New answers tagged newtonian-mechanics
0
In the context of most ideal pulley problems, it's assumed that the string or rope doesn't slip along the surface of the pulley, so the pulley's edge moves along with the string. This could be called a static friction force, however, the value of that force doesn't come into account unless it's the specific focus of some problem. What matters in most ...
2
I believe your mistake is with units, and it is the following:
$$T=\dfrac{M[c_{rms}]^2}{3R} = \dfrac{\left( 1 \text{amu} \right) [11.2 \frac{km}{s}]^2}{3 \left( 8.3144621 \frac{\mathrm{J}}{\mathrm{\text{mol} K}} \right)} $$
This doesn't even cancel out because you're left with a $\text{mol}$ unit. Add avogadro's number.
$$T = \dfrac{\left( 1 \text{amu} ...
2
In addition to the astronomical explanations above, there is theoretical way to create a view like the one pictured above (!) for any actual size of planet and satellite.
Consider the closing shot from Indiana Jones and the Holy Grail, where the victorious archaeologists are riding off into the huge setting sun. The sun subtends about 0.5 degrees as ...
13
The Roche limit applies when the astronomical body in question is held together by gravity rather than electromagnetic forces. This is the case for bodies with a diameter larger than around 500km. Obviously for smaller bodies, like humans, we can get arbitrarily close to the surface, but i suspect this isn't what you're asking about.
For moons much smaller ...
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Firstly, as Peter mentioned, your equations implicitly assume that the id fluinside is causing the acceleration. Otherwise you need to specify that the system is being pulled. I'll deal with both situations here.
If the water causes the acceleration
Unfortunately, we can't directly deal with the individual cylinder caps because the cylinder tube exerts a ...
0
I guess $N_1$ and $N_2$ shown in the figure are the forces acting on the masses, whereas your equation is about forces acting on the vertical part of the rope, and these forces are related to those acting on the masses via the third Newton's law. Therefore, the equation in your book seems correct.
0
See the cylinder as a whole object . The second law of motion gives that the cylinder does not accelerate untill it have a net external force.
So, the external pressures on both sides of the cylinder is not equal.The external agent which accelerates the cylinder applies force ($ma$) on left face towards right so that pressure on left fave exeeds by a ...
3
A constant net force means:
$$\Sigma\vec{F}=\frac{d\vec{p}}{dt}=C$$
where $C$ is some constant. This means that
$$\int \ dp=p=C\int\ dt=Ct+p_0$$
where $p_0$ is the initial momentum. Now, you can easily verify that
$$p_2-p_1=\Delta p=Ct_2+p_0-Ct_1-p_0=C(t_2-t_1)=C\Delta t$$
In particular, you see that $\Delta p \neq \frac{dp}{dt}$, unless $\Sigma ...
0
I think I have figured out the answer, hopefully.
Firstly, let us begin by stating the First Law:
$$\sum \vec{F} = \frac{\delta \vec{p}}{\delta t}$$
When the net force is constant, we it means that there is no change of momentum, in other words,
$$ \frac{\delta \vec{p}}{\delta t} = 0$$
In this case, we know that the function of momentum is a constant, ...
0
You first set $\Delta p = p$ (line 3), then $F = \Delta p$ (line 6), which leads to $F = p$.
Correct is that $F = \dot{p}$. Force equals the time derivative of momentum. Force does not equal momentum. So your equations are incorrect.
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This problem has a recursive flavor that we'll not try to avoid.
Conservation of momentum tells us that
$$m v_0 + (p+n-1)m v(n-1) = (p+n)m v(n).$$
Imposing the boundary condition $v(0)=0$ we find
$$v(n) = \frac{n}{n+p}v_0$$
as claimed.
Let $a_n$ be the time at which the $n$th bullet strike occurs.
We have $a_1=x_0/v_0$ and
$$v_0 (a_n - T) = v_0 ...
1
Your approach ignores the body of the slinky and essentially describes two massive particles coupled by a very light spring, which is not allowed to oscillate or show any of the interesting dynamics a real slinky will exhibit.
Ideally, you should be using some sort of continuum-mechanics approach to this problem, e.g. treating the slinky as a very elastic ...
-1
I have a quantitative answer which is a thought experiment avoiding all but the simplest equations.
An object going from velocity v=0 to v=1 needs to be pushed or pulled in some way. In my explanation I will use the same method to push the object from v=0 to v=1 then from v=1 to v=2, then v=2 to v=3, etc. I will show how the energy of movement embodied in ...
1
That line will get Coriolis acceleration $$\vec{a} = -2 \vec{\Omega} \times \vec{v}$$ ($\Omega$ is the angular speed of the earth's rotation, with a direction pointing into the ground from the view of the south pole). As it's going across the pole, there's a right angle between $\Omega$ and $v$ and the absolute value will be simply $$a = 2\Omega v$$ and the ...
5
The car's engine tries to make the wheels turn. However, the wheels encounter friction against the road so they cannot just spin. As the road has much higher inertia than the car, it will not move when the wheels want to turn. Instead, it is the car that moves.
The end effect is that the engine pushes against the road, just as you do when you push the car: ...
0
Each force causes reaction (3rd law). If move a car from the inside the car moves you as well. That's because you are pushing or pulling. However the engine does not push but converts energy in other directions, usually a rotating one (the same as riding a bicycle). This rotating force has its counter-force which is reaction of ground.
Pushing a car you ...
6
Malicious counter example
The desired object is a sphere of radius $R$ and mass $M$ with uniform density $\rho = \frac{M}{V} = \frac{3}{4} \frac{M}{\pi R^3}$ and moment of inertia $I = \frac{2}{5} M R^2 = \frac{8}{15} \rho \pi R^5$.
Now, we design a false object, also spherically symmetric but consisting of three regions of differing density
$$ \rho_f(r) = ...
2
The block is accelerating at $1\frac{m}{s^2}$ up the incline, since it is stationary with respect to the conveyor belt. What force is causing the block to accelerate? It can't be the normal force (which acts perpendicular to the motion). It must be the frictional force, which counteracts the component of gravity parallel to the incline.
Using Newton's ...
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Tie a rope to the object and attach the other end to a scale. Slowly lower the object into some water, recording the force on the scale and the amount of water displaced at many intervals. Using this data, compute the density of the section of the object that is submerged at all intervals. If the mass of the object is uniformly distributed then the density ...
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If you have a rigid mass distribution sealed inside a black box, then the only things you can observe about its motion are its velocity vector and its angular velocity vector as functions of time. These can be predicted if you know the total force and total torque that act, plus the mass, center of mass, and moment of inertia tensor. So all that can be ...
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If mass is evenly distributed inside the volume occupied by your mass, and you know your mass theoretical density, then if your mass is evenly distributed inside the volume it occupy, it will be exactly equal to the theoretical density, if there is a mismatch between the theoretical density and the measured density, it means that mass is not evenly ...
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The time taken between the N-1 collision and the N collision is $T-T\frac{N-1}{p}=T\frac{p-1+N}{p}$
Edit:
Reasoning:
The difference in T is due to the N-1 collision and is given by: $T\frac{N-1}{p}$
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If this is purely academic exercise then the answer is no, you cannot discern between the different friction forces because they are all acting along the same line of action (through the cg of the part).
0
It explodes with a force of 500N.
This sentence is nonsensical. It can explode and release some energy. It can also explode and impart a very high force onto the fragments (which sill be different for each fragment) for a very short time interval.
Once the explosion takes place, the fragments will not accelerate (they may decelerate due to air drag, ...
0
The microscopic description of the normal force is inherently complicated, not simple, and it cannot be explained in terms of just an electrical repulsion or or just the Pauli exclusion principle.
There is a very general and useful concept of a residual interaction. For example, the force between one neutron and another neutron is fiendishly complicated. ...
1
Now I have got a method to get it directly.And again it came out to be an easy problem.
The answer comes out to be $$I= \dfrac{m}{12}(a^2+c^2)$$
See if we add another such plate along it's side $AC$ , then it comes out to be a parallelogram plate , whose MOI is known, same as rectangular plate
So, by symmetry arguments , both the triangular plates have ...
7
$\hat r, \hat \theta, \hat z$ form an orthogonal, right-handed triad of basis vectors in 3d. Taking the cross-product with them is no more exotic or unusual than doing so for Cartesian basis vectors. $\hat r \times \hat \theta = \hat z$, and so on.
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It's actually not too hard to calculate the moment of inertia (MOI) of a right triangle. And you can make your triangle out of a large right triangle minus a smaller right triangle. So your MOI is just the MOI of the bigger triangle minus the MOI of the smaller one.
Step 1:
Extend line $b$ (move vertex $C$) until you have a right triangle. We'll ...
1
"So what if it is an attraction force? How this should influence our calculations
Because you have written the work--energy relation in an incomplete shorthand. The correct version, $$W = \int \vec{F} \cdot d\vec{x} \quad,$$ depends on the relationship between the direction of the force and the direction of the path.
This relationship is the source of ...
1
For pushing it up, we have to overcome friction(act downwards) as well as the $mg\sin\theta$. So, $$3N=f+mg\sin\theta$$
Now the block is just slipping , so friction is acting upwards, and so does the force applied externally.So,
$$N+f=mg\sin\theta$$
Eliminate $N$ and use $f=\mu mg\cos\theta$.
Solve for $\mu$ you get your answer.
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So let's see from where does the centripetal force comes from.
Imagine we have a body that moves in 2 dimensions. Let's then describe the system using the polar coordinates, such that: $x=r\cos(\theta)$ and $y=r\sin(\theta)$. Let's define two vectors, $\vec{r}$ and $\vec{\theta}$, such that:
$$\begin{cases} \vec{r}=x\vec{e}_x+y\vec{e}_y \\ ...
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The first diagram you drew was absolutely right and the free diagram was also right .
The second rotation would only be possible if the angular velocity is enough to displace the molecules of the air above the circle of rotation such that the molecules below the circle of rotation would push the bob with a force that will be more than the weight of the bob ...
0
This is pretty much well explained in the following Mathematica Simulation
In general for large $\theta$ the frequency becomes a function of Amplitude and non-linearity sets in. Also the phase space plot becomes distorted.
The next error in small angle approx is keeping upto $\theta^3 $ term, which approximately suggests corrections of order $\theta^2/3!$ ...
0
Just try this:
Tie a spring to a wall and pull it. Do you feel a resistive force? Or do you feel a force that is somewhat helping the force that you are applying? In other words, Does the spring pull on you or push you away when you pull on it?
It pulls you, of course, and this is a direct result from Newton's third law, which in this case gives:
...
5
I think the answer is that the second diagram you drew won't happen. I just picked up a string and tried this. What happened is that the first diagram is easy. For the second, I have to twirl the string faster, and I can't quite get it to stay above my hand. The best I can do is to get the mass to swing in a plane almost even with my hand.
Note: it's a ...
-1
The tension vector you are drawing is wrong. It should be pointing in the opposite direction. Also, the weight vector has been misplaced in the first pic. Finally, the centripetal force is not gravity, it is actually tension.
Edit: The part in italix is actually false, I thought the OP drew the rope not parallel to the axis was that he wanted to show the ...
1
The ground will provide all of the static friction. Imagine what would happen if the upper block contributed even a tiny amount to the static friction: It would have to move forward due to the reaction force. Having M2 inch along you pull M1 (which stays stationary) would be very strange indeed.
Static friction always acts to prevent relative motion. It ...
1
If the force exerted by the spring on the attached object / the acceleration of the object is in the same direction as its displacement, you can imagine that the object will continue to go to infinity because there is no opposite force bringing the object back to the equilibrium position. Hence, the minus sign give us the sense that the acceleration of the ...
1
You have to take the mass from $L$ to $x$, since that is the mass which has to be accelerated by the centripetal force due to the tension at position $x$.
The remaining mass of the rod (before position $x$) doesn't add anything to the tension in $x$.
1
Moment of Inertia is defined as:
$$
I={\sum}mr^2
$$
which in this case can be rewritten into an integral:
$$
I=\rho\int_A{r^2dA}
$$
Since the shape of the triangle can't be described by one formula, you would have to split the integral into multiple sections. And I will use polar coordinates, in which case $dA=rd\theta dr$:
$$
...
0
Whether a mass can be considered a point or not depends on the scale at which it is studied. The earth can be considered to be a point mass when we are studying its motion around the sun but not so when we are studying its own rotation. Newton's laws are applied to systems of many particles.
Newton's second law says that the rate of change of momentum of a ...
1
$x$ measures the difference in length of the spring in relation to its relaxed state. If you increase the length (positive $x$), the spring creates a force in the negative $x$ direction, because it wants to return to its relaxed state. Accordingly, if you compress the spring (negative $x$) the spring wants to expand (force in positive $x$ direction) in order ...
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$F=ma$. If $F=0$, and $m=0$, $a$ can be anything. Most physical laws are not "A causes B". They usually say that "A and B can coexist in these conditions". So, it is not necessarily "Force causes acceleration". It is "an accelerating body can coexist with a force if $F=ma$"
The net force on a massless string is always 0 -- it has to be (otherwise it will ...
6
Non-relativistic mechanics can't. Massless objects travel at the speed of light. The only reason to introduce a massless string is so that you can get some effect from the string without having to worry about the string in calculations. As soon as you start worrying about forces on the string causing it to accelerate you've violated the whole reason for ...
1
In the following diagram, is work done by static friction 0 ?, since the point of application is also moving with speed v w.r.t. ground here and is only stationary w.r.t. the block on which sphere is rolling w.r.t. ground here.
Static friction itself is 0. The formula $f_s=\mu N$ defines the maximum possible magnitude of the static friction force, not ...
1
NOTE: This comment was too long so I'll make it an answer.
I would assume that the amount that the spring contracts is negligible compared to the distance that the mass has fallen.
But anyway, regarding your amplitude problem: consider that when you have a mass on a level surface connected to the spring, there is an interplay between kinetic and spring ...
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moment is turning effect produced by a force . while torque is due to rotation of body.
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The second condition is saying that there is no discontinuity in the slope of the rope at the junction. In other words, there is no "kink" in the rope.
Imagine if this assumption were to fail in the following way:
$$
\frac{\partial D_1}{\partial x}(0,t) = -1, \qquad \frac{\partial D_2}{\partial x}(0,t) = 1
$$
Then near the origin, the rope would look ...
1
It depends on the friction of the contact. With a frictionless plane the top would precess around its center of gravity and the contact point will prescribe a circle.
Add friction, and the friction force translates the center of gravity the same way tire traction translates a car. Here you have the cases of a) pure rolling, or b) rolling with slipping.
...
2
Officially, I completely agree with the other answer given. I would like to offer this answer as a simplistic, intuitive answer to the question. No math involved.
I understand where your question comes from. In fact, depending on your current level of education, this question could indicate a high potential for future scientific success.
We all know that ...
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