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First thing you need to do is convert the spherical coordinates into cartesian and then derive a 3×3 rotation matrix from the axis angle information. You can manipulate the rotation matrix using elementary rotations and then get the axis angle for the final orientation. Then you convert the axis angle to spherical coordinates for use in your system. Here ...


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As the ring moves forward, the string unwinds from it. When the ring has completed one revolution, every point on it has moved forward by the distance of its circumference. The string has unwound by an amount equal to the ring's circumference. So while the centre of the ring has moved forward by one ring circumference, the end of the string (where the ...


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Okay, so I figured it out myself. Here's what I think: Take $P$ to be the point on top of the ring, where the string is attached. Now, two things contribute to $P$'s acceleration : the acceleration of the centre of mass of the ring, and the acceleration due to the angular motion. So, $a_P = a_{ring} + \alpha \times R$, where $\alpha$ is the angular ...


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At the instantaneous moment shown in the diagram, we can write: $$2R\alpha_{ring}=a_{disc}$$ as both are in pure rolling. This also tells us that the point on the ring where the thread is attached has an acceleration $=2R\alpha_{ring}=2a_{ring}$ so we find that: $$a_{disc}=2a_{ring}$$ Note that when the string moves to another position this will not be true, ...


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Newtons 2nd law only holds in an inertial frame. If you use the center of mass of the rolling object as your frame of reference then the frame will be accelerating and F=MA won't hold. However if you fix the frame of reference to a relatively stationary point, say the surface of the earth, it will. Even in this case the frame is not truly inertial, but ...


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According to law of conservation of angular momentum by increasing moment of inertia angular velocity should decrease. As given by Iw=constant


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As the puck is moving in uniform circular motion the centripetal force is normal to the velocity vector, no work is produced and the total energy of the puck, which is kinetic=$(1/2)mv^{2}$, remains constant. Pulling the string, even slowly, the string tension becomes oblique to the velocity vector during the transition, work is produced transfered to the ...


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In your scenario, angular momentum $m v r$ is preserved (because your pulling force is radial, with no tangential component). So if you reduce $r$ by half, $v$ must double, and since $\omega = v/r$, it increases by a factor of four. Note this means in a small amount of time that the area swept out by the string is proportional to $v$ and $r$. Since they ...


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$\sum t=dL/dt$ on an inertial system, where L is the angular momentum then if there is a torque there will be a change on the L. In this case L changes direction and because of that trajectory is in approximation a circle. We can say there is a total torque, because of it there is a change on L. If there is a resultant torque, there will be a change on L ...


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The only thing I've figured out, is that the object will rotate but not moving, if the center of mass is on the line defined by the $A$ point, and the normal vector of $\vec{F⃗}$. This is not necessarily true. Look at the free body diagram. Decompose $F$ into $x$ and $y$ components. Newton now tells us that: $$ma_x=\Sigma F_x$$ $$ma_y=\Sigma F_y$$ ...


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The laws describing the movement of the body are: $$\sum \vec F=m\vec a_{cm},$$ and $$\sum \vec \tau= \frac{d\vec L}{dt}$$ wehre $\vec F$ is the external force, $\vec \tau$ the external torque, $\vec L$ the angular momentum and $\vec a_{cm}$ is the acceleration of the the center of mass. As you can see from the first formula, the center of mass will always ...


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The concept of angular momentum only makes sense when we specify a rotation axis. So we will pick the axis passing through the initial center of rotation. When the string is cut, the point mass has a linear momentum $p = mv = mr\omega$. One can define the angular momentum of a particle about an axis as $\vec{L} = \vec{d} \times \vec{p}$ where $\vec{d}$ ...


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I think that what your teacher has told you is that the angular momentum of a body can be split into two components: The spin angular momentum which is an intrinsic property of the body and is independent of the point about which you wish to find the angular momentum. $L_{\text{spin}} = I_{\text{cm}} \omega = \frac v r $ where $I_{\text{cm}}$ is the ...


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If the incline is frictionless then there's only gravity and normal force. That's it. None of these alter the rotation (since they both act through the rotational centre). So the rotation will never stop! the force that causes the ball to come to rest When you say this you mean that the translational motion comes to a rest, only. Because the ...


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Let's say you through a base ball at 90 % the speed of light. Atoms act differently at higher speeds. The atoms stop moving to the side of the ball and start penetrating it emitting radiation, heat and light. Then at higher speeds the atoms ionize losing the bond with other atoms the the ball and then they break or split causing a nuclear explosion. Even on ...


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I assumed that the angular velocity (and so $α$) is the same it I take as pivot the center of mass or the point $O$. If this is the case than parallel axis theorem can be used You are mixing rotation and circular translation. Angular velocity is defined with respect to the (instantaneous) axis of rotation, which you cannot choose at will. It is ...


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The equation of motion $$ \text{torque about stationary geometrical point O} = \text{moment of inertia w.r.t. O} \times \text{angular acceleration w.r.t. O} $$ is valid only if the motion of the body is planar rotation around an axis that passes through O. This is the case if the point O is taken to be point of contact of the body when rolling without ...


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There is an exact solution to the problem. You can consider each point on the sphere by specifying two parameters ($r$ and $\theta$). Now since you know each points velocity vector you can calculate its contribution to the angular momentum by using $dL$ = $dm$($r$ x $v$). Now integrate over $\theta$(0 to $2\pi$) and $r$(0 to R). After doing all this, you ...


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The important thing is the difference in tension between the ropes on either side of the pulley. So you will have $(T_2-T_1) R = I_{\text {pulley}}\alpha$ where $T_1$ and $T_2$ are the tensions on either side of the pulley, $R$ is the radius of the pulley, $I_{\text {pulley}}$ the moment of inertia of the pulley and $\alpha$ the angular acceleration of the ...


3

As an unbalanced force, $\mathbf{f}$ acts to accelerate the disk. Since it is located at the bottom of the disk, O must accelerate as well and is therefore in a non-inertial frame of reference. That non-inertial frame will have a fictitious forces appear that oppose acceleration. We can draw a force $\mathbf{f'}$ that acts through the center of mass in ...


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As friction takes place, it implies that the center of mass is decelerating. To simply apply the Newton's law, one has to stay in the inertial frame. Alternatively, if one discusses the problem in a non-inertial frame of reference, some inertial force $\vec{f}_{int}$ should be properly added to the analysis to restore the law's validity. In the following, ...


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When doing this sort of problem you can add two forces acting at the centre of mass whose resultant is zero. This system of three forces can now be viewed in the following way. The frictional force $f$ is exactly equivalent to a force of the same magnitude whose line of action passes through the centre of mass of the disc (shown in blue) and a pair forces ...


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Use conservation of angular momentum relative to the center of the ring. Since both the ring and the bug were initially at rest, the angular momentum of the system about the center of the ring was zero. Since no external torque is acting on the system, angular momentum of the system must remain zero even when the bug starts moving along the rim. How should ...


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Let's say the roundabout is spinning at a constant angular velocity $\omega$. An object remaining at a particular point on the roundabout a distance $r$ away from the centre of the roundabout will experience a centripetal acceleration of $\omega^2 r$, which means the required centripetal force to enable this acceleration is $m\omega^2 r$. Since $\omega$ is ...


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The centre of rotation of a rigid body is actually rather poorly defined. One sensible definition (indeed, probably the one you want) is to pick the point that has zero velocity once you subtract the mean motion of the object. However, this is only a unique point once you pick a frame of reference (e.g., with respect to the "immobile" ground). If you pick a ...


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What you are referring to is a special case of the Ehrenfest Paradox In its original formulation as presented by Paul Ehrenfest 1909 in relation to the concept of Born rigidity within special relativity,1 it discusses an ideally rigid cylinder that is made to rotate about its axis of symmetry. The radius R as seen in the laboratory frame is ...


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The Lagrangian is in fact an equation of $\vec \Omega$, however, in general it will be a quadratic function of $\vec \Omega$, as the rotational kinetic energy would be given by $$\frac{1}{2}\vec \Omega ^T \mathbf{I}\ \vec \Omega$$ This would give you the desired generalized momenta as a function of the general velocity vectors, as the diagonal entries of the ...


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The 3×3 mass moment of inertia matrix only conveys information about the orientation of the principal axes and not their location. Their location is by definition on the center of mass. It is true that if rotating about an axis parallel to one of the principal axes, but not about the center of mass the MMOI matrix is world coordinates can still be diagonal ...


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Before light speed is achieved the ball made from anything would fly apart, but lets say you could instantaneously spin it to light speed. It fly apart at the speed of light then the balls atoms would not have time to move out of the way of other atoms and particles found everywhere around the ball (even in the void of space) and would cause a nuclear ...


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If you study the problem in the frame of reference of the CM, the contact force of the plane or the rope does work, because the point of contact is moving in this frame. However, note this frame is not galilean, so you may not blindly apply the work-energy theorem (in this exact case it is valid, but take the good habit not to apply it in non-galilean ...


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When youare talking about angular momentum about any axis(inluding one that passes through center of mass), you carry out the "formula":$\vec{L}=\vec{L_{com}}+I_{com}\vec{\omega}$ where direction of $\vec{L_{com}}$ and $I_{com}\vec{\omega}$ is to be kept in mind, during vector summation.While we calculate $\vec{L}$, about COM, $\vec{L_{com}}$ becomes zero ...


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When you talk about a rigid body the distance between any two points is fixed. That implies angular velocity w is same for all points. Now ony moment of inertia changes: I=Icm+md^2 [d is distance between cm and the point] This is the parallel axis theorem. Get I and angular momentum=Iw


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The angular momentum will be different: however, you will be able to calculate it with the parallel axis theorem. Measure the distance from your new axis to the center of mass and call it $d$. Your new rotational inertia $I$ can be calculated from the rotational inertia around the center of mass $I_{\mathrm{cm}}$ using the formula: $$I = I_{\mathrm{cm}} + ...


2

The rotation will not necessarily be parallel to the ground. The general motion will be a combination of rotation in a horizontal plane (the conical pendulum) and oscillation in a vertical plane (the simple pendulum). If the support (pivot) is a fixed point, the motion you get depends on the starting conditions. If you launch the mass horizontally at the ...


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Any force whose line of action is not through the centre of mass of a body can be transformed into the same magnitude and direction force acting through the centre of mass and a couple. A couple (two parallel, non-colinear, equal in magnitude but opposite in direction force) has the special properties that its torque is independent of the point about which ...


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Because the system is in equilibrium only in the CM frame. If you calculate the moment about any other point, you need to consider pseudo force as well, viz., 133.4N pointing to the left passing through the CM. Then you will get the correct answer again. If you do not want to work in the CM frame, then you should notice that the angular momentum is $${\bf ...


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As you mentioned, $\vec {L_o}_⊥$ is proportional to $\omega$. So the middle term means $\vec {L_o}_⊥$ varies by varying angular velocity. You can also derive this. formula like this: $$\frac{d \vec {L_o}_⊥ }{dt}= \frac {d\vec{\omega}}{dt}A \hat{L_o}_⊥+ \vec{\omega}A \frac{d\hat{L_o}_⊥}{dt}= \frac{1}{\omega} \frac{d \omega }{dt} \vec{L_o}_⊥ + \vec{ \omega } ...


2

Yes, there are more unknowns than equations. You do not have sufficient information to solve for the requested quantities. Someone might be playing a prank on you! In reality, each ball and each paddle would have a specific finite stiffness, and one could use this information along with some clever math to determine the final velocities of all the bodies. ...


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Let the body rotate about the $z$-axis, then by the definition of angular momentum $$\vec{L}=\vec{\omega} I_z.$$ where $\omega$ is the angular velocity about the $z$-axis. So we could take the parallel axis theorem and multiply it by $\omega$: $$\vec{\omega}I_{z}=\vec{\omega}I_{cm}+\vec{\omega}ma^2$$ Now ponder the terms in it. If I understand the ...


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You forgot to include angular momentum of center of mass. Thus add $MR^2\omega$ to your answer By the way angular momentum is not $I\omega$, it is $\vec{l_{com}}+I_{com}\vec{\omega}$ It is $I_{contact}\omega$ only for cases when point of contact is the point of instantaneous or fixed axis of rotation.


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I think that I understand what you are asking and I think that it is a good question. If the centre of mass of the disc and the point mass moves a distance $x$ then the work done by the force is $Fx$ in both cases and yet the disc has gained some extra kinetic energy because it is rotating. If the centre of mass of the disc moves a distance $x$ the point ...


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There's one thing in you question that doesn't quite add up and that is when you say exert a horizontal force at a certain distance r from the center of mass. If you are pushing a disk anywhere you push it (on the curved surface at least) is going to be a distance r from the centre of mass because the whole surface is a distance r from the centre of mass. ...


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Everything you have derived is correct. The reason for your perceived paradox is, I believe, a confusion between force and power. The same force can produce more power if it is being exerted at a greater velocity. When you exert a force at a radius r from the CM, the point of application of the force will accelerate more quickly than the CM, allowing the ...


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The weight of the person on the swing is causing gravity to pull them down you could swing in a full circle if you had enough force


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Without having any experimental data at hand, I guess that most planets formed from a uniformly rotating dust disc, and thus their rotational and orbital momentum have the same sign. However, upon random tangential impacts, some of them (Venus, Uranus..) could change their original axis of rotation, and most probably it happened so early we will not find ...


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Here is the beginning of a free body diagram. Each contact force $A$ and $B_y$ acts on the contact normal direction. For $B_y$ this is vertical and for $A$ this is perpendicular to the lever. The weight $W=m g$ acts on center of mass. Now you have to add friction to the contacts as shown with force $B_x$. Here friction is shown with a positive value ...


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When the fan starts spinning, each blade starts from rest. Newton's first law of motion states that unless a force is applied to it, the [velocity][1] of a body does not change. That property is inertia. To speed up, the blades must accelerate. Newton's second law of motion states that the force necessary for an acceleration of a body is proportional to and ...


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You mean go all the way around? It could if you had enough force to overcome gravity and like a tether ball swing all the way although most humans do not have the strength to apply the force needed to push another or them selves to a full revolution around the bar of a swing with out a jerk, but if the chain was replaced with a solid bar to prevent jerking ...


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why it doesn't it swing fully ? I think the question does point to the fact that swings which are usually available in the park does not provide free swinging , low amplitude of the swing and needs constant pushing. All the above is related to energy dissipation of the initial potential energy provided to the swing- and the dissipation is at the hinges ...


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You might be thinking in comparison to a desk or handheld electric fan. As mentioned by @Farcher, $\tau = I\alpha$. $I$, the moment of inertia of a spinning body around a particular axis of rotation, is calculated as follows: $$I = \iiint\rho(x,y,z)||r||^2\ dV$$ Or with uniform density, $$I = \rho\iiint||r||^2\ dV$$ From this formula, you can see that ...



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