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Assume $f$ friction is being applied in direction of F. Direction, now, has no significance as $f$ will come out to be negative if it is opposite direction. $$F+f=ma$$ $$FR-fR=I\alpha$$ symbols have their usual meanings Note that if pure rolling occurs, $f$ is static. Also, $$a=\alpha R$$ You can calculate $f$. If $$|f|> \mu_{static}mg$$ You can ...


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If you have ever seen a pizza being made by hand, you will know that when the baker throws the disk of dough in the air, he makes it spin. As he does so, the pizza "disk" gets bigger because the dough on the outside experiences a larger centrifugal force (in the rotating frame of reference of the pizza. Don't start on "there is no such thing", you asked for ...


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In the situation you gave, it's immediately clear what is meant, and there's no possibility for misinterpretation, so yes, it's perfectly acceptable. (Remember that torque is mathematically defined as a vector for convenience, but the direction of that vector isn't really physical.) The only issue I can see with that is that as you leave the simple ...


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There is a wikipedia article which describes the effect http://en.wikipedia.org/wiki/Equatorial_bulge Basically the bulge is caused by the rotation of the Earth. The centripetal force is given by $F=m\omega^2 r$. Therefore the poles feel a lesser force than the equator which wants to spin out into a disc. This is balanced by gravity which wants the Earth to ...


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I'll answer my own question. Since $\boldsymbol{\omega }=\omega _{z}\hat{\boldsymbol{k}}$ the angular momentum reduces to $ \boldsymbol{L}_{O}=-I_{xz}\omega _{z}\hat{\mathbf{i}}-I_{yz}\omega _{z}\boldsymbol{\hat{j}}+I_{zz}\omega_{z}\boldsymbol{\hat{k}}$. We can split the rod in three pieces, calculate moment(product of inertia for each body and sum up. ...


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As an alternate answer using parallel axis theorem, note that the inertia tensor of a rod pointing in the $y$-direction rotating about its center is $$\mathbf{I}_\text{rod,y}=\left( \begin{array}{ccc} \frac{b^3 \rho }{12} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \frac{b^3 \rho }{12} \\ \end{array} \right)$$ and similar for the $x$ and ...


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The problem is in assuming the force is forwards at a contact point right underneath the axis of the wheel, together with the no-slip condition. This leads to zero friction force, since their is no friction when their is no slipping. A solution can only be found when deformation of the wheel is taken into account, allowing the force to act on a different ...


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Check ✓ except the the speed depends on the center of rotation, not the choice of the origin. The general rule is that if the origin as velocity $\vec{v}_O$ then a point A located at $\vec{r}_A$ has speed $\vec{v}_A = \vec{v}_O + \vec\omega \times \vec{r}_A$. It happenstance that your origin does not move. Check ✓. In general, torques and angular momental ...


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This is what I did and I think is simple and right : Assume $v$ linear speed of centre of mass downwards and $\omega$ angular speed around it. Use the fact the bottom point has no vertical speed to find relation between $v$ and $\omega$. And I am done.


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The factor comes from the moment of inertia of the infinitesimal piece. In the disc method, each piece is a filled flat circle (a disc) of radius $r$, and the moment of inertia of a flat circle is $\frac{1}{2}mr^2$. The $\frac{1}{2}$ accounts for the fact that the mass of the circle is distributed between the center and the edge. But in the shell method, ...


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It is best to read a book on dynamics to get all the details you need, but here is an overview. I assume all the rotation axes intersect at the origin, and the payload center of mass is located at a distance $\vec{c}$ from the last gimbal (in local coordinates). The mass moment of inertia of the payload is ...


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As pointed out in the comments, the 'machine' in the movie Contact is just the gimbal part without the gyroscope. The point of this gimbal is to allow the thing being held at the center to be completely free to rotate in all directions. So, if they are working well they have no effect on the system except in the translational degrees of freedom, and they ...


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1) 'Gravity is not pulling x down' is a rather confusing way to think about it (as it's always there), but you are right. What's happening is the cross-product, which requires two vectors as an argument. The result is a vector that is perpendicular to both initial vectors. Of course being perpendicular to both still leaves two directions (check it ...


2

You'll need to find the individual moment of inertias of the wheels. If they are hollow cylindrical shells with negligible thickness then you can take their moment of inertia about the rotational axis as $mR^2$ where $m$ is the mass of the wheel and $R$ is the radius. Now, as fibonatic mentioned, you will have to assume that the wheel does not slip and ...


2

You can calculate it, but some assumptions would have to be made. Namely the wheel does not slip on the surface, air friction can be neglected and the wheel/surface does not deform big inelastic deformations (which would also dissipate energy). The result depends on de (mass) moment of inertia:


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The translational kinetic energy is simply $\frac{1}{2} m v^2$ where $v$ is the velocity of the center of mass. Rotational kinetic energy is $E_r = \frac{1}{2} I \omega^2$. To solve the problem, we must write the velocity of the rod as function of $\omega$ (or vice versa). Consider the above image. (Note that my convention for $\theta$ is different from ...


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The kinetic enegy at $t=0$ is equal to the friction work done when the wheel has stopped: $E_{kin}=W_f$ that is $1/2J \omega_0^2=F_fs=mg \mu 2 \pi r n $ (with $J=mk^2$) where $r$ is the radius of the bearing bore an n the number of revolutions. Solving for the revolutions gives: $n= \frac{J \omega_0^2}{4 \pi m g \mu r}$ The angular acceleration due to ...


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My question is since both pivots that we have chosen are accelerating, why are not fictitious forces considered? Pivot is accelerating if it is a geometric point defined as the point of contact. Pivot is not accelerating if it is the material point of the small ball; it stands still and has zero acceleration. The description resulting in the equation of ...


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A vector quantity has a direction and a magnitude. The direction of the rotational velocity vector is along the axis around which a body is rotating. Its magnitude is the rotational frequency which tells you how fast the body is rotating.


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There exists indeed a subtle difference, and are often referred to as angular frequency and angular velocity. Both are $\omega$ and have units $\text{s}^{-1}$. The difference lies in the fact that angular frequency is the magnitude of angular velocity, and is hence a scalar-, instead of vector quantity. The direction of the vector is perpendicular to the ...


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A general orientation can be defined by three mutually perpendicular unit vectors. Actually, only two are needed, as the third is derived by a cross product. Two vectors have 6 parameters in general, and to make them unit vectors, you need 4 parameters. Since the vectors have to be perpendicular this reduces the independent parameters to 3. Euler angles is ...


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The body frame is defined by the principal axes of rotation, which are three unit vectors in $\mathbb{R}^3$. If restricted to rotations around the $x$, $y$ and $z$ axes, it takes (in general) one rotation around each axis to align the original three axes to the principal axes of rotation. For example, one can first choose the $x$ axis and use two rotations ...


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If you use ampere's law, you will find that the magnetic field inside the rotating cylinder is uniform. With some fiddling, you will get that the magnetic field is equivalent to a solenoid with an infinite number of loops per length with a $\sigma r \Omega \mathrm{d}l$ current in each loop (a differential amount of current in each loop).


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It cam be simply said that if thumb from right hand rule points outward its +ve and if it points inwards the plane of paper its -ve... direction be k


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It clearly states that "...the body frame co-rotates with the body...". So what you are calling $L'$ is in fact $L$ itself. The torque will be the same in both reference frames.


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The thing about the force exerted by our bodies when muscles contract is that it strongly depends on where the muscle attaches to the joint. Consider the left side of the following image: Notice that the muscle does not attach to the hinge directly, but rather is attached slightly below it. When the muscle contracts, the lower bone swings upwards, similar ...



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