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Provided the bat delivers exactly horizontal momentum impulsively to the second ball, it will not travel as far due to its initial downward velocity, as you say. Dissipating the downward momentum doesn't make much sense in the scenario you described.

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My equation is $$\frac{x^2}{a^2} + \frac{x^2}{b^2} = 0$$ That's not the equation you want for a satellite. That equation describes an ellipse with its center at the origin. You want an ellipse with the origin at one of the foci: $$r = \frac{a(1-e^2)}{1+e\cos\theta}$$ where $r$ is the distance from the origin to a point on the ellipse. $a$ is the ...

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You can actually infer the difference of the approaches just by looking at their names. A vector has direction and an energy (a scalar quantity) does not. Therefore, when you are trying to figure out scalar quantities such as distance and speed, you may find energy method more advantageous; when you look for velocity, acceleration, you have to use vector ...

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I think that you're making this problem more complicated than it has to be in order to simply determine if the assembly will tip over or not. You don't really need the spatial distribution of the forces being exerted by the table or ground on the assembly. All you need to note is that if the pivot point is at x=D1 then the ground will exert whatever ...

3

Momentum is really always conserved. Truly. Throwing something upward (accelerating it with your arm) causes your feet to push harder on the ground. The increased down-force causes the ground under you, and ultimately the entire Earth, to shift direction downward. Fortunately, the rock and the planet attract each other gravitationally, causing the rock to ...

2

If we apply a force $F$ to a mass $m$ and a friction force (drag) $F_d$ also acts on it the force diagram becomes: With $a$ the acceleration the object experiences, the equation of motion becomes: $F=ma+F_d$. As the mass moves towards the right, say for an infinitesimal distance $dx$, an infinitesimal amount of work $dW$ is performed on $m$ by $F$: ...

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The definition of force is under the assumption that there is no friction involved in the system. This means you can imagine the surface it is measured on as some kind of "super slippery ice" that has no frictional properties whatsoever. About the question whether the forces required to produce the same acceleration on different surfaces are the same: No ...

2

If there is no torque, then $$\sum \tau = \mathbf{r_1}\times\mathbf{F_M}+\mathbf{r_2}\times\mathbf{F_m}=0$$ Therefore, $$\mathbf{r_1}\times M\mathbf{g}+\mathbf{r_2}\times m\mathbf{g}=0\tag{1}$$ where $\mathbf{r_i}$ denotes the position of the center of mass of the combined system relative to the force applied. If we give the box dimensions $h$ and $l$, the ...

2

Current trend today in MBD is towards writing code, doing simulations for some practical problems. That is not entirely true and it mostly depends on the actual areas and topics you are dealing with, as for all the other subjects. Of course, due to the industrial applications, the practical side always has more money and more academic positions, but ...

2

A free body diagram on the $2m$ mass would have $2mg$ down and $T$ up. This would give a Newton's 2nd Law equation, assuming up to be the positive vertical direction, of $$T-2mg=2ma_{2v}$$. The $m$ mass free-body diagram would yield two downward forces, $T$ and $mg$ with a Newton's 2nd Law equation of $$-T-mg=ma_{1v},$$ assuming the tension magnitude in the ...

1

Yes, the cart will move, due to the force applied by the string to the pulley. To solve, calculate the string tension while the weights are moving, and then note that the pulley has to provide an opposing force in order to change the string's direction. The reaction to that force acts upon the cart, accelerating it. Momentum is conserved, because the ...

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This is a very good question that addresses a common misconception. Short answer: The box $m_1$ does NOT slide backward. The friction force in the $+x$ direction causes the box 1 to slide forward, but not fast enough to keep up with the larger box 2. There are no backward force in the x-direction - there is only the forward friction force in the +x ...

1

When you hit the falling ball, friction between bat and ball will momentarily stop the side of the ball that is hit from moving down. However, since this force of friction is not applied at the center of mass of the ball, this will not result in a complete arrest of the vertical motion: instead, the ball will acquire some spin. However, it will not lose all ...

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Time of flight is determined only by the vertical component of velocity - it is the time interval between when the projectile was released ($y=y_0$) and when it reaches the ground ($y_t=0$). As the collision is with a vertical wall it acts in the horizontal direction (assuming no friction during the short duration of the collision) and so has no effect on ...

1

You've computed the acceleration of the block once it is moving. However, if the block starts stopped there is a minimum force needed to get it started that is larger than the force needed to maintain it's motion. That minimum force to start the motion is computed using the maximum force of static friction and the assumption of equilibrium. The working ...

1

Nothing is stored, actually. Here, on Earth we are used to the fact that things "naturally" stop moving after a while (if we do not make them continue the movement somehow) "by themselves". But this is only apparent, as there is always certain force that prevents movement. This force is gravity often coupled with friction - gravity pulls the ball and the ...

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The following diagram should give you some insight: The black dot is the center of mass. If the center of mass is below the center of curvature of the bottom, then when you tilt the doll the c.o.m. will be displaced relative to the point of contact with the surface such that there is a torque that will attempt to right the doll again. If you place the ...

1

Everything we can measure is relative, because measurement is comparision with a standard. Hence, there are, in a strict sense, no absolute frames in experimental science. The task is rather to make it clear in our mind against what we measure and specify our physical quantities, in order to be able to share our results and to build a common knowledge. Take ...

1

You need to balance the moments about point P. The horizontal force F time $L_3$ will equal the mass $M$ times the horizontal distance $L_4$ between P and M. $$F \times L_3 = M \times L_4$$ This calculates the force required in the current position you've shown. Worst-case, an infinite force will be required after rotating $90^{\circ}$ from that shown, ...

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If the lift angle is $\theta$ (shown at zero in the diagram) then the payload lever arm is $$x_1 = \tfrac{L_1}{2} \cos \theta+L_2 \sin\theta$$ The force lever arm is $$x_3 = L_3 \cos\theta$$ Static balance exists when $$\left. \vphantom{\int } (M g) x_1 = F x_3 \right\} \\F = \frac{x_1}{x_3} M g = \frac{\tfrac{L_1}{2} \cos \theta+L_2 \sin\theta}{L_3 ... 1 While not quite an inherently "physical interpretation," the technique of inertial imaging allows one to use higher order mass moments for characterization and identification of biological molecules and molecular complexes. (See: Inertial imaging with nanomechanical systems) The basic idea here is that sticking ("adsorbing") a molecule ("analyte") onto a ... 1 The caveat here is that the second law is stated that net force is equal to the change in momentum. Assuming you and your buddy are not too wasted and are able to synchronize throwing the bottles off with the exact same force, exactly in opposite directions and through the center of mass, the net force is zero, and therefore there is no change in momentum ... 1 Here is my derivation of this result. I hope you find it helpful: Say we have n different forces F_1, F_2, F_3... F_n, applied at n different points. Now we pick two centers P and Q, and express the radial vectors (1) from point P to each of the n points (where forces are applied) as r_1, r_2, ... r_n (2) from point Q to each of the n points ... 1 Torque? Why do you think you need to think about torque? Is the center of mass over the base of support? It is if 0 < M D_1/2 + m D_2 < D_1. I.e., if \frac{-M D_1}{2 D_2}< m < \frac{D_1}{D_2} (1-M/2). 1 Newton's Third Law tells us that the momentum imparted to fragment #1 is equal and opposite to the momentum imparted to #2. So if I take as my system all the atoms of the original object, we see that the momentum of the system hasn't changed at all. 1 From your description, I think the quantity you are talking about is similar to the concept of inertia where the mass is abstracted into a matrix through its distributions in space. Mathematically, you can understand the matrix properties as following: For a positive semi-definite matrix, the Eigen values are always real and non-negative which makes sure ... 1 This may not be the ultimate answer as I don't know the quantitative relations between variables. But I can say the following: The ultimate velocity is determined by the power (P) of the car (or other objects, let's use car for example) and the friction (\mathbf{f}). Now that$$P=\mathbf{v}\cdot \mathbf{f} It means that all the power of the car is ...

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