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1

When you are at the bottom of a roller coaster about to move upwards (in say a circular motion), the net force you experience is given by, $F_{net} = \frac{mv^2}{r}$ The forces acting on you are the normal force of the track $N$, and gravity, $mg$. Therefore, $N - mg = \frac{mv^2}{r}$, and thus, $N = \frac{mv^2}{r} + mg$ This normal force is the force ...


1

Depends on the nature of the collision. If there is a mechanism that takes energy from the system, i.e. a deformation, than energy is lost. You could think of your example as the center of mass of your rod as a point mass that starts rotating on a massless string once is passes the pivot. As usual, energy and momentum are conserved. You could do the ...


0

First be aware that, so far, you have been dealing with projectile velocity as a 2-dimensional phenomenon. This is going to have to change. In order to conform to standard notation, you'll need to refer to altitude (vertical motion) as the z-coordinate, while horizontal will be handled by x and y coordinates. For ease of calculation, you can assume that you ...


2

In the systems you describe, each string connects always two masses. The tension force exerted on these two masses by the string is equal in magnitude and opposite in direction. Hence, the work done by each string on the two masses attached have opposite sign. As a consequence, if you sum up all contributions from each string and each mass, the net work ...


1

Say, you want to these objects of equal masses and let them rotate on circular orbits (just to understand the scalse). You will thus have a force between them \begin{equation} F={Gm^2\over d^2}, \end{equation} where $d$ is a distance between them. Thus you'll get from $v^2/(d/2)=F$ \begin{equation} v^2=\frac{Gm^2}{2d}. \end{equation} So if you have, say, ...


0

Indeed. It is due to the law of conservation of angular momentum. The angular momentum of the rotating element within the motor will exactly cancel that of the rest of the motor, thereby giving zero net angular momentum, as with the initial conditions.


3

It is certainly possible in principle. That said, you should look up the Hill sphere and consider the amount of mass required to make the arrangement stable in the face of perturbation from the rest of the solar system. If I wanted to plan this as a technology project in the near future I would probably propose the Earth-moon L4 or L5 point at the venue.


0

Allowing the bar to pivot about point G, you will see that R1 (if it was the only force) would rotate the bar clockwise and R2 (by itself) would rotate the bar counterclockwise. The direction of rotation is determine by the direction of the force (up or down) and where it acts in relation to the pivot (right or left).


0

The casing will spin in the opposite direction. That is the principle of reaction wheels.


0

They are the same. If the force that A put on B were different than the force that B put on A, then the system would be in violation of Newton's third law. The friction equation tells us that $F_F=\mu*F_N$. $F_N$ is the normal force, or the force that each of the materials applies on each other. Because the frictional forces must be the same, and the ...


2

The angular momentum of the earth && mouse wheel system does not change. When the earth pulls on the mouse, the mouse pulls on the earth, so no net moment is seen any arbitrary point in the universe.


20

In this case, gravity is still an external force. In a zero-g environment, the mouse would also begin to move around the inside of the wheel, opposite the rotation it causes in the wheel, which would keep the angular momentum at zero. This would happen because the only way for the mouse to exert a force on the wheel and rotate it is for it to push itself in ...


0

You need to measure the torsion constant by timing the oscillation before adding the two big spheres. If the big spheres are present the system is no longer a simple harmonic oscillator because the restoring force is now a complicated function of the rotation angle.


0

We can call the first object P, and the second object Q. Then we can say that the location of the particles can be notated as $ <P_x, P_y>, <Q_x, Q_y> $ Therefore, the distance, $d$, between A and B is $d=\sqrt{(A_x-B_x)^2+(A_y-B_y)^2}$. So, because $x(t)=v_x*t+x(0)$, and $v=<U_1*\frac{B_x-A_x}{d},U_1*\frac{B_y-A_y}{d}>$ Combining all ...


0

I feel like there are three main factors to this question you are trying to highlight... The strength of each person The height variation of each person The amount each person leans when pulling the rope Let's look at this as an ideal situation to your proposal. The strength of each individual are the same, the height variation is a steady slope, and the ...


-1

The drawing has it completely wrong, because it doesn't take into account the leaning that players do. Leaning is beneficial to the team, because it allows you to use gravitational force as well as your muscles to pull the rope and removes the height differences in the rope that would put shorter players at a greater disadvantage than they already have ...


0

So the x part of the drag is $$\begin{align*} f_x &= f \cos\theta \\ &= (k v^2) \cos\theta \\ &= k v (v \cos\theta) \\ &= k v v_x \,, \end{align*}$$ and $v$ is dependent on the $y$ component of motion as well as the x-component. Similar consideration, of course, apply to the y-component of drag. So the independence is explicitly lost and ...


0

You could change your orientation by moving your limbs on a rotational manner, like the reaction wheels used on satellites. Then when well positioned if able to eject something that would work for speeding you down.


1

I am going to assume that you have not yet studied linear algebra, sorry if it seems as if I am talking down to you at any point. You are correct in that we can split a vector into two components in the plane. This is because any two linearly independent(not parallel or anti-parallel) vectors form a basis(a set of vectors from which you can "build" other ...


4

If you were to hit yourself (to be specific, let's say you're using your hand to hit your chest) then your hand would indeed produce a force on your chest, but your chest would also produce a force on your hand that is equal in strength and opposite in direction (Newton's Third Law). The combined effect of these two forces (when looking at your body as a ...


1

In a reference frame where the center of mass of the Earth-Sun system is at rest, we will have $$ m_\text{Sun} \vec{v}_\text{Sun} + m_\text{Earth} \vec{v}_\text{Earth} = 0 \quad \Rightarrow \quad \vec{v}_\text{Sun} = - \frac{m_\text{Earth}}{m_\text{Sun}} \vec{v}_\text{Earth}. $$ In particular, if this is true at some initial time, then it's true at all ...


2

I suspect this is an example of the spinning-egg problem, in which a prolate spheroid (such as an egg) spun on a table about one of its "short" axes will tend to "stand up" so that it's spinning about its long axis. A few explanations have been proposed for this phenomenon, most notably: H. K. Moffatt & Y. Shimomura, "Spinning eggs — a paradox ...


0

Your entire analysis is entirely correct and complete (and in my opinion the best way to go about it). There's no more information that need to be derived about this issue. Other analyses are just a different representations of the same facts.


1

Is this the correct way to find the derivative of kinetic energy? $$ K=\frac{1}{2}m v^2 \\ $$ So: $$ \frac{dK}{dt} = \frac{1}{2} \left(\frac{dm}{dt} v^2 + 2mv \frac{dv}{dt} \right) $$ If the mass does not change over the time, then $$\frac{dm}{dt}=0$$ And finally $$ \frac{dK}{dt} = \frac{1}{2} \left(2mv \frac{dv}{dt} \right) $$ So simplifying: $$ ...


1

Answer to the question in the title? Two vectors are only equal to each other if they are the same (this is a general rule: equality means the things compared are identical). That means having the same direction as well as the same magnitude. So how could changing the direction of motion not be acceleration? Don't get hung up on fact that in 1 dimension ...


3

Here is the procedure: $KE = 0.5mv^2$ $\frac{d}{dt}KE = 0.5m\frac{d}{dt}v^2$ So the question becomes,how do we find the derivative of $v^2$ with respect to time? One can easily see that $\frac{d}{dt} = \frac{dv}{dt}\frac{d}{dv}$ (Notice how the $dv$ cancels top and bottom) Therefore, $\frac{d}{dt}v^2 = \frac{dv}{dt}\frac{d}{dv}v^2 = \frac{dv}{dt}\times ...


2

The time derivative of $v^2$ is $2v \frac{dv}{dt}$ not $2v$. You must use the chain rule.


3

Yes, there are an infinite number of solutions, though your teacher will want you to choose the most obvious one. When the force does work on the mass, that work can be converted into two forms: the potential energy of the object the kinetic energy of the object If you apply a force of $800g$ then once the object has been raised the 2.4m it will still ...


0

Very briefly said: A change of velocity $\mathbf{v}$ in time means that $\frac{\textrm{d}\mathbf{v}}{\textrm{d}t} \neq \mathbf{0}$ The acceleration $\mathbf{a}$ is defined as $\mathbf{a}=\frac{\textrm{d}\mathbf{v}}{\textrm{d}t}$ Therefore $\mathbf{a} \neq \mathbf{0}$ or in words if your velocity changes you will have an acceleration. Velocity also ...


1

This becomes easier to understand if you think of velocity being made up of perpendicular components. For example, let $v = v_x\hat i + v_y\hat j$. That is, velocity is made up of an x-component, $v_x$ and a y-component, $v_y$. When there is a change of direction, the $v_x$ and $v_y$ components will change. This means there must be some horizontal ...


0

Good question. It is very didactic to think that atomic bonds behave like springs. see for details When you push something, you are actually unbalancing the equilibrium position of the atoms in that solid because you are compressing them. Then, they exert a force to go back to their normal position, in which the potential energy is minimum. The same ...


1

Your last equation is a quadratic in $t$. The $a$ is simply $\sin(\theta).g$. You can then solve it with the usual formula for a quadratic equation. There are two solutions to a quadratic, and that's because if you go into negative time you'd be pulled down by gravity any get to the new X position. This solution, of course, wouldn't apply to your ...


0

Two hints: Constant of integration! Redo your calculations, and remember that when you integrate, you introduce a constant $C$ whose value is to be determined by the boundary conditions that you've stated. $v=\frac{dx}{dt}$


2

The black machine is a weight lifting machine. It is self contained with no power source. If it can lift an external weight and return to its original state as shown below, it is a perpetual motion machine. Suppose the blue weight is water. We could add a water wheel and generator on the right. You start at the top and work your way to the bottom. Then you ...


0

Power is defined as the derivative of the work done onto something with respect to the time, namely $P=\dot{W}$. Integrating over the duration of the time interval gives you the work done by the engine on the car: $$ W=\int_0^t dt' P(t') $$ which in turn is the differential form associated to the force vector field, that is $$ dW = F_xdx + F_ydy + F_zdz $$ ...


0

Use the definition of work done over a period of time $t$ , and you have : $$ E_k = \int_0^{t} \vec{F} \vec{dx} = \int_0^{t} \vec{v} d (m\vec{v}) = \int_0^{v} d\bigg{(} \frac{mv^2}{2} \bigg{)} = \frac{mv^2}{2}$$


0

If the centripetal force is greater [than the tangential force] the resulting [force] vector is near the perpendicular [or centripetal] [force], if the centripetal force is in perfect balance [with the tangential force] the resulting [force] vector should point exactly at 45°. Can you please explain why this doesn't apply to an orbiting body? Who ...


1

But we also know that a perpendicular force always causes an acceleration according to the rule of addition of forces. All forces cause acceleration. Perhaps you mean specifically tangential acceleration (changes in speed)? If the centripetal force is greater the resulting vector is near the perpendicular, if the centripetal force is in perfect ...


6

Your intuition is correct. For the ball to change its angular momentum (to go from "backspin" to "forward spin"), there needs to be a net torque acting. There are two forces on the ball: gravity, and the normal force of the slope. Both these forces act through the center of mass - so neither force adds torque. Without torque, there is no change in angular ...


2

The direction of the motion at any time $t$ is the direction of the velocity vector $\textbf{v}(t)$ as derived by solving the equations of motion; likewise $\omega(t)$ gives you back the direction of rotation according to the right hand rule. friction is the force that causes rotation is not entirely correct. Any force with non-zero torque generates ...


1

Let me recall what Newton's laws are, to start with: 1) In the universe exists at least one reference frame (that we call inertial) where $\textbf{v}= \textrm{const.}$ whenever no external interactions act on the particle. All other reference frames (if any) moving at constant speed wrt this very one will be inertial as well. 2) In the above reference ...


3

What matters here is how the value of $c$ compares to the value of $k$. Let us choose a $\zeta = \frac{c}{2\sqrt{mk}}$ One can show that when $\zeta =1 $ the system is critically damped, and will not exhibited any oscillations and will return to the origin in the shortest possible time interval. When $\zeta > 1 $ the system is over damped and will take ...


1

What data do you have for linear motion? Your equations are correct if you have the acceleration as a function of time and the orientation is constant. The angular accelerometer can give you the angles as a function of time with integration. Unfortunately, drift can be a problem. The received wisdom is to use an accelerometer (linear or angle), integrate ...


2

I had this kind of question myself for a long time because before studying Physics I've studied Math and in Mathematics we do things quite differently. In Mathematics, a general procedure is to give some definitions, probably by specifying some axioms, then we derive and prove theorems from this. On Physics there is a similar procedure, but it goes a little ...


2

Equations of motions need not be proven: they are such because they are experimentally true and there is no basic reason for that (at this point I wonder how you proved that $\dot{\textbf{p}} = m \textbf{a}$: you must have done so just re-writing a different form of the same equation, or any other starting point which you assumed to hold true). In the ...


-2

You can see horizontal rain in a tornado. I saw some this week while I was parked in a parking lot, probably an F1 tornado. It ripped trees apart mid-trunk--3-4 foot circumference trees. The roads were littered with trees, it took 1 hour to find a way home because all of the roads were blocked by felled trees--a 5 minute drive usually. When we came out ...


1

Law of inertia states an object in uniform motion continues in its state(i.e moving along straight line with uniform velocity) unless a force impressed thereon.. Mass is the measure of intertia of a body,i.e the massive the body greater the force required to change its state of motion... in the absence of ext.force the momentum (mv) of a object is ...


1

You are correct, they are different. $k$ is not a property of the material, its a property of the entire object. Imagine having a small amount of a fairly tight spring. It takes a lot of effort to extend it even a centimeter or two. Now without changing the material, connect a few hundred of the springs together. Extending it a centimeter now will take ...


1

There is deceleration caused by friction and drag, which is complicated as far as computing is concerned (as someone noted above). It can be determined empirically with help of some controlled experiments and curve fitting. The simplest is off course to assume that it is constant deceleration. Depending on the application, it may suffice. If so, the ...


1

The usual approach: write down conservation of angular momentum, linear momentum, and energy. Assume the impact is elastic and infinitely short duration. In that time the spring didn't move and the third particle didn't come into the equation. That means the problem can be reduced to two simpler problems: two particles that hit elastically (after collision ...



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