I was doing this problem for self study in rotational dynamics:     Two discs are mounted on thin, lightweight rods oriented through their centers and normal to the discs. These axles are constrained to be vertical at all times, and the discs can pivot frictionlessly on the rods. The discs have identical thickness and are made of the same material, but have differing radii $p$ and $q$. The discs are given angular velocities of magnitudes $x$ and $y$, respectively, and brought into contact at their edges. After the discs interact via friction it is found that both discs come exactly to a halt. Which of the following must hold? Ignore effects associated with the vertical rods.   Answer choices:   $(x^2)p = (y^2)q$ $xp = yq$ $x(p^2) = y(q^2)$ $x(p^3) = y(q^3)$ $x(p^4) = y(p^4)$   I got the answer $x(p^4) = y(p^4)$ by setting the angular momentum of the two discs equal to each other (because of conservation of momentum) and solving, but the answer is $x(p^3) = y(q^3)$. How do I solve this problem?
123572
Problem #25 from the 2009 $F=ma$ contest:     Two discs are mounted on thin, lightweight rods oriented through their centers and normal to the discs. These axles are constrained to be vertical at all times, and the discs can pivot frictionlessly on the rods. The discs have identical thickness and are made of the same material, but have differing radii $r_1$ and $r_2$. The discs are given angular velocities of magnitudes $\omega_1$ and $\omega_2$, respectively, and brought into contact at their edges. After the discs interact via friction it is found that both discs come exactly to a halt. Which of the following must hold? [...] Ignore effects associated with the vertical rods.   Diagram here:     Probably conservation of angular momentum is useful here, but I am not clear how it can be applied. There are two distinct axes of rotation, one for each disc, so we cannot talk about conservation of angular momentum about one axis. If the two disks rotated about the same axes then the solution is straightforward, but this is not the case.
36565
If I take r as the radial vector of the moving object and v as the velocity vector of the moving object in central force field.   Then r should be perpendicular to r×v . So that depicts that r.(r×v) = 0 . So how does this say that the particle is lying on a constant plane ?
65676
Consider a cylinder that rolls without sliding on an inclined plane. If it's placed at the top of the plane, with its center of mass at a height $h$ from the bottom, it will have a potential energy $mgh$ (considering the bottom of the plane as the zero point). Then, the cylinder will start rolling down the plane, as its weight does work on it, causing its potential energy to be transformed into rotational and translational kinetic energy. I'm confused about the origin of the rotational kinetic energy of the cylinder. Since the static friction and the normal force are applied at the point of contact with the plane, neither of them can do work on the cylinder (which is why its mechanical energy is conserved), since the cylinder is, by constraint, not sliding. This seems to imply that the cylinder's rotational kinetic energy comes from the work done by its weight. However, I don't understand how this happens, since a rigid body's weight can be seen as being applied on its center of mass, which means it can exert no torque on it. Instead, the only torque exerted with respect to its center of mass comes from static friction. Moreover, if the inclined plane was frictionless, there would be no torque exerted on the cylinder, so it would not rotate. Yet, I don't see how this situation could be distinguished from the previous one by only looking at the cylinder's mechanical energy (since the normal force would still be doing no work). So, my question is: how is gravitational potential energy converted into rotational kinetic energy, and what is friction's role on this?
72400
A cylinder at rest lying on a rough ground is given an impulse which imparts a translational velocity (no angular velocity) to it. The question goes ahead with finding time after which rolling starts etc. etc.  But I have another question: The friction from the ground will decrease the translational velocity but increase the angular velocity (as the torque of friction will be in the direction of rolling). After some time, the translational velocity will become equal to angular velocity and hence pure rolling will start.  After this time work done by friction will be 0 as the point of contact is not moving and hence translational velocity will be constant. Now my doubt is that, when translational velocity is constant, it implies angular velocity will also be constant (as translational velocity is a multiple of angular velocity). But the torque from friction (about the centre of mass) will not be zero which means angular velocity will not be constant. In some book where there was the same problem it was written friction will vanish. How can friction vanish? How can anguluar velocity be constant even when there is a torque by friction?
28276
I have searched everywhere on the web for a way to do this, but I can’t find anything. Someone have any ideas for doing this without using physical simulations? The operation should be this: the red wire passes through the blue object (which is static) and is anchored to the yellow object, which can rotate When the yellow object rotates, the wire wraps around the cylinder in a similar way to a fishing rod.
88781
They say that for a rolling body, the velocity of the contact point is zero. I'm not getting this. How can it be zero when it's in continuous motion?
55609
In a situation where a disk is rolling WITH slipping on the ground i.e velocity of centre of mass is greater than $r\omega$, is angular momentum conserved about a point on the ground.  What confuses me is that friction decelerates the disk to make $v_{com} = r\omega$ and in this process some velocity is lost so according to formula of angular momentum $L=mvr$,about a point on the ground $L$ decreases as $v$ decreases.  But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change.  edit:this question is about applying conservation of angular momentum in rolling with slipping.
39374
When you spin an asymmetrically weighted, 2D disk-shaped top, the heavy part actually rises to the top. Why is this?        In the second video it shows that this is because of friction but trying this on a frictionless surface won't do much: The disk will go up and down and still the heavy part will be on the top.
82212
Consider two blocks, one on top of the other on a frictionless table, with masses $m_1$ and $m_2$ respectively. There is appreciable friction between the blocks, with coefficients $\mu_s$ and $\mu_k$ for static and kinetic respectively. I'm considering the fairly routine problem of determining the maximum horizontal force $F$ (say, to the right) that can be applied to the top block so that the two blocks accelerate together.  The problem is not hard to solve symbolically. If the two blocks move together, their accelerations are the same, and the top block doesn't move with respect to the bottom block, so only static friction is in play. In a standard coordinate system (with $x$ oriented to the right), the sum of horizontal forces for the top block is  $$F-F_{sf}=m_1a$$  and for the bottom block  $$F_{sf}=m_2a$$  where $F_{sf}$ is the force of static friction. Solving for $a$ in these two expressions, and then equating them, gives  $$F=\frac{(m_1+m_2)F_{sf}}{m_2}$$  The maximum such force will therefore be achieved when $F_{sf}$ is maxed out at $\mu_s m_1g$, so  $$F_{max}=\frac{m_1}{m_2}\mu_s(m_1+m_2)g$$  I understand this solution, but conceptually I don't have a response to the following nagging question: $F_{max}$ is clearly larger than the max static friction force $\mu_sm_1g$ (because $\frac{m_1+m_2}{m_2}&gt;1$), so why doesn't the application of a force of magnitude $F_{max}$ to the top block cause kinetic friction to take over? This line of reasoning would suggest that applying a force $F$ of magnitude greater than $\mu_sm_1g$ would cause the top block to start moving with respect to the bottom block (in which case the blocks no longer accelerate together, as in the above solution). I'm at a loss, conceptually, to say what's wrong here. I suspect it has something to do with being careful about reference frames, but a clear explanation would be much appreciated.
89384
I recently wondered whether there is a simple proof that circular wheels maximise mechanical efficiency. By this I mean:     Show that for a wheel with a given width and cross-sectional area, the   circular wheel reaches the bottom of an incline faster than a wheel   with any other shape.   This sounds like a rather elementary problem in classical mechanics but I haven't found a proof of this fact in any of the classical mechanics texts that I have read.   Note 1: I assume that the distribution of mass is uniform across the wheel so comparable wheels will have equal mass.   Note 2: Dry friction is assumed to be present.
67162
The following quote comes from:      Consider an ice skater that has just gone into a spin with arms   stretched out.  If it helps, the skater is wearing lead bracelets on   each wrist.  As the skater spins, the angular momentum and angular   kinetic energy are constant;  friction with the air and with the ice   can be eliminated.  The skater must exert an inward centripetal force   on the bracelets to keep them rotating in the same circle, but this   force does no work, since the bracelets are not changing their radius   of rotation.      Then the skater pulls in his arms and the bracelets, closer to his   axis of rotation.      Several interrelated things happen:           since there is no external torque, the angular momentum of the system stays the same;   Since the various masses of the skater are all moving towards the axis of rotation, the total moment of inertia of the system   decreases;     Combining 1 and 2, the angular velocity of the system increases;  if the moment is halved, the angular velocity doubles   Combining 2 and 3, with some quantitative treatment, the angular kinetic energy increases;  if the moment is halved and the   angular velocity is doubled, the kinetic energy doubles;   The skater does work;  his arms are now exerting an inward force through a distance as his arms draw the bracelets inward;  the work done can be rigorously shown to be identical to the increase in   kinetic energy...      However, it only covers a case when work is done to pull arms inwards. It is understandable that the work would be done in such case.  But what is happening in the opposite case when the skater has their arms inwards and lets them go outwards? S/he doesn't do any work but kinetic energy (and total energy) is lost. Why is it lost?  E.g.  I = 2kg * (1m)^2 w = 3.14/s L = 2kg * m^2 * 3.14/s Ek = 9.869  After shifting 2kg by 0.1m outwards: I = 2kg * (1.1m)^2 = 2.42 kg * m^2 L = 2 * 3.14 = 2.42 * w w = 2.5963 Ek = 8.1566   So, where did 1.7124 J go?  The quoted answer describes the easy case and omits the hard one.
42352
I have been told [Warning: I leave this because it's what I asked and allows to understand the dialogues in the comments, but Azad, whom I thank, has pointed that the formula does not hold in general in the form it is expressed] that the angular momentum of and rigid body with respect to any point $P$ can always be expressed as $$\mathbf{L}_{P}=\mathbf{r}_{cm}\times M\mathbf{v}_{cm}+\big(\sum_im_iR_i^2\big)\boldsymbol{\omega}$$ where $\mathbf{r}_{cm}$ is the position of the centre of mass with respect to $P$, $M$ the mass of the body, $R_i$ the distance of the $i$-th point, having mass $m_i$, composing the body, and $\sum_im_iR_i^2=I$ its moment of inertia with respect to the instantaneous axis of the rotation around the centre of mass of angular velocity $\boldsymbol{\omega}$.  I know that the velocity $\mathbf{v}_i$ of each point $P_i$, having mass $m_i$, of a rigid body of mass $M$ can be see as the sum of a translation velocity of one of its points $C$ plus a rotation velocity around that point: $\mathbf{v}_i=\mathbf{v}_{C}+\boldsymbol{\omega}\times\overrightarrow{CP_i}$. If we chose $C$ as the centre of mass I see that $$\mathbf{L}_{cm}=\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_{i}=\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_{cm}+\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})$$$$=\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i}) $$because, if I am not wrong, $\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_C=(\sum_i m_i\overrightarrow{CP_i})\times\mathbf{v}_C=\mathbf{0}$ since $\sum_i m_i\overrightarrow{CP_i}$ is the position of the centre of mass with respect to itself, which is $\mathbf{0}$.  How can it be proved that $\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})=(\sum_im_iR_i^2)\boldsymbol{\omega}$? I have searched a lot on the Internet and on books, but I find nothing. To give some background of mine, I have studied nothing of analytical mechanics. I find the formula very, very interesting both in itself and because, if the moment of inertia does not depend upon time, $\forall t\quad I(t)= I(t_0)$, the above expression can be differentiated to get the formula of the resultant torque with respect to the centre of mass $\sum\boldsymbol{\tau}_{cm}=\frac{d\mathbf{L}_{cm}}{dt}=I\boldsymbol{\alpha}_{cm}$ where $\boldsymbol{\alpha}$ is the angular acceleration around the centre of mass. I heartily thank you for any answer!    Some unfruitful trials: by using the "BAC CAB identity" as suggested by Azad, whom I heartily thank, $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=(\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c}$, I can see that$$\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})=\sum_im_i\|\overrightarrow{CP_i}\|^2\boldsymbol{\omega}-m_i(\overrightarrow{CP_i}\cdot\boldsymbol{\omega})\overrightarrow{CP_i}$$which, by decomposing every $\overrightarrow{CP_i}$ into an axial component $\mathbf{A}_i$ and a radial component $\mathbf{R}_i$, whose norms respectively are $A_i$ and $R_i$, with $R_i$ as the distance from $i$ to the axis of rotation, becomes $$\sum_im_iR_i^2\boldsymbol{\omega}+\sum_i m_i A_i^2\boldsymbol{\omega}-m_i(\mathbf{A}_i\cdot\boldsymbol{\omega})\overrightarrow{CP_i}$$but I cannot prove that $\sum_i m_i A_i^2\boldsymbol{\omega}-m_i(\mathbf{A}_i\cdot\boldsymbol{\omega})\overrightarrow{CP_i}=\mathbf{0}$.
87296
Let us say you are in an inertial reference frame with a circular planar disk. If you take your meter measuring rods (or perhaps tape measure) you can find the diameter and circumference of the disk. If you divide the circumference by the diameter, you will get exactly $\pi$. Now you start rotating the disk. A book I have claims that now the ratio will be different (lets call it $\pi_\circ$ to avoid ambiguity.) This has caused to reevaluate my understanding of special relativity (this is a set up to general relativity, but the problem itself only requires the special theory.) My problem is that, sure, the measuring rods (or tape) would contract on the circumference, but wouldn't the circumference contract a similar amount, therefore giving a measurement of $\pi$ in every reference frame according to your own measuring rods. Of course, this reasoning goes on to make me question how I understood all of special relativity up this point.
37432
So Tenser's floating disc "follows you so that it remains within 20 ft" if you are more than 20 ft away.  So let's get a bit of carpentry skill together, and (whilst keeping the total weight under 500-1000 pounds) rigging up a cart that is 25 feet long. The underside of the carriage is 3 ft tall, and the wheels are 3 ft diameter.  The wizard casts Tenser's Floating Disc, and the party's barbarian picks up the carriage, and puts the rear end on top of the floating disc (let's strap it in with some ropes for good measure).  The wizard then climbs on the carriage, and moves towards the front.  As soon as he gets > 20 feet away, what happens? What speed does the happening occur at?  Bonus question: Assuming the above answer is that the carriage moves, at a high enough speed to be useful, any ideas how to steer such a device other than the wizard moving back within 20 feet to stop it and turning it manually like a peasant?  To alleviate concerns that the carriage is an obstacle to the disc, an alternate construction could instead use a 50 ft plank, not touching the ground, with the disk in the middle, the wizard at the front and an equal weight counter balance on the other end. That way, the disk is carrying things, but it not touching anything other than the things it is carrying.
50984
Exercise:  A disk of radius $R$ and moment of inertia $I_1$ rotates with angular velocity $\omega_0$. The axis of a second disk, of radius $r$ and moment of inertia $I_2$ is at rest. The axes of the two disks are parallel. The disks are moved together so that they touch. After some initial slipping the two disks rotate together. Find the final rate of rotation of the smaller disk.    Attempt: $L_{1_0} = L_1 + L_2 \rightarrow I_1\omega_0 = I_1\omega_1 + I_2\omega_2$ $\omega = \frac{v}{r} \rightarrow v = \omega r$ $\omega_1 R = \omega_2 r \rightarrow \omega_1 = \frac{r}{R}\omega_2$ $I_1\omega_0 = I_1\frac{r}{R}\omega_2 + I_2\omega_2 \rightarrow \omega_2 = \frac{I_1\omega_0}{\frac{r}{R}I_1 + I_2}$ $$\omega_2 = \frac{I_1\omega_0}{\frac{r}{R}I_1 + I_2}$$  Request: Is my solution correct? If not, where and why?
80356
There is a ball rolling down an incline, with no slipping. If we consider the point of contact between the ball and the inclined plane to be the pivot point (for our torque calculations), then I have noticed that friction is no longer in play, and gravity (and normal force) is the reason for the torque. However, if we consider the center of mass of the ball to be the pivot point, friction is the only factor, since both the normal force and gravity effect the center of mass (so the net torque caused by them is 0, since the pivot is the center of mass). However, both of these produce different torques.. for different reasons. So my question is,   Is gravity the reason a ball rolls, or is friction? And why?  [Note: This is not homework, it is a genuine question I had while looking at some physics problems.]
32430
So I've been trying to create a mathematical model for an electric motorcycle and began to wonder about the maximum possible torque that could be supplied to the rear driven wheel without having the bike begin to lift up on one wheel.  I found ways to calculate this value online; however, the basic concept as to how the bike actually lifts escapes me.    My problem started when I began to think about what axis the bike will rotate about when it is doing the wheelie.  My first intuition was that the frame and front wheel, together, rotate about the rear axle.  But when I drew a free body diagram of the frame/front wheel system just at liftoff (see below) I made note that the only forces acting about the rear axle, point O, is the force of weight.  This means that an increase in applied torque and subsequently, the applied force Fa, should not effect the rotation about the rear axle.  I know that the applied force on the back wheel is indeed correlated with the propensity for a bike to wheelie, so I considered that the axis of rotation I was choosing was wrong.  If we take the free body diagram above and sum the moments about the center of mass, we would find that an increased applied force would in fact cause the solid body to rotate.  The problem with this understanding is that during rotation, the center of mass of the drawn system should actually rises relative to the surface the bike is moving across.  If the bike were to be truly rotating about its center of mass, then the back wheel would begin to dip below the surface of the road like you might see in a glitchy video game.    So I suspect that the bike is in fact rotating about the back axle, but I don't understand why, please help!  edit: I added the external torque from the back wheel to the frame, which would allow the bike to rotate about the back axle  edit 2:  I suppose that the torque acting on the frame via the engine should not effect the rotation as the movement of a motorcycle can be perfectly replicated by applying a force at the back axle.  A third possible axis of rotation might be the lowest point of the back wheel.