# Tag Info

0

I'm not familiar with "Modern Analytical Mechanics" by Pellegrini & Cooper so I can't comment on that one but I'm very familiar with the other two books you mentioned. Landau's books are generally excellent but tend to be shorter in length and sometimes very dense. Nearly every paragraph has some profound insight that you'll miss if you don't ponder ...

1

Let me expand on (and correct a minor error) in what I said in the alluded-to thread. Unfortunately, I do not know the mentioned article (although "The Moon's Twin", published in 1989 in "The Magazine of Fantasy and Science Fiction", discussing the Jupiter/Io system, might be it). Therefore, I can't directly address it, but I think I can say three major ...

1

When you first start moving the end of the rope, there will be points along the rope that are not yet moving, and have not experienced the greater tension that results from your motion. It stands to reason that as these points first "feel" the wave, they will move towards the source of the wave (where the tension is greater) as well as transversely. If that ...

1

The two body problem can be made equivalent to the one body problem. Say you have a mass $m_1$ at $\vec{r}_1$ and a mass $m_2$ at $\vec{r}_2$ interacting gravitationally. Now focus your attention on the center of mass $\vec{r}$ and the difference vector $\vec{r}_{12}=\vec{r}_2-\vec{r}_1$. Then the kinetic energy of the system can be written as ...

1

The force due to sliding friction causes the center of mass to decelerate - since it is the only external force acting on the ring, and we know the center of mass moves as though all forces act there. You might find this earlier answer of mine and the links therein helpful. As for the "in the real world the ring will stop completely" part of your question - ...

1

If the chain is folded up on itself with the initial horizontal separation then yes it can be ignored. The picture is misleading because it shows a gentle bend connecting the two sides when in idealized reality you have a discontinious sharp bend at the bottom. Remember ideally there is not flexular rigidity (no resistance to bending) and inflexibiliy (no ...

2

Now, why can't it be reached? The short answer is that, due to the nature of the equation of motion, if the particle velocity $v$ is the terminal velocity $v_t$ at some time $t$, then $v = v_t$ for all time $t$. That is to say, a solution to the equation of motion is $$v(t) = v_t$$ in which case, the acceleration is zero and thus, the velocity is ...

0

It's an aymptotic behavior. To understand the statement rigorously, you need to solve the differential equation to find velocity as a function of either time or distance travelled (to do the latter, write $\dot{v} = v\,\mathrm{d}_x\,v$ and you have a DE for $v$ as a function of $x$). However, physically, witness that as the object gets nearer and nearer to ...

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Because it's only a limit for velocity, to which it will be continuously approaching by decreasing the difference between the actual velocity and this limit (and decreasing overall acceleration (pace of changing this velocity) toward zero).

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Check the principle of relativity of Galileo (there are no absolute velocities, only relative ones). Both scenarios are the same and the only difference is how you choose to a stationary frame. One usually assumes that the surface/wall doesn't move at all, given the eventual very high mass (compared to the ball), to simplify the analysis.

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The lever-approach to the problem is rather naive $$\frac{F_a}{F_b} = \frac{r_b}{r_a}$$ This implies that $F_a l_a = F_b l_b$ and equal work needs to be done. because this ignores the fact that the mass at a is different from the mass at b: $m_b > m_a$, therefore $$[F_b l_b] = m_b*a*l_b > [F_a*l_a]> m_a*a*l_a$$ Moreover, it has been noted ...

0

Tractors commonly have larger rear wheels for several reasons. One of these reasons is weight distribution. The larger the rear wheels, the further the wheel will have to sink into the ground to become stuck, which, on the common automobile, is a little over halfway between the center of the rim (wheel), and the contact point to the ground. This is the point ...

0

Be careful - you are applying the reasoning about a static situation to a situation that is not static. This rod will be rotationally and linearly accelerating, so you can no longer assume the net force or net torque is zero - $\Sigma F_y = m a_y$, and $a_y \ne 0$. If the rod is instantaneously horizontal and at rest, it's not too hard to find $F_n$ at ...

1

When you have a lightly damped oscillator, there is a small correction to the resonant frequency. This is derived in detail on the wiki page for the harmonic oscillator. The form they give is $$\omega = \omega_0\sqrt{1 - \zeta^2}$$ Where the $Q$ (quality factor) of the oscillator is given by $Q=\frac{1}{2\zeta}$.

5

You see, when you have a pendulum with friction you account it by including a force $\vec{F}_r=-b\vec{v}$. Then your differential equation for the pendulum is $$ml\ddot{\theta}=-mg\theta-bl\dot{\theta}\iff\ddot{\theta}+\frac{b}{m}\dot{\theta}+\frac{g}{l}\theta=0$$The solution of this differential equation depends on the values of $b$, $m$ and $l$ but since ...

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There may tension come into the picture first if the block shear. otherwise friction will come into the picture before tension because there must be some displacement of the block for tension to act. if block get sheared then even without relative displacement between block and surface tension will act.

1

Yes, the force points along the vector of the relative velocity between the object and the air. Quadratic drag is an interesting phenomenon. You have to calculate the net velocity vector (which includes a horizontal and vertical component) and compute the force along that axis; when you then decompose it into horizontal and vertical components you will find ...

0

If you open a door by gripping it near the hinge, you apply GREATER energy for a SHORTER time than when you grip it near the outside edge, which requires LESSER energy for a LONGER time. As the friction of the hinge and the weight of the door are equal in both cases, the total energy applied is the same. It's only the time applied and the intensity of ...

0

Your force has a component along the slope, so yes, the object will move along the slope. It will not leave the surface though, if that's what you mean by "lift" If you find it counterintuitive why the object has a vertical acceleration component despite your applied force being horizontal, you must think about the normal force. This is always, as the name ...

0

The key idea here is that Static friction is self adjusting.... It will oppose and nullify effect of your applied force F , provided maximum friction value is not exceeded. If you imagine or draw a Free body diagram of forces on the body , you'll see that friction due to ground on Body acts in opposite direction to F. So, Tension in string is not developed ...

2

Answer to the first question: This depends to some extent on the 'models' used for the forces of friction and tension. A typical model for string tension is as a restoring force that obeys Hooke's law: $$T = - kx$$ at least for a small positive extension $x$ in the length of the string, or equivalently, displacement of the block along the length of the ...

0

V1 : Velocity in front of panel V2 : Velocity behind the panel Force acting on the plate (Newton second) : F=m*a The accelleration is found by assuming a decrease of velocity from V1 to V2 over a distance s. The acceleration is thus found as: 2*a*s=v1^2-v2^2 . The mass transport over the distance s is m=rho*A *s By insertion: F=rho* A * s* (v1^2-v2^2)/ ...

0

See: The string will not develop any tension until force is applied beyond limiting friction, that is, both string and the frictional force b/w the surface will start opposing the motion only when the friction is kinetic. Hence, Static friction acts first, then comes both kinetic frictional force and the force opposing string's elongation. Thanks

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When the force you apply is less than $\mu mg$, the tension is not affected and does not affect the motion of the block. But if you apply a force larger than $\mu mg$, then the tension in the string increases. Obviously, the block will start its motion when the string breaks (if it is a real string). Finally, the friction direction is always perpendicular ...

2

The perfect head on collision is a special case where we don't need to worry about any relative angles. We can solve it using Physics 1 conservation of momentum and energy, all in the lab frame. For equal mass particles $$m u_1 = m v_1 + m v_2 \implies u_1 = v_1 + v_2,$$ and $$\frac{1}{2} m {u_1}^2 = \frac{1}{2} m {v_1}^2 + \frac{1}{2} m {v_2}^2 ... 0 I think deriving that same result in the "head-on" case where one the scattering angles are 0 and \pi ends up dividing by 0.$$\cot \frac\theta 2 = \frac{1}{0}$$-1 It is slightly more complex than it looks. There is friction, both because of the hinges and the air drag. Friction caused by hinges is (somewhat) irrelevant of speed. Air drag gets higher the faster the door is moving. That means that you need to move the door as slow as possible to conserve energy. Suppose you move it really slow, speed nearly 0. Then ... 1 Your target may not stop a very dense, compact bullet: It may exit, departing with some (possibly large) fraction of its initial momentum. If your bullet does not penetrate a target as well, as a less dense rubber bullet might (depening on circumstances including its speed and properties of the target), it might be stopped completely (either stuck in the ... 6 Assuming an ordinary hinged door (M = 3Kg, L = 1m), would it take more energy to open it when applying force in the middle of the door (point b: r=50cm), rather than at the end of the door (point a r=100cm), My own answer is no, since the change in force would be proportional to the distance required to open the door and therefore the total ... 0 Static friction does not produce or consume work in most of the times. For example for a solid body that rolls without sliding the velocity of the base point A is \vec v_a = \vec v_{cm} + \vec v_{tangential} \Rightarrow v_a = v_{cm} - \omega R = \omega R - \omega R =0 which implies that x_a = 0. The static friction is a force that acts on A so W_T = ... 6 The physics 101 answer is no: it takes more force, but it is compensated by the smaller displacement so the energy stays the same. If we start with a static door, and we end up with a door rotating at some speed, the energy into the door is the work done by the force and it must be the same independently from the point where the force was applied But let's ... 0 Torque=(R) x (F) Energy req. to rotate=(T).(Theta), Ta=Tb, ONLY Fa is less than Fb. & T at Hinge=0, it'll not rotate there. The force required will increase from A to hinge point. Energy needed hence is const from A to hinge(except hinge point). [Ta means torque applied at A] 0 It doesn't. The work is done by the active force(ex. A human trying to pull a bull.). This work is converted into frictional energy(ex. Heat generated b/w surfaces) 1 As @Gautham said, W = Torque * angular displacement which is also equal to the energy needed. here the angular displacement is same as we know and torque will be also same(Large distance implies less force needed,less distance implies large force needed but torque will be same) So energy needed will be same in both cases. 3 The answer is NO. Change of energy is work i.e W = \Delta E and here the work done is$$W = \text{Torque} \cdot \text{angular displacement}$$which is equal in both the cases. The only change is one needs to apply more force to achieve the same amount of torque at a smaller radius.$$\text{Force at "b"} > \text{Force at "a"}$$but not the work done or ... 12 You are right. To open the door during the same time interval (for pushing at a and b), you should induce the same angular acceleration. Since rotation in both cases is about the same axis, this means you need the same torque, this gives$$ \frac{F_a}{F_b} = \frac{r_b}{r_a} $$where r is the distance from the point of contact to the axis. However the ... 2 There are two senses to v_{\mathrm{min}}. Firstly it means that the speed of the particle must be greater or equal to v_{\mathrm{min}} at \theta = \pi or else it will not complete the cycle. Instead it will follow a ballistic trajectory until such time that tension comes back into the string. Secondly, the speed of the particle is not constant: it is ... 4 The question is so ambiguous that it allows a resounding yes. This is because "balance" is not defined, neither are the dimensions and the material used for the pencil, nor the location of where the "balancing" is to happen. The material and the shape of the surface to balance the pencil on, are not specified, nor the length of time it should stay balanced. ... 4 Here is how you deal this this problem as a system of equations. For each contact pair assign a normal direction \hat{n}_k and and impulse J_k. The possible contacts are AB, AC, and AD. We can introduce symmetries and simplifications later. The initial velocity if body A is v_A along the horizontal axis, and after the collision it is v_A + \Delta ... 5 The problem is equivalent to 4 spheres colliding simultaneously, where top sphere center is at 60^o relative to the x'x axes (same goes for bottom sphere): We'll name them: sphere A (dark blue), and spheres 1, 2, and 3. During the collision the spheres will behave like springs with an infinite hook constant. The forces on the spheres will be ... 2 Let me be the firs to answer 'Yes' (more or less). As the saying goes: In theory there is no difference between theory and practice. In practice there is. What I'm getting at is that there will always be differences between theory and practice, and that it is up to the physicist to decide which assumptions/simplications are suitable and which ones are ... 0 Its pretty much solvable. Lets name the sides of the Hexagon in the order 1,2,3,4. When the hexagon A collides with B,C,D the 2-3 side collide with C and C moves in a direction of A and it will not effect all the other hexagons. Same way the side 1-2 will collide with D and give it motion in a direction perpendicular to side 1-2 and same for side 3-4... ... 2 Altitude can indeed have such an effect. As your linked article explains, one can get a rough sense of the aerodynamic force on a spherical ball by neglecting viscosity (i.e., model air as a bunch of ballistic particles that do not drag on one another), in which case the formula is1$$ F = \frac{16\pi^2}{3} C_l \rho \omega v r^3.  The important point is ...

13

No. The weight of the pencil is roughly 1 Newton, and the area is about 500 square picometer (5 * 10-22) which means the pressure on the tip is around 2 ZettaPascal. That's quite a bit more than what graphite (or diamond) can withstand (that's masured in GigaPascal)

1

By definition, the work done by a force is $W = F\cdot d$, so how can static friction do work ? Can this force move the body a distance of $75~\text{m}$ ? Friction does negative work on the truck, slowing it down and does not move it forward. What does positive work on the truck, accelerates it and makes it translate $75~\text{m}$ is the engine ...

0

I believe that yes, it could. However you must also take into consideration that air density may not be the only apprehension that a player is dealing with throughout a game. As to answer your other question multiple world stadiums are covered at a sea level however the highest was located at Estadio Da Baixada, Curitiba which was 920m (3,018 ft). Depending ...

4

No. Firstly, the point of the pencil is generally not sharp enough to have just one single atom. People attempt to make that kind of tip in STM's. Even if you somehow did manage to get it sharp enough, graphite is so soft the the weight of the pencil will crush the tip. It won't stay a single atom wide. So there is no way to balance the pencil on a single ...

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TL;DR: there are many factors that prevent a pencil remaining perfectly balanced. The most important of these is the uncertainty principle that will make the pencil fall over in less than four seconds. For details, read on... Short answer: NO. The first photon of light that hits it would disturb your perfect equilibrium. The moon's tidal forces (which are ...

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No. To balance perfectly, the pencil would have to be perfectly upright and perfectly still. The uncertainty principle limits how well you can do both at the same time. Momentum and position form a conjugate pair. $\Delta x \Delta p \geq \hbar$. Angular momentum and angular position form one too. $\Delta L \Delta \Theta \geq \hbar$ This doesn't guarantee ...

-1

When the figure is symmetric about y-axis, $x_{cm}=0$. As to $y_{cm}$, I don't know an easy way without integration. Actually the integration is not so hard and I just checked that it is indeed $-r/\pi$. For a half circle with radius $r$, $y_0=\frac{4r}{3\pi}$. There are three half-circles with mass: 4, -1, 1 and corresponding $y_{cm}$: $-y_0$, $-y_0/2$, ...

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