New answers tagged

1

It is a method that is generally used for conservative unidimensional problems (problems with only one degree of freedom, here your angle $\theta$ or cartesian coordinate $x$). You'll notice that it is equivalent to using Newton's second law in this case : let us write the total energy $E = \frac{1}{2} m v^2 + V(x)$, $V$ being potential energy. The problem ...


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The interesting thing is that in the diagram shown above the frictional force is zero so there is no way a frictional force can do any work. If there was a frictional force acting on rolling object then there would be a torque about its centre of mass and so there would be an angular acceleration. Also if there was a frictional force acting on the rolling ...


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The point in which the friction acts is not actually moving, but rather as the object rolls past any given point, a new point of friction takes over. See in the animation below how a point on the circle touches briefly:


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Mass is not a function of time, so the only thing inside the integral that needs differentiating is the position. Let me try to make this clearer. The integral over $m$ is really the integral over the volume: $$\int_m r \cdot dm = \int_V r \cdot\rho(r')~ dV$$ Here I deliberately distinguish between $r$, the vector from the origin of the coordinate ...


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You need to read the rest of that page you quoted. In fact, there is only a perceived momentum transfer in the Earth's frame of reference, but since the Earth is rotating, that's a lousy (so to speak) frame to work with. Further, as the page states, a day later the pendulum's back where it started. The only way to effect an actual momentum transfer would ...


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How can we detect Earth's spin? Apparent motion of Sun You will have observed that the sun reappears every 24 hours. There are two common explanations for this. One of them is that the earth rotates with a period of approximately 24 hours - this is the only explanation supported by the scientific evidence. The main alternative had a rather convoluted ...


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Perhaps the least convenient but the most direct is go to the Moon and observe the Earth. The (average) length of day as measured by timing stellar transit to stellar transit, sidereal day, differs by 4 seconds from the length of day defined by timing noon on one day to noon on the next day, solar day. A satellite launched East requires less energy to ...


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This is not a simple problem. The interactions of the earth's atmospheric fluctuations and the rotational patterns are a matter of research: Atmospheric loads (= air pressure), e.g. during a high pressure weather system, can change the shape of the elastic Earth by up to two centimetres and can also alter the Earth's gravitational force. In this ...


0

The left hand diagram is in the plane of the couple produced by the weight of the rod and disc and the reaction force at the pivot point. The right hand diagram is my attempt at a 3D schematic. The key to understanding what is going on is first to realise that the direction of the external torque $\tau$ is at right angles to the direction of the ...


1

The impulse is the change in total momentum of the body of mass. The momentum $P$ of a rigid body is the product of the velocity of the centre of mass $v$ and total mass $M$. A free rigid body's centre of mass translates at constant velocity, while the body itself performs a rotation about the centre of mass. Thus, after the impulse $J$ has been applied, ...


0

Intuitively the motion of a rigid body is split between, translation of the center of mass, and rotation about the center of mass. The intrinsic inertia about those two motions are the mass and the mass moment of inertia. To rotate a body about any other point away from the center of mass means that the center of mass has to translate in addition to the ...


1

Yes you have to take the frictional force F into account so this exerts a negative impulse moment RFdt on the right hand disk with radius R as well as rFdt on the left hand disk with radius r. So the delta of the rotational moments have the ratio r/R. Assuming the left hand disk spins with rim velocity v0 before impact and after impact they have equal rim ...


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In order for the spinning disc to set the stationary one in motion, friction forces need to act between the contacting surfaces: That requires a Normal force $F_N$ to act between them. Using the kinetic friction coefficient $\mu_k$ we can then state: $$F_F=\mu_k F_N$$ The friction force on the $m,r$ disc causes angular acceleration $\alpha=\frac{d ...


2

We generalize the result from the second answer you linked: $$\vec{L}=\sum_i \left(\vec{r}'+\vec{{r}_i}' \right)\times \vec{p_i}=\vec{r}' \times \sum_i \vec{p_i}+\sum_i \vec{{r}_i}'\times\vec{p_i}=\vec{r}' \times \sum_i \vec{p_i}+\vec L' $$ Now $\vec{r}'$ is some arbitrary vector, not just the vector to the center of mass. Thus when the system is moving ...


1

When you drop a stationary disc onto a rotating one there must be a time when there is relative motion between the discs as you cannot have an infinite acceleration. If there is no friction then nothing much happens and the spinning disc carries on spinning and the other disc just sits still on top of it. To get an interaction between the discs you need ...


1

yes i think your assumption is correct based on the fact that in equilibrium the discs shouldn't slip on each other. but then you need to change your equation for angular momentum. use parallel axis theorem to write down the moment of inertia for the stationary disc to get the answer in the equation. since you have chosen your axis to be passing through the ...


0

In a frame outside the rotating frame you surmise that the normal reaction provides the force which produces the centripetal acceleration. In the rotating frame the normal reaction (inwards) is equal on magnitude to the centrifugal force (outwards). You now have a statics problem with a net force of zero on the mass.


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You should also note that static friction acts on the body when it is about to topple. Friction acts along the ground so it's torque about A is also 0. Torque due to F is clockwise and torque due to mg is counter-clockwise. Since torque= lever arm * force (or in vector form $\vec{\tau}=\vec{r}x\vec{F}$), torque due to mg is $mg\frac{a}{2}$. So net torque is ...


0

One way to see that $\vec{A}(t)$ rotates around $\vec{\alpha}$ is that its tangent (or derivative) is always perpendicular to $\vec{\alpha}$ and $\vec{A}$. In order to give more definitive proof it might be easier to write the differential equation is matrix form $$ \begin{bmatrix} \dot{A}_x \\ \dot{A}_y \\ \dot{A}_z \end{bmatrix} = \begin{bmatrix} 0 & ...


0

The equation is quite relevant in physics. Think of the precession of a magnetic dipole in a magnetic field, NMR etc. \begin{eqnarray*} \frac{d\mathbf{A}}{dt} &=&\mathbf{\alpha }\times \mathbf{A} \\ \frac{d\mathbf{\alpha \cdot A}}{dt} &=&\mathbf{\alpha \cdot \alpha }\times \mathbf{A}=0\Rightarrow \mathbf{\alpha \cdot A}\;\mathrm{const} \\ ...


2

Your claim is false (in $\mathbb R^3$). If $\vec{a}$ and $\vec{b}$ are fixed vectors (I assume they are not co-linear and satisfy $|\vec{a}|=|\vec{b}| \neq 0$) there are infinitely many rotations $R \in O(3)$ such that $R\vec{a}=\vec{b}$. One is the rotation $R$ of the angle between $\vec{a}$ and $\vec{b}$ performed around an axis orthogonal to the plane ...


0

The volume of the disk is $z\frac{\pi r^2}{2}$ not the weight. Otherwise you have done everything correctly! :) Here's how I solved your problem, it's very similar to your solution but keeps track of the directions involved (and doesn't require any integration). Apologies if this does not clarify things. The rotational force or torque ($\tau$) is ...


2

There's another component to the solution: nonuniform friction. The blade, even if untaped, has different coatings than does the shaft. The butt end may or may not have a tape roll applied. All this means that, even if you throw the stick with zero spin applied, when it hits the ice it's almost guaranteed that there'll be a nonzero net torque due to the ...


1

It is called conservation laws, conservation of momentum and conservation of angular momentum. Because friction on the ice is very small, the geometry of the stick is a line with non uniform mass, there will be angular momentum and linear momentum that will be transferred to the ice at the points of contact. To only rotated there should be no linear ...


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For a rotating body, the kinetic energy is given by $$E=\frac12 I \omega^2\tag1$$ The angular momentum is $$L= I\omega\tag2$$ This means that the energy could be written as $$E = \frac{L^2}{2I}\tag3$$ But the expression you wrote makes no sense. As for your question, equation $(2)$ above shows that angular momentum depends on both angular velocity, and ...


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Anwser: Moment of inertia doesn't depend on angular velocity. From the introductory textbooks, we know that $I=\sum m_is_i^2$, where $s_i$ is the distance $m_i$ to the axis of rotation. And in the situation of rotation around a fixed point, $I$ is a tensor, more complicated. Even so, the moment of inertia only depend on the positon of axis. In the situation ...


0

This may be a result of bad wording and/or misunderstanding. Let me try to clarify - suppose a particle is tied to a string of length $l$ and orbits around its center with some angular velocity, $\omega$. That is, in polar coordinates, its position is given by $(r,\theta) = (l, \omega t)$. From this, we can find the velocity in this coordinate system - ...


0

Expanding on the comment, here is an answer. Getting all the microcosmic force correct is a bit outside of the scope of this answer, so let me answer the question with a toy model instead. Assume $N$ particles on a ring, attached to each other with perfect springs. The force of the springs is then $F=kx$ where $x$ is the displacement from equilibrium. We ...


1

When there is curvature, tensile forces (in a ring, in a string, ...) will give rise to a net force as shown in the following sketch:


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Question: Gearboxes offer a significant benefit in that they affect the inertia ratio by a factor of the gearbox ratio squared. Answer: • Consider a motor running an inertia load JL via a mass-less, loss-less gearbox. The effective inertia (J) or its impedance (Z=sJ, where s is a complex variable), as seen through a gearbox of ratio G (where G > 1 for a car ...


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There are several factors that may be taken in account, but the more important is the energy used deforming the tire. Suppose a deflated tire. As you move forward and the tire rotates, the part of the tire that is starting to touch the ground has to be deformed (since the tire is flat). You have to use an important amount of energy for that. Note that the ...


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I studied the subject for 5 years and came to the conclusion that in average the toppling movement of the earth occurs each 12,000 years in average. The mass of the meteorite hitting the earth must be equalor more then 10 12th kg See for details in www.couldthesunriseinthewest.com/- Johan Leupen


1

You can calculate angular velocity by taking origin ofyour frame on semicircle centre.unknowns as torque and friction force can be calcuted by balancing forces and writing torque equations knowing mass of beetle and semicircle.


0

That's a sloppy explanation. It's more simple than that. Angular momentum and angular velocity are parallel, if your axis lies along one of the eigen-axes of the object: of course, symmetry axes are usually eigen-axes (defined by the eigenvectors of the moment of inertia tensor). That's obvious from the definition $\vec{\Gamma}=J\vec{\omega}$ which is an ...


3

Intuitively you can think of it this way: Moment of intertia measures how hard is to change the angular momentum of the object. Now, the angular momentum of a point single point of mass $i$ is: $$ \vec{L_i} = r_i \times m_i \vec{v_i} $$ To make this point rotate faster, you need to apply some torque. The mass part is easy, and one of the $r_i$ is just ...


2

When two forces of equal magnitude, opposite in direction and parallel act on an object, the object will rotate without having translational speed. This condition is not enough. In rotational mechanics the position is also important. The point that makes this simple to overview is, that changes in translational motion is described by forces and e.g. ...



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