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A few clarifications first: I will assume that the bigger cylinder (radius $R$) is not moving, and I will be doing the problem in this frames. I will assume that both $\theta$ and $\phi$ are positive in the clockwise direction. This is against convention but it'll make a few negative signs disappear, and you many find it easier to think this way given how ...


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To keep the balance rotating with uniform angular velocity, you must ensure that gravity (being the only force) must not do any work on any part of the system and the system as a whole. The way to do this is to keep the center of mass of the system always at the pivot throughout the motion so that torque due the balance's own weight is zero always. That is, ...


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Actually with the lever sticking up above the balance scale shown; balance axis point; it adds gravitational weight to whichever side is initially positioned downward, and also since this side is down, it should weigh more being closer to the ground... and should tend to keep this side down... However is doesn't... If you put a weight, on some point of a ...


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The problem with attempting to do the analysis with the forward point of contact on the box when it is sliding is that the box is accelerating. This makes a non-inertial frame and there's more moving parts. Besides the force of gravity on the center of mass, there will be fictitious forces. First, lets assume friction is zero. If so, we can calculate ...


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Take a look on the forces diagram in the first picture. Let's ignore the force $N$ for now, it can only contribute to the tipping (due to its orientation), but since it is going to be confined to acting only on the red point as soon as the tipping occurs (if it does), it won't have any influence at all. E.g. it produces zero torque about the red point. ...


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Comment to the question (v1): Besides what John Rennie wrote in his correct answer, note that the velocity profile of an irrotational/free vortex falls off as $$\tag{1} v~\propto~ 1/r,$$ while a galactic rotation curve may actually increase with $r$, and in any case, it never falls off faster than$^1$ $$\tag{2} v~\propto ~1/\sqrt{r},$$ so the analogy ...


1

It's part and parcel of the definitions of angular momentum and torque. Consider a system of particles, not necessarily a rigid body. Without loss of generality, we can use an inertial frame that is instantaneously co-located and co-moving with the system center of mass. The angular momentum of the system with respect to the origin of this frame is defined ...


1

The physical explanation for why torque increases with r is that the longer the lever arm is the greater angular acceleration you can cause for a given force F. If a screw is stuck because it was screwed in too hard (ie with too much torque), you need to get a longer wrench. With the longer wrench (ie, larger r_w) you can generate greater force at the edge ...


1

Why would there be no radius in torque? A real torque is a real force that acts on a real rotating body (rigid or not) at a real radius. If you look at e.g. rotating machine parts, they all have a finite diameter. That diameter is of enormous importance for the design of a part, because together with the material constants it determines just how much torque ...


2

If we take the Milky Way as an example, the black hole at the centre, Sagittarius A$^*$, has a mass of about 4 million times the Sun. However the mass of the Milky Way is somewhere around a trillion Suns. So the central black hole makes up 0.0004% of the total mass. While the central black hole may have been important in the formation of the Milky Way, its ...


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Can a ball stay still while laying on a inclined plane? In freshmen physics, the inclined plane and ball are perfect and the ball moves, so for your purposes, no. If either the surface or the ball have imperfections, we can tip the plane and the ball won't move until gravity exceeds the sum of the normal forces. To imagine those normal forces, we look ...


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This is a very late response, but there is no accepted answer as of yet, and none of the answer quite hit the mark. Regarding the magical collision hypothesis, that smacks of being rather non-scientific. Scientists as well as Missourians are wont to say, "Show me!" Other than the fact that Venus's rotation is anomalous, what, exactly, is the evidence for ...


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For the yo-yo question: what is R (path of center of mass)? Think about how a yo-yo works - there's a string attached to a center axis and wound around itself. This means that your string attached to the ceiling will be applying tension T at some distance r away from the yo-yo's center of mass. But because the string is not fixed but wound around ...


1

If the object is not spinning, then moving it will take the same effort whether the mass is spread out, or all at one point. Imagine that you are on a roundabout, but you are sitting on a disk that it itself able to spin. Initially you are facing North, and you are exactly in the middle of your disk (although it is off axis of the roundabout). Now the ...


3

General remarks. That's right. Torque is defined as $\mathbf r\times\mathbf F$ where $\mathbf r$ is the position where the force is applied, and $\mathbf F$ is the force being applied. The so-called Law of the Lever can then be derived from the following fact (which itself can be derived from Newton's Laws) about systems of particles: The net external ...


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As you say, torque can be defined as $ r \times F $. Then the law of the lever is derived simply from the condition that at static equilibrium all forces on the lever and all torques about the fulcrum sum to zero.


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As others have already pointed out, the moment of inertia depends mostly on the mass distribution. As an example, imagine two cylinders with exactly the same mass. Suppose that you cannot see what is inside them, it's covered on both ends (and the covers mass is negligible). The first is a solid cylinder. The second one is a hollow cylinder. Ka-boom! ...


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If you are just interested in the rate of precession, then $r$ just has to be the distance from the support (tip of the top) to the center of mass - this is the distance that gives rise to the torque through $mgr\sin\phi$ However, if you want to do this as a vector equation, then you have the instantaneous angular momentum (vector) of the top $I\vec\omega$ ...


3

The "missing" kinetic energy is still there ... it's now in the rotation of the yo-yo.


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For me the second case was more intuitive. The friction and gravity both induce torques of the same sign on the body. Gravity acts through the center of mass, which in this case aligns with the center of rotation. So gravity is not inducing any torque. All the torque is coming from friction. If you imagine the same scenario where the ramp is ...


0

Friction always opposes relative motion of the point of contact. To know the direction of friction assume that no friction is present and see the direction in which point of contact is moving: friction is opposite to that direction .


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If the sphere is not spinning, and the axis about which you are taking angular momentum stays through the same position in the sphere, then the relative velocity of the sphere to the axis will always be zero regardless of translational motion. Therefore, the angular momentum will be zero.


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If you are looking for quantitative values, look to further than the equations of motion $$ \begin{align} \sum_{i} \left( \vec{F_i} \right) & = m \vec{a}_{cm} \\ \sum_i \left( \vec{\tau}_i + (\vec{r}_i-\vec{r}_{cm}) \times \vec{F}_i \right) & = I_{cm} \vec{\alpha} + \vec{\omega} \times I_{cm} \vec{\omega}\end{align} $$ where the left hand side is ...


0

The angular equivalent of the impulse-momentum theorem states The change in angular momentum of a system is equal to the product of the (average) external torque time the time it is applies In math that is (finite version): $$ \langle \vec{\tau}_{ext} \rangle \,\Delta t = \Delta \vec{L} \,,$$ or (infinitesimal version): $$ \vec{\tau} \, \mathrm{d}t = ...


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I would say three degrees of freedom, and you can use, e.g., the Euler angles (http://en.wikipedia.org/wiki/Euler_angles ).



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