Tag Info

New answers tagged

1

You've almost got it! The constant thrust comes from a mass rate $\mu$ of fuel being expelled at a velocity $v_f$ as opposed to the speed of the rocket itself $v$. Therefore the equation is instead: $$ (m_0 - \mu t) \frac{dv}{dt} = \mu v_f - \alpha \frac {(m_0 - \mu t)}{r^2},$$ where $\alpha = G M.$ Hence the gravitational term you wrote as $G m_e (W_0 + ...


0

You are assuming constant thrust $T$ during flight, presumably until the rocket runs out of propellant. You are also assuming (or neglecting) any air drag to be zero. The resulting drag force could be very significant at high speeds and assuming your rocket is launched from the Earth's surface. With those limitations in mind the balance of forces of the ...


0

The g-force experienced by the pilot's spaceship is no different from that of of the pilot of a fighter plane (on Earth) or a racing car driver (on Earth) with exception of (perhaps) magnitude. The pilot will experience three types of g-forces: During linear accelerations: The pilot will experience an inertial force opposite to the sense of acceleration ...


1

I don't have a good knowledge of physics but the basic answer is yes, g-force is pretty much an acceleration force. For example 1g (Earth gravity) is basically an acceleration of 9.8m/s2 towards the Earth, you don't accelerate because the ground resists this force. In terms of whether some one could pass out then yes you could. In space the weightlessness ...


-2

When a elevator is freely falling with the a>>g(a=accerelaration, g=accerelerarion due to gravity) the gravitational force acting on the body looses its effect which help it to be in contact with the floor so the person experiences super weightlessness which resulting his head to be crash against the ceiling of the elevator and he looses his contact from the ...


-1

Your calculation of final momentum after the collision has a sign error in it. The pulley serves to change the direction of the motion. This means that a mass moving upward on the left side of the pulley is given a mathematical sign of "+" for the associated velocity. As the string goes over the pulley, the direction of the motion changes such that a ...


1

The ceiling is applying a total upwards force on the pulley of magnitude $2T$, in order to counteract the force of equal magnitude applied by the masses. So the net force on the system (consisting of three masses and the massless pulley and string) is not zero. Since $\int {Tdt} = mV = mv/3$, $\int 2Tdt = {2 \over 3} mv$ which is the amount of momentum lost. ...


0

In my opinion, the statement regarding the conservation of momentum says that "The momentum of a system remains conserved if no external force acts on it". I think that the mistake you have committed is that you tried to apply momentum conservation principle along the y-direction, along which gravity(an external force for the system) acts. So according ...


2

According to me, the momentum of the mass on the other side shouldn't be negative because the system considered is a connected system, i.e, the momentum transfer to the mass hanging on the other side of the pulley is momentum transferred through the pulley's string. I agree with @BowlOFRed about the contribution of the momentum transferred to the ceiling, ...


0

If it helps, you're mixing up cause and effect, and missing one part of the force diagram. Your diagram applies to any vehicle on a banked turn. However, it is missing a force applied horizontally to the left, which opposes $N sin{\theta}$. This is, of course, the centrifugal force produced by the motion of the vehicle, and has the value $\frac{mv^2}{R}$. ...


0

When the Earth pulls one of its inhabitants closer to its center of mass, that causes the Earth to spin a little bit faster. When the inhabitant hits the ground he is moving into same direction as the ground (to the east), but faster than the ground. Let us consider the angular momentum of two masses at the opposite sides of the Earth, at the equator, on ...


4

The pulley (and the attachment to the ceiling) are part of the system here. Because of this, you cannot simply use conservation of momentum on the three given masses. If the final velocity were $v$, then the total energy of the system would have increased since both the pan and counterweight would be moving and the other mass would not have slowed. You ...


4

this question is really severely damaged: the title (top/bottom quarks) does not match the question being asked (up/down quarks plus tau electrons), and the question literally being asked has a meaningful typo (tau selectrons) which invokes ideas from the still-speculative physics of supersymmetry, which is even crazier. To answer the question you literally ...


0

According to Wikipedia the predicted top quark lifetime of $5 \cdot 10^{-25}$ seconds is too short to attract other quarks to form a baryon, let alone find an object to orbit to form an atom. Various charmed and bottom baryons have been observed, but no top ones.


0

Approximation Slightly better and more elegant than the first. Let's make some assumptions. Let's assume that on impact, a fraction $\lambda$ of the ball's Kinetic Energy is transferred to the liquid. Let's also assume that all of the energy transferred will go into shooting water up. This should give the upper bound on how high the water "tower" will be. ...


1

Your moment of inertia is incorrect. You must calculate is based on the individual masses and their distances from the pivot: $$\mathcal{I}=\Large\Sigma \large\left( m_ir_i^2\right).$$ If you do this you should get an answer that agrees with what @ChrisDrost did, 2.47 s And you shouldn't assume that the center of mass is 1/3 of the way below the rotation ...


1

Let me first do this the way that I know is correct: with Lagrangian mechanics. This says that all of the physics you need is contained in the Lagrangian, which is the kinetic energy minus the potential energy. Your three masses Left, Right, and Bottom make the kinetic energy $K = \frac 12 m (v_L^2 + v_R^2 + v_B^2),$ where $m = \text{1 kg}.$ Defining ...


0

Condition for traversing the whole circle: Let the particle of mass $m$ is attached to an inextensible string of length $R$ which provides the necessary centripetal force along with gravity. Let the initial velocity be $u$ at the lowest point of the vertical circle. After time $dt$, it moves to some other point transversing angle $\theta$. The height at ...


1

1."The particle will complete the circle when at the highest point if the string doesn't slack at the highest point when θ=π." - Is it because if the rope slacks, then the mg component will pull it inwards and so the particle will not move as a circle? Yes. Gravity (the weight $W=mg$) is then strong enough to pull it back from the "swing". The rope is ...


0

"Friction" as we understand it is the force that resists lateral motion between two objects that have "apparently smooth" surfaces, and some normal force between them. If we look closely at the mechanisms of friction we typically find that "apparently smooth" is not the same thing as "smooth" There can be attractive (Van der Waals) forces between two ...


0

In the special case that the system has no initial angular momentum about the center of mass -- that is, the velocities of the particles (which may be zero) are along the line they lie on -- then the particles will eventually collide at the center of mass, as you describe. Otherwise, the particles will undergo elliptical motion.


0

Torque and moment are essentially the same thing and are calculated in the same way - it's really the context that determines which word is used. 'Torque' is usually used when we're talking about the twisting effect on a shaft and 'moment' is usually used when we're talking about the bending effect on a beam. If you're using a spanner to tighten a bolt, ...


0

You're not wrong; if you had two isolated particles that both started as completely stationary (which, on a non-classical note, is impossible due to the Heisenberg uncertainty principle), they'd collide at very high speeds (limited by radius of the objects themselves). However, remember that gravity is a central force. If two objects are moving in a circle ...


0

Since this is a thin-walled tube, the cross-sectional area of the tube is $(2\pi r)w$ where $w$ is the tube thickness, and $r$ is the tube radius. It follows that the stress is: $$ \sigma = \frac{T}{(2\pi r)w}$$ where $T$ is the force. If you double $w$, the stress is cut in half. Contrast with solid cylinder For a solid cylinder, the cross-sectional area ...


0

If we take the bicycle as a whole the only force that making the bicycle accelerate is the frictional force applied on the wheels. For the moment lets forget the front wheel. Then, $$F_f = Ma$$ Since the wheel is not slipping, for the back wheel $$F_t R_g - F_fR_w = I \alpha = IR_w a$$ With the use of these equations $F_f$ and $a$ can be quatified


1

Faced by the same question and a background that includes courses in vector calculus, I have sought a simpler answer. My answer is much that same as to why one can easily balance on a typical bicycle. Bicycles are constructed so that the point where the extension of the front fork pivot would hit the ground is in front of the point where the front tire ...


0

Gert's answer covers the physics of the motion well, but glancing through I didn't see it address the issue of why one spin travels farther (rather than just in a different direction) so consider this a supplement. I see you specified in your definitions of draw and fade that they are for right-handed golfers. This is important because due to the geometry ...


0

1. Frame of reference: The golf ball is at the origin of an $x,y,z$ coordinate system (see Fig.1). $x$ and $y$ are in the horizontal plane. The golfer wants to project the ball in the $y$ direction. 2. Rotational and translational motion: The ball is hit in such a way that it acquires an initial speed $v_0$, at an angle $\theta$ (see Fig.2). The ...


0

Yes so-called pseudo forces do work and if they were to be describable as a conservative force, then yes the corresponding mechanical energy would be conserved. The best example I can find is the gravitational pull we feel at the surface of the Earth. It is in fact the sum of the "true" gravitational force owing to Newton's law of gravitation and the, ...


0

You don't need angular velocity to find the linear (translational) speed. If you have answered question d.ii then you have an equation like $$T_{cen}=ma_{cen}$$ where $a_{cen}$ is the central (radial) acceleration, and $T_{cen}$ is central (radial) force towards the center. You now need to know the following formula for a circular orbit: ...


0

Use Fresnel coordinates: $$ ma_n=m\frac{v^2}{r}=F_n\\ ma_t=\frac{dv}{dt}=F_t\\ $$ Now $F_n$ denotes the total forces in the radial direction, and using a diagram you find: $$ F_n=P_n+T_n=T\sin{(30)} $$ Now for the tangential direction do the same thing to find: $$ F_t=P_t+T_t=-mg+T\cos{(30)} $$ Note that the speed is constant so $dv/dt=0$: $$ F_t=0\\ ...


1

While not necessary to solve this problem, I want you to know that... Concept # 0: the angular velocity of circular motion is directly proportional to the linear velocity of motion, $$ v = \omega r $$ where $v$ is the linear velocity, $\omega$ is the angular frequency, and $r$ is the radius of circular motion. Concept # 1: whenever an object exhibits ...


0

I'm not going to answer your question directly, but maybe can point you in the right direction. First assume you have some type of electromagnetic actuator that provides a linear force proportional to current. By shorting the windings of the actuator coil you will basically have a damper mechanism, a device that will resist changes in displacement. But such ...


0

You are calculating an unreal force by using Newton's second law. Remember that Newton's laws are valid in inertial frames of reference. And Newton defined inertial frames as those frames where an object continues to be at rest or in constant motion unless acted upon by a real physical force.In your example the train or you is not an inertial frame. Hence, ...


0

I think the real reason is because if you change your velocity then everything that used to be at rest should now continue just as they did before you changed your velocity. You can call it indifference, nothing cares how you move unless you interact with it. Or you can call it relativity. What this does is reduce the case 1) of things at constant motion ...


0

It's not a force that would break the glass, it's an uneven force (i.e., a baseball smacking the center of the glass). A great illustration of this is air pressure. Air pressure has a force of about 100,000 $\frac{N}{m^2}$ - so a normal glass case big enough to hold a fire extinguisher or something would have about 10,000$N$ of force evenly distributed ...


1

On an intuitive level if $a\ll R$ then nearly all of the deformation will occur close to the surface. Imagine for a bit that R is radius of the earth and you're pushing on some dirt with base ball so there's a circular contact patch with a radius of about 1/2 an inch. Now if earth were half as big would the forces/stresses/strain/contact area be any ...


1

Consider this previous exchange on the Cosmic Microwave Background The crucial bit is this: However, the crucial assumption of Einstein's theory is not that there are no special frames, but that there are no special frames where the laws of physics are different. There clearly is a frame where the CMB is at rest, and so this is, in some sense, the ...


1

If this is a correct description of what happens, can we conclude that g does same work on P and on P'? Yes. This is correct. If g acts perpendicularly to the velocity, it performs work of magnitude zero. This is also correct. The reason the two statements above are not contradictory is that the work done by the gravity changes the direction of ...


0

I cannot comment on the full design because the question is lacking on details, but I can explain more about the situation where you hit a mass with a ball and a spring reacts to it (like the sketch shown) Consider the friction force as $f=\mu m g$ and the equations of motion $$m \frac{{\rm d}^2x}{{\rm d}t^2} + k x \pm f = 0$$ The sign of $f$ depends on ...


1

I propose redefining this problem as follows (because I'm not sure it has a solution the way the OP has defined it). Let $y=f(x)$ be some symmetrical (around $y$) function like $x^2$. Let the point mass experience a friction force acc. to the usual simple model $F_f=\mu F_N$, with $F_N$ the Normal force acting on the point mass in the point $(x,y)$ ($N$ ...


0

Define the $y$ axis as the vertical axis and the $x$ axis as the horizontal axis. A projectile flying through the air has a velocity vector $v$, let's say at an angle $\alpha$ to the horizontal: The velocity vector $v$ can be decomposed into two components: The $x$ component: $v_x=v \cos\alpha$. The $y$ component: $v_y=v \sin \alpha$. It's important to ...


14

You probably got voted down cause this can easily be google searched, but the simplest way to explain it is that a tide happens because the lunar tug on one side of the ocean is measurably more than on the other side of the ocean and as the earth rotates the tidal "bump" follows the moon so you get 2 high tides and 2 low tides a day. A tide is effectively ...


0

Gravity, as you know it from your daily experience, is an everywhere parallel force, pointing downwards. In this field, each body is accelerated downwards, regardless where the body is, an where it is moving. But as you know, gravity is not parallel on a larger scale, the force always points to the center of the earth. Imagine your bullet is shot in space ...


0

What other answer? Yes an object falling accelerates due to gravity and gains KE. An object at height X has potential energy. It took work to raise the object to height X because of gravity.


2

At the risk of sounding like a broken record... it is a good idea to draw a diagram for all but the very simplest problems (and even then): You can immediately see that the normal force is made up of two components: $F_c \sin\alpha$ and $F_g\cos\alpha$. The friction results from the combination of both of these. In your approach, you ignore the normal ...


3

We integrate $x dm$ to say "for every piece of mass $dm$, sum up the amount of mass times the coordinate $x$ of that mass". It makes sense to integrate over $m$, because every piece of mass has an $x$-coordinate. Likewise, in kinematics, it makes sense to integrate over $t$ because for every $t$ we have a velocity $v(t)$. However, calculus doesn't care ...


0

Assume the ball slides down the slope without friction and that it starts from stationary ($v=0$) at a height above the horizontal $h$. During the friction free slide the object's potential energy $mgh$ is converted to kinetic energy $\frac{mv^2}{2}$, so that: $mgh=\frac{mv^2}{2}$ and: $v=\sqrt{2mgh}$. This speed vector is of course oriented parallel to ...


4

The reason why a gyroscope does behave in this strange way is that if you try to rotate it's axis in some direction, the "endpoints" of this axis have to be pushed perpendicular to what our first intuition would say. In order to verify why the axis starts rotating in this strange way, let's make some simplifications: the gyroscope consists of two identical ...


2

I will simplify this problem by assuming that the only forces come from Newtonian gravity and by limiting the masses of the moons to much smaller values than that of the planets, such that the moons exert a much smaller gravitational forces on all other bodies and therefore can be neglected; so the only sources of gravitational forces are the two planets. ...



Top 50 recent answers are included