Hot answers tagged

49

Other answers don't mention the fact that no single impulse (e.g, like being fired from a gun) can launch a projectile into orbit. A purely ballistic projectile fired from a gun must either crash back into the planet, or it must escape from the planet altogether. In order to achieve orbit, at least two impulses must be applied to the projectile. The first ...


28

Anything launched into orbit by such a gun needs to travel at orbital velocity (in fact above orbital velocity) in the lower atmosphere. That's generally undesirable, to put it mildly: there will be really serious heating.


15

Aside from the interior ballistic aspects of these various projects, it was quickly realized that any satellites launched by gun would have to withstand high g-loadings during firing of the gun and the size and mass of the satellite would be greatly constrained by the dimensions of the bore of the gun and the maximum impulse which could be provided by the ...


10

I think the heart of the question is whether one could arrange a continuous combustion of propellant along the length of the barrel. In that way the acceleration occurs along the length of the barrel in a more gentle way. Since the expanding gases from the propellant in a shell casing expand and the pressure of the expanding gases declines along the way it ...


4

If the pipe is cylindrical, there is no equilibrium height. All the air that goes in at the bottom must come out at the top, so the force balance will not depend on the height of the ball. In a ball flow meter (rotameter), the pipe is slightly conical, so that the gap between the ball and the pipe increases as the ball rises. If you want a very coarse ...


3

The rod exerts an upward shear force on mass on the left. This goes along with @lemon's comment regarding bending, although the shear force is not a tension. There is also a bending moment applied by the rod to the mass, with axial tension within the top half of the rod, and axial compression within the bottom half.


2

One more nasty factor: What is the expansion speed of your propellant. Take the Jules Verne approach and your spacecraft falls far short no matter how much powder you put in the gun because the expansion velocity is too low. Your craft will never exceed the expansion velocity of the propellant. Note, however, that you don't have to use explosives (or ...


2

I want to focus on one thing, here. The nature of the normal force. You write Doesn't normal force oppose any other force? which is a easy impression to get when you are introduced to the normal force in the context of things sitting on other things in a gravitational field, but that's not the best way to think about it. The normal force keeps ...


1

When rotating, the force needed to keep the two masses in circular motion is the centripetal force, (in this case, the tension). Now, you get it.


1

Wanted to give a different angle to thinking about this. Since you mentioned you wanted to consider this from the inertial frame (non rotating), then there is indeed no (fictitious) force. But in that frame, we can consider the water is indeed falling; however, the rate at which the bucket is also "falling" is such that the two stay together - in other ...


1

In the bucket experiment when the bucket reaches the top of the circle why will it have a normal force acting on the water downwards? The normal contact force is exerted by the bottom of the bucket as explicitly mentioned by the author. So, it is acting downward when the bucket is inverted. Doesn't normal force oppose any other force? There is no ...


1

Imagine a scenario where the bucket is rotated at just the right speed so that the centripetal acceleration required to keep the water on a circular path is exactly 9.81 $ms^{-1}$. Then at the top of the rotation, all the centripetal acceleration is supplied by gravity. However the bucket might be rotating faster in any given scenario but it still rotates ...


1

But the acceleration is not a partial derivative! Its a total derivative, $\frac{\mathrm dp}{\mathrm dt}$, with a $\mathrm d$ instead of a $\partial$. Anyway, I guess you might want to read about the Hamilton-Jacobi equation.


1

I would look at this in a slightly different way. Rearranging it: $$ m \ddot{x} = -(a|\dot{x}|+k) x = -k_{eff} x$$ If you look at it that way, it is really a variable, non-linear stiffness $k_{eff}$ that depends on the velocity, rather than a damping that depends on the position. In this respect (assuming $a > 0$), the stiffness coefficient has a lower ...


1

Let's look at what happens right when the falling jumper passes that equilibrium point (EP). At that point, as you correctly pointed out, the force on them (up, from the bungee) is equal to the force down (gravity). So, there is a total net force of zero. However, they already have momentum from falling. Newton's first law tells us that if there's no net ...


1

Put a coordinate system on the center of mass and place each leg i at $$\vec{r}_i = \pmatrix{x_i, & y_i, & z_i}$$ where $z_i = z_{c}+\theta_x y_i - \theta_y x_i$ describes the vertical deflection of the point, given the center of mass vertical position $z_c$ and the two tilt angles $\theta_x$ and $\theta_y$. Add vertical loads the each point ...



Only top voted, non community-wiki answers of a minimum length are eligible