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31

Sure. It would happen all the time with propellers and rotors if we didn't design them to avoid it. This is something that is generally avoided in aerospace applications because the shock wave causes pretty significant drag. A lot of work goes into making sure this doesn't happen through careful selection of the blade cross section, adding sweep the ...


13

It's very possible to spin something fast enough to create a sonic boom, but engineers usually try very hard to avoid it. Several aircraft have been built, on purpose, with supersonic propellers. One production aircraft is the Tupolev TU-95 Bear long-range bomber. Eight counter-rotating 4-blade props turn fast enough so the prop tips are supersonic. Rumour ...


8

I want to add to the other answers that when an object is rotating at a supersonic speed, an observer will be hit by a rapid series of sonic shock waves, as the shock wave is an ever-expanding spiral. This is what makes supersonic propellers so terribly loud. The images below depict the process. The red circle in the middle is the trajectory of a propeller ...


8

. After the collision, according to the conservation of energy, the speed of both balls should be 35 m/s. ... Now, obvisouly, 35 m/s in R2 is 85 m/s in R1, not 79m/s. Where does the discrepancy come from? The collision is an artificial example: it is considered inelastic, but there is no loss of energy. If it followed the ordinary rules, in ...


7

The stroke of genius The laws of Nature are wonderful: extremely simple - make a tremendously complex world, a few symbols can be extremely powerful and beautiful all roads lead to Rome, you may go from London to Rome following many different routes, only a fool would get there via Moscow. But an eagle can fly over the Alps in a straight-line. I'll try to ...


7

If I drop a ball from a height $H$ and the ball rebounds from the floor it will bounce back up to a height of $e^2h$ where $e$ is the Coefficient of Restitution $C_R$ of the collision between the floor and the ball. Why is this the case? Because, as you can see in the link I added, it is defined as $$ (C_R) = \frac{v}{V_o} $$ and in the ...


5

(a) Write down the potential energy of the rope as the function $y(x)$. You're almost right, up to a minus sign in the limits of the integral: $$V=\dfrac{mg}{l}\int_{-x_0}^{d-x_0}y(x)\sqrt{1+y'^2}\ dx$$ (b) Since the problem is static, interpret the potential energy as the Lagrangian and find the Lagrangian density. The Lagrangian is usually given ...


4

Since the force is based on the wetted perimeter, any configuration that would make the perimeter very large in a very small area would be overwhelmed by the surface tension of the water droplets connecting nearby perimeters. So the effective perimeter would be much lower. So you are sunk!


4

Because the equations for linear and angular motion are very symmetrical In Newtonian mechanics, linear momentum is a vector while angular momentum is pseudo-vector which hints at its true nature as a higher rank tensor object. In relativistic mechanics, four-momentum is a four-vector while angular momentum is a (rank 2) four-tensor. So, the ...


3

The total force acting on a raindrop equals $g$ minus air resistance which increases with velocity. In other words, as the raindrop speeds up, air resistance increases which decreases the acceleration (until eventually the acceleration equals zero and terminal velocity has been reached).


3

Your fractal pattern will fail for reasons already given. However, given a large enough lake, you should be able to stand on a frame which is supported by a very long (perhaps circular) wire. All you need is for the force per meter (assuming uniformly applied) caused by your body weight and the structure itself to be less than the surface tension force ...


3

Negative energies are totally fine, because you had to pick a zero-point for energy. In your calculation you picked it to be at infinity. You could have chosen the zero-point for potential energy in such a way that your system had zero energy, or whatever. Only changes in energy are meaningful, in general. Consider this: what happens if you add energy to ...


3

Suppose I drop a ball from a height $H$ above a flat ground, how can I determine how far it will have traveled by the time the ball hits the ground for the 3rd time $L=x$?. I know how to calculate the speed $V$ when it first hits the ground, using constant acceleration formulas, I can also work out how fast the ball is travelling ($v$), after ...


3

Since he is falling at terminal, hence constant, velocity, he is experiencing NO net (total) force. There are (at least) 2 forces acting on him though, which are his weight (900N) and the air resistance. Because he has constant velocity, i.e. he is not accelerating, these forces must balance, i.e. the air resistance is 900N upwards whilst his weight is 900N ...


3

On an intuitive level, the initial speed $v_0$ can be considered to have two effects: one on the horizontal velocity, and one on the vertical; the former affects the range in a direct sense, and the latter increases the time the projectile is in the air. The combination of both of these gives an overall $v_0^2$ contribution. If you were to increase the ...


2

the speed of both should be according to the conservation of energy If both balls have the same speed after the collision, the collision is inelastic, i.e., kinetic energy is not conserved. If the balls are identical, then conservation of momentum requires that $$\mathbf v'_1 + \mathbf v'_2 = \mathbf v_1 + \mathbf v_2 = 150 \mathrm{\frac{m}{s}}$$ If ...


2

Let a sphere be dropped from height $h$ to a fixed horizontal plane. If $u$ be the velocity of the sphere just before striking the plane, then $$u^2 = 2gh \implies u = \sqrt{2gh}$$. If $v$ be the vertically upward velocity with which the sphere rebounds to a height $H$, then $$v = \sqrt{2gH}$$. Since both $u$ & $v$ are perpendicular to the horizontal ...


2

First, let's review the basic ideas of simple harmonic motion (I'm assuming an early university level). Starting with Newton's equation: $$F=ma$$ and using Hooke's law $$ma=-kx$$ then recognizing that acceleration is the second derivative of position x $$mx''= -kx$$ We know that simple harmonic motion is sinusoidal, so we substitute $x=\sin(\omega t)$ ...


1

The answer is "yes", for speed of sound, and "maybe" for the sonic boom. It is not the speed a problem, but the apparition of different pressures along a surface or on sides of a surface and the shock waves. The extremities of the turbine inside of a vacuum cleaner can function at speeds above speed of sound if the vibrations caused by pressure are limited. ...


1

$m_1 = m = 10$ kg $m_2 = 2m_1 = 20$ kg $F = 60$ N; $ma = F - \mu mg-T$ $2ma = T - 2\mu m g$ $\Rightarrow ma+\mu mg = F-T$ $ 2(m a+ \mu mg)=T$ $\Rightarrow 1/2=(F-T)/T \Rightarrow T = 2F/3$


1

Yes you can use the motor to slow and stop the disk, but you have to have (1) Alignment of the motor and its load angular momentum with the disk's angular momentum, and (2) enough motor load (moment of inertia) and motor speed to fully transfer the disk's angular momentum into the motor and load. The process of momentum transfer is done all the time on ...


1

$F=\frac{Gm_1m_2}{r^2}$ is wrong because you are essentially summing up the magnitude of the gravitational forces of each point on the ring without considering their direction. This is wrong since force is a vector. To find the force, take 2 small elements, diametrically opposite on the ring each of length $dl$ and mass $dm$. Draw the direction of the force ...


1

Calculating the gravitational force on the axis of a ring is equivalent to calculating the gravitational force of a pair of opposing small portions of the ring in which the full mass $M$ of the ring is thought to be concentrated. The result is an axial force equal to $$F = G M m\frac{cos(\theta)}{S^2}$$ Where $\theta$ is the half-angle between the two ...


1

Instead of integrating the force due to each mass element, which requires you to compute the component in the x-direction, you can calculate the gravitational potential, which is a scalar quantity. The force is then minus the gradient of the potential. The gravitational potential energy is: $$V(x) = -\frac{m_1 m_2 G}{\sqrt{x^2 + R^2}}$$ The force in the ...


1

The fastest way is to compare kinetic energies in the two cases: \begin{align*} KE &= \tfrac{1}{2}I_{\text{cm}}\omega^2_{\text{cm}} + \tfrac{1}{2}M(R\omega)^2_{\text{cm}} \\ KE &=\tfrac{1}{2}I_{\text{inst}}\omega_{\text{inst}}^2 = \tfrac{1}{2} (I_{\text{cm}} + MR^2)\omega^2_{\text{inst}} \end{align*} So $\omega_{\text{inst}}=\omega_{\text{cm}}$. The ...


1

Suppose I drop a ball from a height $H$ above a flat ground, how can I determine how far it will have traveled by the time the ball hits the ground for the 3rd time $L=x$?. It is very simple, supposing the $C_R=.56$ At each bounce will travel $l = H * C_R^2$, and if you substitute $H $ with $h_1,h_2...$ at each bounce you get the formula for the ...


1

The problem lies in your simplistic assumption that the perimeter is the only thing that matters. The actual force can be no greater that the weight of displaced water (see for example a capillary) and as the force you try to exert, so the amount of water displaced will increase. That doesn't mean you could not use surface tension to "walk on water" - just ...


1

If the motor is fixed on a moving rotating disc there can only be one result. If you attach the rotor from the engine to the platform, then either the rotor or the platform (or the engine) will tear itself apart. It doesn't matter if that platform is moving or not. See my drawing; if I glued the rotor disc to the platform, it cannot rotate. If it is forced ...


1

The impulse momentum theorem would be your best answer: $p(t) = \int_0^tf(t')dt'$


1

The other answers are very good so I will concentrate on the more physical meaning, on intuition rather than mathematics. Imagine you have a large rod of a big mass M.it is difficult to rotate it. Now considered the same mass M compressed to nearly a point.Now we have your question about the middle of the rod but on the extreme of it having a huge ...



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