# Tag Info

0

I am not sure If I understand your question correctly. The gravitational force in this problem acts only "downwards". However, to define a direction in two dimensions one needs two coordinates. In the conventional coordinate system the force acts in the $(0,1)^T$ direction, where $0$ and $1$ are the coordinates. In the problem presented by you another ...

0

You are considering an object (cannonball) of mass $m$. Once it is launched at speed $v$ at an angle $\theta$ from the horizontal direction, the cannonball feels only one force, its own weight $m\vec g$. $g$ is the gravity acceleration ($g\simeq9.81\,\mathrm{m.s^{-2}}$). Newton's equations state that $m\ddot{\vec r}=m\vec g$. The mass can be removed from ...

0

This is a variation on the semi-well-known helium balloon experiment. Sit in a car while holding a He balloon so it floats freely. Accelerate hard & observe which way the balloon moves.

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This is really a continuation of my comment, but it got a bit long for a comment. As I mentioned in my original comment, if the acceleration is constant then we get a static pressure gradient just like we get a pressure gradient in the atmosphere. Incidentally, the Earth's gravitational field is approximately constant up to say the stratopause, so it's a ...

2

Mass 1 is given an initial velocity $v_1(0)=v_o$. You want the velocity $v_2(t)$ of mass 2. One way to do this is to break up the motion of mass 2 into the motion of the center of mass of the two-mass system, and the motion of mass 2 relative to the center of mass: ...

4

If you look at this problem in 2D you have the following parameters at some instant which describe your trajectory (position and velocity) around a celestial body with gravitational parameter $\mu$: radius $r$, radial velocity $\dot{r}$ and angular velocity $\omega$. There are also a few others, but these do not really matter in this problem, due to ...

3

If you try jumping on a trampoline, you will notice that when you jump up, the trampoline bends and stretches underneath you. It stretches some even if you stand still, but it stretches extra when you jump. The trampoline is elastic. When it's stretched, you can feel it pulling back towards its normal shape. Thus, just before you jumped, the trampoline was ...

2

The string contacts the point on two infinitesimally close points with different slopes. Imagine a small pulley end the two points are the entry and exit point of the string. If the string is between points A on the left and point B on the right (with B lower) then we call the angles of the string from horizontal $\theta_A$ and $\theta_B$. If the mass is ...

2

Consider a mass-spring system executing simple harmonic motion. If I draw the displacement, velocity and time graph, it would look something like this: You may see that when t=1 second, velocity is maximum and acceleration is zero. Another way to explain this is by using the definition of acceleration. Acceleration = change in velocity/time = gradient ...

1

As taught in calculus, to find the local extrema of a function, one solves for the values of the argument where the derivative of the function is zero. So, at the maxima and minima of velocity, the (time) derivative of velocity is zero. But acceleration is the (time) derivative of the velocity. Do you see the answer now?

1

It took me quite some time to clearly understand the experiment you're describing. Actually, pouring a full bottle in a container is a quite intriguing thing. Consider the following starting configuration : This of course is an unstable situation, as the pressure $P_0$ cannot be at the same time the pressure of the air in the bottle, and the atmospheric ...

0

Not sure i understand fully your question but in general friction can be seen as a force (vector) pointing in the opposite direction of motion with (if there is a motion). Moreover, the force is tangent to the surface of contact. Thus for an object with spherical symmetries (like a pulley, cylinder, sphere...), the force is perpendicular to the radius.

0

See the diagram for guidance : http://picpaste.com/p012-Lrn6pmdn.jpg Draw the gravity vector and the centripetal vector, then the resultant of the two, the banking should be normal to this resultant for no side forces. m = car mass in kg g = local gravity rate in (m/s)/s. v = velocity in m/s. r = radius to centre of gravity in metres. A = banking angle in ...

1

Actually I think I disagree with the answer by BMS (the group of asymptotic symmetries of asymptotically flat spacetimes?). However I am not sure to have understood BMS'answer completely. In my opinion, there is no difference between the definition of work in pure mechanics and work in thermodynamics (I stress that I am speaking of thermodynamics and not ...

0

One can consider the quantities $\int F_x\,dx=\int m\ddot{x}\,dx=\frac{1}{2}m(\dot{x}_f^2-\dot{x}_i^2)$ The $y$ version of above The $z$ version of above Are these what you're after? These three quantities aren't usually considered in standard problems, but they seem valid to me. Your "Result 2" is the sum of the three bulleted equations here.

1

You are right in thinking that the car's acceleration is what keeps it in place, but it is important to remember that an object moving at a constant speed in a circle is accelerating (despite not speeding up). The reason for this is that acceleration is defined as a "change in velocity," and velocity is a vector quantity (i.e. it has magnitude and ...

1

Assume that the liquid has a uniform density, $\rho$, and that the diameter of the U-tube is large enough to preclude capillary effects. Tha acceleration of gravity, $g$, is the same for both arms of the U-tube. Pick a reference point at the bottom of the U-tube, where the absolute pressure is defined as $P_0 \text{ }$. Move from this point up each arm ...

2

If I understand your question correctly you assume that the spaceship is driven by some kind of engine giving it the necessary speed to revolve around the earth. As the astronaut does not have such an engine you believe he should fall back on the earth. If this interpretation is not correct, maybe you could make your point a little more clear. There are two ...

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Remember that for the astronaut's spaceship to be in a steady orbit, it must be moving around the earth at the appropriate velocity $v$, where $\frac{GMm}{R^2} = \frac{mv^2}{R}$ (i.e. the gravitational pull of earth is matched by the force needed to accelerate the astronaut in a circular orbit), where $M$ is the mass of the earth, $m$ of the ship, and $R$ is ...

1

Here is one issue where thermodynamics and mechanics could differ in the definitions of work. In mechanics, a non-careful, ambiguous, but common definition for the work done by a force $\vec{F}$ is $\int\vec{F}\cdot d\vec{s}$. The problem with this is that we're not told which infinitesimal displacement $d\vec{s}$ to use; one could use (1) the infinitesimal ...

1

As I cannot post any comments I have to post this as an answer although the essential points were already given: In classical mechanics energy itself was no meaning. Only energy differences have a physical interpretation. Thus in the classical case energy is only defined up to an arbitrary constant. So any fixed state's energy can be set to zero (but of ...

0

This question of the causation of gyroscopic torque (and its magnitude and direction) troubled me all my life until I had a 'Eureka' moment about 20 years ago. I have written up the explanation of this phenomenon on my website, www.newtontime.com, using nothing more than Newton's Laws of Motion and without any need for 'fancy' mathematics. The analysis is ...

6

As weird as it sounds, the answer is "yes." Take, for instance, a satellite in gravitational orbit around some heavy body. It's energy is given by $$H=\frac{p^2}{2m}-\frac{GMm}{r}$$ Clearly, there are solutions to this equation which have $0$ energy (look at a slowly moving particle that's really far away), but those solutions necessarily involve a ...

0

No. According to the energy-momentum relation, the magnitudes of the energy and momentum of a particle of rest mass $m$ are related through the equation $E^2=p^2+m^2$. Obviously $0\leq p^2\leq p^2+m^2=E^2$, so if the energy is zero we have $0\leq p^2\leq0$, or $p=0$.

0

(By trial and error using personal excel calculator.) g assumed at 9.82 (m /s)/s Unsure if your initial angle is 60 above horizontal or 30 above horizontal. Anyhow, here's both : a) Initial angle 60 degrees above horizontal, time taken to reach - 30 degrees = 11.75866 seconds. x = 587.933 m , y = 339.443 m b) Initial angle 30 degrees above horizontal, ...

1

It's up to you whether or not to include elastic potential energy. You'll get the same answer as long as you're careful about what you define to be inside and outside of your system. The basic idea to use here is the work-energy theorem when only mechanical energy is of concern: $$W_\text{net,external}=\Delta K_\text{total}+\Delta U_\text{mechanical}.$$ ...

0

To state it simply, friction is the resistance to motion of an object within a system, in this case a ruler on a desk. As you suggest in your question the normal force to the surface is important to friction, the equation is: Coefficient of friction = force required to maintain constant velocity / normal force however turning the ruler on its side does ...

-2

Suppose the cylinder is very wide. Then certainly centrifugal force would cause the fluid pressure to be higher at the perimeter than it is at the center. So if the hole is near the periphery, there is a greater "head" there, so fluid should be ejected at higher velocity. Ignoring viscosity, the velocity should be proportional to square root of pressure. ...

1

Take the scalar product of $(1)$ and $(4)$ to get $$\mathbf{F\cdot U}= \gamma^2(u)\left(\frac{dE}{dt} - \vec f\cdot \vec u\right)$$ In the proper frame, this becomes the rate of change of the rest energy, so that for a rest-mass preserving force, $\mathbf{F\cdot U}= 0$. Hence the four-force in this case must be of the form $$\mathbf{F} = \gamma(u)(\vec f, ... 1 It's quite easy to get confused by differences in notation: in this post of mine I defined v_\parallel and v_\perp as the velocity components parallel and perpendicular to the coordinate acceleration \vec{a}. In your question however, you're interested in the angle between the velocity and the force vector \vec{f}, and that's a different angle: ... 0 It's easy to see without doing any math, but just by looking at the picture. Let's consider first the case of low k_{12}. In this case, m_1 and m_2 basically don't notice k_{12} because it is so weak that it is drowned out by the other springs. So the low k_{12} case basically gives the same value as the k_{12}=0 case for all three frequencies ... 0 You have the right ideas, but you don't dare to put them in maths ! Momentum conservation is vectorial. Here, you have a 2D system, so let's write it with 2 component vectors. I will denote \vec{P}_{i/f} the initial and final momentum respectively. So the momentum conservation yields  \vec{P}_i = \vec{P}_f  I choose to represent the x-axis as the ... 0 Momentum conservation says \vec{p}_A = \vec{p}_B + \vec{p}_C, we can split this in components:$$p_{A,x} = p_{B,x} + p_{C,x} \\ p_{A,y} = p_{B,y} + p_{C,y}$$Some of these momentums are 0. Which one? For the block B - the only one with a motion not parallel to one of the axes' - you have to use trigonometry. To get the velocity, once you have the ... 1 In the lab frame there is no centrifugal force. The ring goes outside because it lacks of a centripetal force. Let's take a step back: if the rod spins slowly then the ring does not slide out because of the friction. In this case the friction is a centripetal force, this means that it is responsible for keeping the ring on a circular motion. If you increase ... 2 Frankly, until you bring along an accelerometer as well as tracking the bus's speed thru corners (I'm assuming you mean you sway during a turn, not just going down the road), I'm going to remain skeptical of your claim. Centripetal force is centripetal force. Now, if you happen to counteract turning force unconsciously, it may be that you feel more ... -2 restoring force is refer to the system which bring force back to it normal position with out the effect of distance which the force existed to it Constance force. F1+F2= total force. F= kx 4 The affine Galilean structure is assigned by the first principle of Newtonian dynamics, i.e. by giving the class of inertial reference frames in the spacetime G^4. On the one hand it assigns the structure of an affine space to the spacetime, on the other hand it selects a subclass of permitted transformations between reference frames. A reference frame ... 3 Sorry it is impossible that, if both \psi, \phi belong to the domain of a self-adjoint operator A the identity$$\langle \psi| A \phi \rangle = \langle A\psi| \phi \rangle $$fails. The point is that your function T\psi, where \psi({\bf x}):= e^{-r}, does not belong in turn to the domain of the self-adjoint operator T so that:$$\langle T\psi| ...

1

You didn't specify in what direction the force of hand is applied, so for simplicity I assume that you are applying the force perpendicular to the desk. Now there are four forces on the book: 1) Gravity ($mg$) is trying to take the book down; it has a component $mg\cos\theta$ that is perpendicular to the desk and a component $mg\sin\theta$ that is parallel ...

-1

At the point the bullet leaves the gun it is traveling 800m/s. You need gravity to run; I guess he could be running in a vacuum. This is with random values. You would have to construct a bullet decription for yourself in a Ballistic Calculator.

-1

because a bullet does not travel at a constant speed, it would most likely shoot ahead of you (more than 400 m/s) but than even out with you untill there is a change in velocity. (this is just a theory from me, and i have no education in the field of physics, im in 9th grade)

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At low velocities like this you can ignore special relativity and simply add the two velocities. This is really easy to see if you imagine yourself standing still and the Earth moving under you. Relative to you the gun should fire just like you were standing still. This is called an inertial frame of reference. You see the bullet leave at $400\: ... 0 The incomparable Chris Hadfield did a related experiment on the ISS using water on a cloth. You can see that the water does not fly off the cloth. To simplify the experiment consider water on a flat surface: The air/water interface has an energy per unit area, so increasing the area of the air/water interface takes energy. This is also true for the ... 1 When you hit the obstacle (if you don't destroy it) your speed goes down to zero in a quite small time. That gives you the acceleration. It starts when you start hitting and it ends when you come to a complete stop. This acceleration is due to the force that the obstacle generates on the car. Think about:$\displaystyle a=\frac{\Delta v}{\Delta ...

1

You want to use energy conservation still. The total energy of the system is still $\frac{1}{2}kd^2$. The difference here is that you will have an extra term in your kinetic energy due to the rotation.

3

You should realize that the first equation you write gives you the value of $a$. In these kinds of problems, you are always given some force, and you are expected to apply Newton's laws to the problem. Therefore $F_{\mu} = -mg\mu \underbrace{=}_{\text{$2^{nd}$Law}} ma \quad \quad \to \quad \quad a = -g\mu$ Now you know the acceleration, you can find how ...

1

Remember the law $F=ma$; you already know the force from friction $F_{\mu}= -mg\mu$. Hence you can get $a=-g \mu$, and one of your unknowns is gone.

1

If you only consider viscous dissipation within the droplet, this should indeed go to zero in the vanishing velocity limit: the (local) dissipation rate is quadratic in the velocity, so that decreasing the velocity by a factor of $\lambda$ reduces the (local and global) dissipation rate by $\lambda^2$. Of course, the process takes $\lambda$ times longer, ...

2

The centrifugal force on the ring is the pseudo force when in the ring's reference frame, which causes it to move outwards, given by $$\vec{F} = m\frac{v^2}{r} = mr\omega^2$$ Where m is the mass of the object, v is the tangential velocity of the object, and omega is the angular velocity To find the time required for the ring to fall off, you need ...

0

No, that isn't right. Think about the gravitational field of a sphere. Equi-potential surfaces form concentric spheres about the original sphere. An object in the gravitational field of a sphere will follow the field lines (or lines of action of force) - which in this case are radially inward. Now if you imagine a field with two spheres, they will be ...

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