is downward, ...


2

When you move the pen at constant velocity, you are correct that you apply $F=\mu_kmg$. This is true regardless of what constant velocity you choose. The difference is you have to supply more force initially to accelerate the pen to a faster speed, but once it is up to speed you only need $\mu_kmg$ to keep it at that speed. So it really isn't the force ...


1

If we are dealing with the gravitational field of the earth, then because of the inverse square law, you should, in theory, need a tiny bit less force to raise an object from 50 m to 100 m, as you do lifting it from ground level to 50 m. But this difference in force is minute, I don't know offhand if we have instruments to measure it. But you can work it ...


1

I thought about this in the molecule level, but then if molecules are at constant velocity then there is no "pushing" force that they apply on each other. To clarify my misunderstanding - I imagined the system at constant velocity at molecule level as if it was accelerating so molecules of A push molecules of B and also the opposite. If we are speaking about ...


1

If the blocks initial velocities are zero (i.e. the blocks start to move from rest), then it is impossible for block B to move with constant velocity. Because the only horizontal force acting on it (if there was) is friction force due to block A. We have: $$\Sigma F_B=m_Ba_B$$ If block B moves with constant velocity, then $a_B=0$ Thus, friction force acting ...


1

A properly drawn free-body diagram will have a tension force vector acting along the line of the rope, toward the pivot point and a gravity vector acting straight down. If you establish a coordinate system which is instantaneously parallel and perpendicular to the rope, you then will decompose the gravity vector into two components ( $mg$ times trig ...


1

If you think of it in terms of conservation of momentum and collisions, the simplest version works just the same as tossing a handball at a on-coming freight train. The interaction is elastic, and the ball returns with the same speed it had going in in the center of momentum frame, but the center of momentum frame is moving in the ground frame, so the ball ...


1

For inelastic scattering, the initial momentum is $m_b v_{b_i}$. After collision, both $m_b$ and $m_c$ move together, with a velocity $v_{b_f}=v_{c_f}=v_{cm}$. By conservation of momentum $m_b v_{b_i}=m_b v_{b_f}+m_c v_{c_f}=(m_b +m_c)v_{cm}$, whichyield the equation that you are looking for


1

The systematic way to set up the equations is to draw a free body diagram for each mass. The FBD shows all the forces acting on that particular mass. You can then use Newton's second law to get the acceleration from the resultant force. The weight of mass 2 acts on mass 2, not on mass 1. Of course the weight of mass 2 will affect the motion of the system, ...


1

You must look at all forms of energy. Just before the explosion, the projectile has gravitational potential energy GPE, kinetic energy KE, and also chemical potential energy CPE stored in the dynamite. Just after the explosion the 3 fragments all have the same GPE as before. The CPE has disappeared in the explosion. As Jim says, we must assume that it ...


1

You definitely don't need to use General Relativity to answer this question. It depends upon what you mean by "feel". If "feel" means "detectable by sophisticated instruments" then, yes, it can be "felt". But your body is not a very sophisticated detection instrument. According to what I've read elsewhere, the Earth speeds up by $1000$ $m/s$ as it moves ...


1

There is extensive (100s of articles) lit. on the web regarding atmospheric drag and drag in general. A spherical bob and a very thin rod makes the problem very simple. One may use this graph: http://eis.bris.ac.uk/~memag/Teaching/Multi/dragcurve.pdf , and then calculated the Reynolds number to find the drag force using the formula given by the first ...


1

There are two reasons. Inside the Sun, the force is no longer an inverse-square law. It actually grows linearly with $r$. The second reason is that Goldstein (as well as any other classical mechanics book) is interested in orbits with a non vanishing angular momentum with respect to the center of the Sun. An oscillation along a line passing through this ...


1

You have two questions. Here is the answer to your first question, "So why stone does not come back/reach to the centre of circle?" The answer is that the stone is accelerating toward the exact center of the circle. Velocity is composed of speed AND direction. As you noted, the velocity is always changing, however the speed is not. This is related to your ...


1

The above isn't drawn correctly. And it is far too complicated in my opinion. I think I was able to solve this using high school physics: This bugged me a long time. I am surprised no one ever posted an answer on the internet. The length of his rope was 60.8 meters, assuming a drop above the target building of about 12 meters. The difference in height ...



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