Hot answers tagged newtonian-mechanics
7
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 ...
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 ...
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) = ...
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: ...
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 ...
5
$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 ...
5
You can always decompose a motion like this into two parts: (1) rolling without slipping and (2) slipping without rolling.
What is slipping without rolling? It means the object moves uniformly in one direction along the surface, with no angular velocity about the object's own center of mass. For instance, a box that is pushed along the ground can easily ...
5
This is something I played with while testing a n-body code I wrote during college. Unfortunately I don't have any animations, or even the original code anymore - but I can report qualitative results.
Removing Jupiter and Saturn does indeed have a significant destabilizing effect -- an a chaotic one at that (i.e. depending on precise initial conditions, ...
4
In polar coordinates, the velocity being tangent to the circle, it is directed along the $\hat{e}_{\theta}$ vector. The centripetal force is directed along the $\hat{e}_r$ vector.
So
$\frac{F}{m}\hat{e}_r = \frac{d\vec{v}}{dt} = \frac{d(|v|\hat{e}_\theta)}{dt} \underbrace{=}_{\text{Chain Rule}} \frac{d|v|}{dt}\hat{e}_{\theta} + ...
4
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
$$
Then near the origin, the rope would look ...
3
I think you just forgot that the $\int_A^B F\,dl$ is not a scalar expression. Rather it should be written in a form $\int_A^B \vec{F}\cdot d\vec{l}$. Then it comes to the sign of the scalar product:
$$\vec{F}\cdot d\vec{l}=F\,dl\,\cos\theta$$
where the angle $\theta$ is taken between the vector $\vec{F}$ and the direction of the tangent to the integration ...
3
A constant net force means:
$$\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$$
In particular, you see that $\Delta p \neq \frac{dp}{dt}$, unless $\Sigma ...
2
Because I have a closed loop, should there be some sort of "back action" from the tension in one leg of the loop on that on the other leg?
Cool question. About the "back action." Imagine, instead of a smooth cylinder, a cylinder with a thin, rectangular bump directly opposite the point that the load hangs beneath (on the top part of the cylinder.) ...
2
This problem has a recursive flavor that we'll not try to avoid.
Conservation of momentum tells us that
$$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 ...
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 ...
2
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
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
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 ...
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 $dA=rd\theta dr$:
$$
...
1
For pushing it up, we have to overcome friction(act downwards) as well as the $mg\sin\theta$. So, $$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$$
Eliminate $N$ and use $f=\mu mg\cos\theta$.
Solve for $\mu$ you get your answer.
1
If the force exerted by the spring on the attached object / the acceleration of the object is in the same direction as its displacement, you can imagine that the object will continue to go to infinity because there is no opposite force bringing the object back to the equilibrium position. Hence, the minus sign give us the sense that the acceleration of the ...
1
$x$ measures the difference in length of the spring in relation to its relaxed state. If you increase the length (positive $x$), the spring creates a force in the negative $x$ direction, because it wants to return to its relaxed state. Accordingly, if you compress the spring (negative $x$) the spring wants to expand (force in positive $x$ direction) in order ...
1
The ground will provide all of the static friction. Imagine what would happen if the upper block contributed even a tiny amount to the static friction: It would have to move forward due to the reaction force. Having M2 inch along you pull M1 (which stays stationary) would be very strange indeed.
Static friction always acts to prevent relative motion. It ...
1
Now I have got a method to get it directly.And again it came out to be an easy problem.
The answer comes out to be $$I= \dfrac{m}{12}(a^2+c^2)$$
See if we add another such plate along it's side $AC$ , then it comes out to be a parallelogram plate , whose MOI is known, same as rectangular plate
So, by symmetry arguments , both the triangular plates have ...
1
In the following diagram, is work done by static friction 0 ?, since the point of application is also moving with speed v w.r.t. ground here and is only stationary w.r.t. the block on which sphere is rolling w.r.t. ground here.
Static friction itself is 0. The formula $f_s=\mu N$ defines the maximum possible magnitude of the static friction force, not ...
1
"So what if it is an attraction force? How this should influence our calculations
Because you have written the work--energy relation in an incomplete shorthand. The correct version, $$W = \int \vec{F} \cdot d\vec{x} \quad,$$ depends on the relationship between the direction of the force and the direction of the path.
This relationship is the source of ...
1
Basically , it means at each instant the bottom most point has $0 $ velocity , it doesn't mean that the point has no acceleration . But at an instant it has $0$ velocity . And because of that at each instant $v_{cm}=\omega r$ for the bottom most point , and if this doesn't happen , then static friction acts to make it $0$ .
Its like suppose you are walking ...
1
The relative speed of the point of contact of the rolling body w.r.t. the surface on which it rolls is zero.
If the surface is at rest then the velocity of the point of contact of rolling body and surface is zero.
Mathematically:
$$v_1 -\omega R=v_2$$
Also we can get the reltation in accelerations ..... Differentiate the above eq.
...
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.
...
1
When the player hits the ball with top spin, it makes the ball, well, spin.
By spinning, the ball will modify the airflow around itself and thus create an air pressure profile which will deflect the ball : this is the Magnus effect.
So by applying top spin on the ball the way tennis players do, the ball is rotating in the direction of the trajectory. This ...
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