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To open the door with a constant angular acceleration say $\alpha$ we need a constant torque.However the force required to produce this torque varies as the perpendicular distance of line of action of force from pivot point varies. Since , $$\tau =F \times l$$ where F is Force , and l is perpendicular distance of line of action of force from pivot point. ...

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This is a recursive method, and what you have for each link is a force balance, entirely equivalent to $\sum F = F_i-F_{i+1}=m\,a_i$ but with screw quantities. Do a free body diagram for the {i} link and write out the NE equations of motion $$\boldsymbol{F}_{i}^{\{i\}} -\boldsymbol{F}_{i+1}^{\{i\}} = \left(\boldsymbol{1}-[ad_{v_i}] \right) \mathbf{G}_i \... 0 To understand gyroscopic precession intuitively, without using moment vectors, consider this orbiting satellite analogy. Watch “TheHue’s SciTech” video from 25 seconds until you are comfortable with predicting which way a gyroscope will precess when a rotational moment is acting upon it. https://www.youtube.com/watch?v=n5bKzBZ7XuM . Now consider Red Acts’s ... 2 The most general case is that of a solid rigid body attached to the end of the pendulum. The initial potential energy is converted into kinetic energy at the bottom of the swing. Kinematically the speed of the center of mass is linked to the rotation of the pendulum$$ v_f = \ell \omega_f$$The solid object has not only mass m but mass moment of inertia ... 0 Consider first a slippy surface on which a wheel is rolling. Because the surface is frictionless the wheel just rotates in place, there is no rolling. Now assume a rough surface, so there is friction and so the wheel rolls forward and it never slips. Here, the translational velocity is the velocity of the axle, as the axle is above the point of contact then ... 0 I found the solution. I leave an answer here so that anyone else who wants to know what wasn't working can read this. Every piece of the plate is subject to an apparent force, that is NOT applied on the center of mass, but on the piece itself. If this force was the same for all pieces with the same mass, then we could consider it as applied one the center of ... 1 To point 1 Does slipping mean zero angular velocity or is it just the v velocity not being equal to wr? The geometric bond v=\omega r tells the speed v of a particle a distance r from the wheel centre. Within the wheel, this equation must hold true for every particle - otherwise the wheel would break apart because particles would "skew" ... 0 Angular momentum depends on the axis chosen. If the axis is chosen at the point where the friction force acts on the rolling body, then the perpendicular distance of the friction force from the chosen axis will be zero at all times and hence the torque due to the friction force shall be zero. As no other external force is mentioned in the question, the net ... 1 Here are the answer to your doubts in order:- Slipping means there is relative motion between two surfaces. Only when v=r\omega, there is no relative motion between the surface of the rolling object and the surface on which it is rolling and hence no slipping. In other cases, when v\not=r\omega, there is relative motion between the surfaces and hence ... 2 Here are answers to your questions in order: Slipping means v \neq w r . As soon as it accelerates, v > w r  and so it will not roll past the starting position, as w will remain the same. If there's enough friction to keep it from slipping, yes. It is not conserved relative to the center of mass of the wheel if there is friction. You can use ... 2 this animation is result of simulation the equation of motion, I used Euler- Lagrange with non holonomic constraint equation (rolling condition). Ring rolling on paraboloid surface Paraboloid parameter equation$$\boldsymbol R= \begin{bmatrix} x(\lambda~,\vartheta) \\ y(\lambda~,\vartheta) \\ z(\lambda~,\vartheta) \\ \end{...

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Notation often emphasises one aspect more than another. Its often an aspect that is under consideration. For example a tangent vector can be written as: $v$ - it's position is left implicit $v_p$ - it's position is explicit And this is matched by the notation for a tangent bundle $TM$ - we are considering the whole tangent bundle $T_pM$ - we are ...

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The integral is similar to a sum, as $dx$ approaches zero. $\int x dm$ does add up to zero at the centre of mass. If we take each $dm$ element as $1$, then it's similar to $1+2+3+(-1)+(-2)+(-3) = 0$ (depending on what the $x$ and $dm$ weightings are, but the point is they cancel out). $\int x^2 dm$ is different, it's simlar to adding like this $1^2+2^2+3^2+(-... 1$\int xdm=0$by definition of center of mass, you are right. However,$\int x^2 dm\neq 0$since it is a summation of positive (or rather non-negative) terms,$x^2\geq 0$. Example: imagine two points with mass$m$located at$x=\pm1$. Their center of mass is at$x=0$. Then$\int xdm=-m+m=0$, and$\int x^2dm=m+m=2m$. 2 In this post, I found in centripetal force depends on the coordinate system where you write the equations in, but it in the solution, it was used that centripetal force is same in all frames. So, what made it that we could equate centripetal force of two different case in this case? In the cited post it says "both frames do agree on the magnitude and ... 1 It is quite similar to drawing perpendiculars to tangents of a circle in order to find its centre. ICAOR is an axis about which the body can be considered to be rotating at a given instant. From a geometrical point of view, this definition refers to a straight line about which particles on the rigid body are performing circular motion, where planes of these ... 1 Sorry for being a bit late here. What I find to be the magic part is: rotation of a body on an axis can always translated to rotation on a different axis + a linear motion. Let's consider your case, viewed from above, before the split: You can see the linear velocity vectors on multiple points as well as the CM. Now, after the split, the vectors are the ... 0 So, there should be some force acting radially out to cause static friction acting radially in, but even in that case sum of forces acting in radial direction is zero since static friction balances outer force trying to move us. If so, there is no net force acting radially in to allow rotation. Where have I gone wrong? There is a force acting radially out. ... 0 We need three relations for this: $$v=\omega r\qquad\qquad a_c=\frac{v^2}{r} \qquad\qquad F_c=ma_c$$ The first is the geometric bond that ties angular velocity$\omega$(which you want to keep constant) to speed$v$and string length$r$. The second is the tie between this speed and both string length and centripetal acceleration$a_c. The third is Newton's ... 0 There is no requirement for the centripetal force to be the same in both cases. I think the problem wants you to assume that the clown wants to exert the same force before and after for the simple reason that the clown just wants it so -- perhaps his arm is not strong enough or the string will break if the force gets any bigger. 0 Short and simple: friction opposes relative motion between surfaces. You do not need an opposing force for there to be friction. As a simpler example, consider the case of a block stacked on another block. You gently push on only the bottom block so that the system starts moving as a whole. The static friction acting on the top block is the only force acting ... 0 With the merry-go-round rotating at constant angular velocity, for an object fixed at the edge of the merry-go-round, in the inertial frame (observer on the ground) the object is accelerating due to the change in its velocity vector even though its speed is constant (acceleration is a vector). The force that causes the acceleration is radially inward and is ... 2 Friction is the centripetal force in this scenario and aside from gravity it is the only force acting on the person on the Merry-go-round. In an inertial reference frame (say you're standing on the ground looking at the Merry-go-round) and you are staring at person B on the Merry-go-round, you will see Person B moving in a circular motion. From Newton's ... 0 Although the OP states that the 'surface is fixed', it is tempting to imagine, for instance, that the blue surface is a relatively small flat surface with the topmost part located at the earth's north pole (corrected from the related comment mentioning south pole) so that the initial velocity causes the rightward acceleration due to the Coriolis effect (... 0 Since the plane is horizontal and the rotation axis is vertical, the point of contact between ball and plane is on the axis of rotation. Supposing an ideal point contact (which is of course not realistic), there would not be a relative velocity between ball and plane in that point. Hence, there would not be any cause for the ball to move sideways. \vec v=\... 1 Is the surface rough or smooth? If smooth, then friction is not present and plays no role - and I doubt you would see such skewing off from the initial direction. Such a situation would be akin to gravity-free motion through empty space; no rotation could alter the linear direction in such a scenario. (Unless we are dealing with e.g. the Coriolis effect - ... -1 The friction force on an object which is sitting on a rotating turntable acts toward the center of rotation and provides the required centripetal acceleration. 2 The center of mass will follow the same trajectory regardless of the internal forces and motions of an object (as long as the external forces applied to the object don't change as the object changes). Even if an object explodes mid-trajectory, the CM will still follow the original trajectory (although in practice for explosions, the air resistance will ... 0 The inner cylinder doesn't affect the outer one because of the lack of friction. The only force on the outer cylinder is the friction with the inclined floor and the force due to gravity, so if floor-friction is large enough, the outer cylinder undergoes pure rolling motion. Similarly, the outer cylinder doesn't affect the inner one. The only force on it is ... 1 First, let's comment on I know sliding is when the Vcm goes faster than the point of contact on the ground. Actually, sliding is when the point of contact on the ground has any speed (let's call this Vb from "bottom" except 0. The cases are: a. Vcm=Vb: only sliding b. Vcm>0, Vb=0: only rolling (let's call this perfect rolling) c. Vcm>0, ... 0 I got it solved , I was actually conserving Angular momentum wrt Centre of Mass before and after Collison but the point is , Centre of Mass is actually accelerated between collision so it was an inertial frame of refrence only before and after Collison and not during the collision ! 1 It depends on what v is. When v is the velocity of a particle as viewed by an inertial observer then yes where v=0 on the plane is where the axis of rotation goes through (in 3D it is a bit more complex because you can have translation parallel to the axis of rotation also). Also note that the axis of rotation is instanteneous which means it can move ... 1 I have always thought that if any point on a rolling body has v=0, then it must be the instantaneous axis of rotation. This isn't correct. Consider a cylinder rolling without slipping on a level surface amd moving with constant translational velocity v. This means that the axis of rotation must be moving with translational velocity v. It also means ... 0 I'm guessing you are referring to a problem involving conservation of momentum. Most of the times the required data are given. If the problem states that you simply drop the block on to the disc, then you have to take it's velocity as zero because it doesn't have any angular velocity at first. But just as when that block comes into contact with the disc, it ... 2 The rotation rate in the equation is the rotational velocity of the body relative to the non-rotating world as measured in a non-rotating frame instantaneously coincident and aligned with the body's frame. I've recently posted a more detailed description of how this equation works on the Robotics Stack Exchange site, https://robotics.stackexchange.com/a/... 0 The reason is that the period will be dependendent on the radius of curvature R but not on the length of the hoops, provided the oscillations are small and the hoop hangs at its center of curvature (i.e. with the same amount on mass on the two sides). To prove this, we will go step by step: first we consider a full ring and solve its motion, then we ... 0 You can use the formula work = force \times distance to imagine it intuitively. When the pushing force is close to the pivot, the force moves through a small distance (for a given angle of rotation), the work done is low, so the kinetic energy gained by the door, related to the speed, is low. When pushing with the same force further from the pivot, the ... 2 Imagine a 100 metres wide door. It'll take a loooong time to close this door via pushing at the edge because you'll have to run ~300 metres. Your force only moves the door a small but for each second, which is why it is easier to do. But also slower. If the door is not very heavy (just styrofoam e.g.), you might be able to close the door much faster by ... 0 I think this is the situation\vec{M}_O = \vec{M}_{\rm ext} + \vec{c} \times \vec{W} = \vec{M}_{\rm ext} + \pmatrix{ -\tfrac{a}{3} m g\\ 0 \\ 0}$Here$\vec{c} = \pmatrix{0 \\ \tfrac{a}{3} \\ \tfrac{a}{3} } $is the center of mass relative to O, and$\vec{W} = \pmatrix{0 \\ 0 \\ -m g}$the weight acting through the center of mass. The mass moment of ... 1 My result is equivalent to that from JalfredP, but I considered just a segment of the hoop measured to an angle$θ_m$on either side of the top position. I'm looking for the angular frequency of oscillation:$ω^2$= mgL/I where L is the distance from the pivot point to the center of gravity, and I is the rotational inertia about the pivot point. For the ... 3 I was in the middle of writing an answer when I saw 2 perfectly good answers, but I think at least one important point remains unanswered. Also, just in case you don't know about vector products and/or Lagrangians, here goes my explanation. Angular momentum is not just any combination of coordinates and velocities. It's the one that doesn't change when the ... 4 Historically, this was mostly derived from observation (for example, conservation of angular momentum appears straightforwardly in Kepler's 2nd Law for the motion of planets). Nowadays, however, conservation laws (of momentum, angular momentum, energy…) can be found directly from symmetries of the problem, through Noether's wonderful Theorem. Simply put, ... 6 Using Newton's law one can easily check that$\vec L= \vec r \times \vec p$is a conserved vector for a motion of a particle in a spherically symmetric potential (while axial symmetry implies conservation of one component of$\vec L$). If you try to modify this definition, you will not obtain a conserved quantity. If you are familiar with analytical ... 1 So the instant of release the problem changes and is now just a free-moving body with certain initial conditions. Just before the release, the rod has rotational velocity equaling the orbital rotation, linear (tangential) velocity. After the release, there are no external forces/torques acting so the body continues with the center of mass on a straight line ... 5 The string is terminated by a small loop around the axle. The loop is just a bit larger than the diameter of the axle so that, when the yoyo is fully unwound and spinning rapidly, the yoyo can continue spinning with its axle sliding on the lower part of the small loop. The tension due to the weight of the yoyo stops the string being swept around the axle. ... 0 Problems like these are always solved with the following steps Description - Decide which DOF parameters to use to fully describe all the configurations of the system. In this case DOF=1 and the angle$\theta$suffices. Kinematics - Track the position of the center of mass as a function of the DOF parameters and evaluate position, velocity and acceleration ... 0 The forces on the axle (from the pendulum) must be equal and opposite to the forces on the pendulum from the axle. I would start with the radial and tangential forces. Radial: T – mg cos(θ) = m(L/2)$ω^2$. (The tension,T, acts on the axle and the center of mass.) Tangential: F(L/2) = Iα. Where F is the force (from the axle and perpendicular to T) which ... 1 I have little experience with dental drills (although I have indeed just come from the dentist), but I have written a few milling programs on a CNC mill, so that I think: I can answer this one! It's not entirely clear to me whether milling or grinding is generally used in dental treatment. But the considerations are similar. Let's start with milling: A ... 1 Teeth are brittle, which I painfully know since a school "mate" broke my incisor off during a fight (sudden floor contact). Hence, trying to grind big chunks out of it might cause more damage than intended (e.g. invisible cracks that cause caries years later). It is because bigger chunks require higher forces, no more complicated than that. ... 1 Dental enamel is a hard material so the drill needs to exert a large force on it in order to break it. The force is proportional to both the drill radius (r) and it angular frequency ($\omega$, rpm), simply because the speed of the drill material at the interface is$v=\omega r$. So you could go for large$r$and moderate$\omega\$ as for concrete. However ...

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