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Because a rigid body acts like its entire mass is contained within a point called center of mass. This greatly simplifies solving problems such as those in your examples. But in principle you can adopt "your approach" given in your example, where you consider that both sides have mass, although this unnecessarily complicates the solution. For ...

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Hints: If you look into The moment of inertia of a triangle through the 'centroid' The parallel axis theorem and use fact that The co-ordinate of the centroid is $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$ Then maybe you'll prove the result used in the video.

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You can reduce a continuous mass to its centre-of-mass point when the extend of the object isn't important. When dealing purely with inertial forces and gravity and more, then the rotational effects of all points on either side of the centre-of-mass will typically cancel out. Then their locations do not matter, and it is much easier to model the object as ...

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At first, coin is spinning in own fixed reference frame (along axis which goes through the center of coin). So given this fact, it not differs much from spinning humming-top rotated by $90^{\circ}$. Both systems would be stable due to angular momentum conservation $\vec L = \vec r \times \vec p = \text{const}$, if no dissipation forces would be acting upon ...

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There is a straightforward description of spinning tops (or other spinning objects) via Hamiltonian dynamics: Our generalized coordinates are the orientations of the three principal axes of the spinning object relative to an arbitrary axis. For a top under the influence of gravity, it's convenient to choose the axis along which the force of gravity points (i....

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Presumably you are talking about the 'spin' angular momentum. This would be zero if the satellite always faced a distant star. If such a satellite contracted then its rotation wouldn't increase. Why it's so, is a deep question. There was Newton's bucket. Newton considered that rotation is relative to an absolute space. A satellite always facing the earth ...

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The angular momentum vector of Newtonian mechanics and the element of $\mathfrak{so}(3)$ in Hamiltonian mechanics represent the same quantity. $\mathfrak{so}(3)$ is canonically isomorphic to $\mathbb{R}^3$ where the bracket is the cross product. This isomorphism is given by the hodge star, which identifies the vector $\mathbf{L}$ with the antisymmetric ...

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You need a simulation loop which integrates $\ddot{\theta}$ to get $\dot{\theta}$ and $\theta$ at a later time. So at each time steps follow these steps Known vectors of joint speeds and angles $\dot{\boldsymbol{\theta}}_n$ and $\boldsymbol{\theta}_n$ at time $t_n$, or initial conditions for the first step. Calculate coefficient matrix $\bf A$, and vectors $... 2 This setup needs to be thought of in three distinct phases: before the collision, a small time$\Delta t$around the moment of the collision, and finally the motion after the collision while the ball is rising. Your question is not 100% clear about which phase you're concerned with, so I'll give a brief description of all three. Before the ball hits the &... 1 Angular momentum is in general conservated about any point, as long as all influencing torques are included. In your case, you can include the torque due to the pointy-edge-normal-force, and the torque due to gravity and the torque due to any other pushing forces etc. If all is included, then angular momentum is conserved. Now, note that since we choose the ... 1 the angular momentum of the ball is made up of two components. The spin. The normal contact force on the sphere acts perpendicular to the surface of the sphere. It therefore passes through the middle of the sphere and has a torque of zero. The force of gravity and the normal force from the ground also act through the middle of the sphere. Since there is ... 0 Torque$\vec M$is the first moment ("angular equivalent") of force$\vec F$and angular momentum$\vec L$is the first moment ("angular equivalent") of momentum$\vec p$. Just as force changes momentum, torque changes angular momentum. With no force momentum is constant, with no torque, angular momentum is constant; specifically,$\vec ...

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Take into account the following and knowing that the repeated indices are implicitly summed over: \begin{equation} |\vec{\omega}\times \vec{r_i}|^2 = e_{k \alpha m}\omega_{\alpha}r_{im}e_{k \beta n} \omega_{\beta} r_{in} \end{equation} Knowing that Levi- Cevita symbol contraction from an equation with Kroneker symbol \begin{equation} e_{k \alpha m}e_{k \...

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From your question, you seem to think that the components of $\mathbf{r}$ are labelled by Greek letters, such that $\mathbf{r} = (r_{\alpha},r_{\beta},r_{\gamma})$, making $\gamma$ the third component. This is, however, not what is meant here. Could it be that you are unfamiliar with Einstein's summation convention? The $\alpha$ and $\beta$ are not supposed ...

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The point is not that a unconstrained rigid body rotates around its COM. In reality, for any time $t$, a rigid body (constrained or not) always rotates around any of its points with an angular velocity $\boldsymbol \omega (t)$. Suppose $3$ generic points $P_0, P_1, P_2$ that belongs to the rigid body. And lets $P_0$ be by hypothesis the center of rotation. ...

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There are two ways to look at this problem that can be useful: 1. Using forces: The work $W$ done by moving something along curve C is defined as follows $$W = \int_C \mathbf{F} \cdot\text{d}\mathbf{s}$$ where $\mathbf{F}$ is force and $\text{d}\mathbf{s}$ is element of the curve. If the force is constant then simpler formula applies: $$W = Fs \cos \alpha ... 3 The kinetic energy increases for the dancer after pulling her arms in (for example). That would decrease her moment of inertia, but from conservation of angular momentum, increase her angular velocity \omega.$$K.E. = \frac{1}{2}I\omega^2$$so the kinetic energy has increased. The energy has come from the work she has to do to pull her arms in, i.e. the ... 1 The concept of torque is not merely "necessary". It simply is. Torque is axiomatic, in the same way as "left" and "up" or "beneath" and "more than." Argue all night about the specific words used in any given language and still, how could "necessary" come into this, any more than it does for other ... 7 If I understand correctly, the crux of your question is: Why not use F = mr\alpha to explain the change in an object’s rotational motion instead of torque? Reason #1: A more general case of concentrated mass There are a few reasons, but the first good reason is that the r in \tau = Fr and the r in mr^2\alpha are not always the same r! For ... 5 Why not use F = m r α to explain the change in an object’s rotational motion instead of torque? There are three major conserved mechanical quantities: energy, momentum, and angular momentum. Each of them are separate quantities and they are separately conserved. Because they are conserved, we are interested in the rate of transfer of each of them. Force is ... 31 The concept of torque is not necessary as such. But it is useful. When things are useful, someone might eventually start to use them. Then, sadly if you will, everyone else will have to understand them as well in order to follow what is being done. In the case of torque, it is useful because it is a "rotational version" of force. You can't simply ... 2 Consider a stiff but massless bar with a frictionless axle at one end. A mass (m) is mounted on the bar at a radius (r) from the axle and a force (F) pushes on the bar at a radius (R). The work done by the tangential component of the force pushing the bar through an arc length (S) will be transferred by the bar, accelerating the mass through an arc length (s)... 9 One big reason for using torque is that it creates a simple way of understanding when an object is in equilibrium. When the sum of the FORCES on an object is zero, then we know that the object does not have translational acceleration. Similarly, when the sum of the TORQUES on an object is zero, then we know that the object has no rotational acceleration. ... 3 Because from F = m r \alpha, it can be seen that for knowing what magnitude of force was applied to a rotating object with given angular acceleration - you have to know distance \vec r vector, i.e. position where this force was applied. And in the second (linear) Newton law, force does not depend on distance. So it must be that angular acceleration has ... 13 In principle you could always use force and not use torque. But, the use of torque greatly simplifies the analysis of the motion of a system of particles. For example, for rotation of a rigid body about a fixed axis the rotational motion is simply described as \tau = I \alpha where \tau is the net torque, I is the moment of inertia, and \alpha is the ... 0 The formula for PE is not F /s it's Fy. That subtle difference is quite important because if you move something perpendicular to gravity, you get no change in PE. It then follows that there is no rotational PE unless the rotating body is connected to a spring (or other energy storage). 0 So this question cannot be solved using conservation of energy alone. To explain this note that the system has two degrees of freedom. We can choose the independent generalized coordinates of this system as \theta (rotation angle of the cylinder) and h (displacement of the block). The configuration of the system is completely specified if (h,\theta) is ... 0 Thanks to BioPhysicist for the hint. I decided to complete my answer now, As mentioned in the post, the work energy equation we get is:$$mgh=\frac{1}{2}mv_1^2+\frac{1}{2}\omega ^2+\frac{1}{2}mv^2$$Also, as mentioned in the post (Equation 5) we have:$$g=a_1+a \space\space\space-(1)Using Newton's speed equations, as the system was released from rest, ... 0 You already have the expression for the force, but you need the part the force is resisting, the weight on the front wheel. This weight m g depends on the rider, the position of the center of mass, and how the weight is distributed between the front and rear wheels. 1 In the case of a gyroscope, it is normally free to swivel about on its x, y, and z axies: three degrees of freedom or independent ways in which it is free to move. One of these axies corresponds to its own rotational axis leaving two axies to display the airplane's attitude in pitch (nose up or down) and roll (banked left or right). In the specific case of a ... 0 Torque is equal in all the point of axis of rotation but what about other exis say if body is rotating on z-axis 60 ft lb. Whether it will be same on Y or X- axis. 0 starting with the kinetic energy (only rotation) \begin{align*} &T=\sum_{\rho=1}^{N}\,m_\rho\,\frac{\mathbf{v}_\rho^{2}}{2}\\ &\text{with}\\ &\mathbf v=\mathbf\omega\times\mathbf r\quad \Rightarrow\\ &T=\sum\,\frac{m }{2}\,(\mathbf\omega \times\mathbf r )^2= \sum\,\frac{m }{2}\,\left[\omega ^2\,r ^2 -\left(\mathbf\omega\cdot\mathbf r\right)... 1 The derivation of the mass moment of inertia tensor from the summation of particles has the form of a symmetric matrix. Each particle of mass m_i located at \boldsymbol{r}_i contributes to the MMOI tensor by an amount that equals \mathrm{I} = \sum_i m_i \left( (\boldsymbol{r}_i \cdot \boldsymbol{r}_i) \mathbf{1} - \boldsymbol{r}_i \odot \boldsymbol{r}...

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Moment of Inertia of thin lamina is used having same cross section as the object of interest, like for a cylindrical rod, MOI of the corresponding circle is taken. It isn't the mass moment of inertia that is used, rather it is the area moment of inertia which is mass MOI divided by the density, i.e. instead of $r^2dm$ we have $r^2dA$. Now to understand how ...

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In classroom demonstrations the gyroscope wheel is almost always released gingerly. By gingerly I mean the demonstrator doesn't remove support fast (allowing a sudden drop), instead the support is taken away slowly. Another scenario is when the spin rate of the gyro wheel is very high. Then the frequency of the nutation is high, with corresponding small ...

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If the man also walks towards the cylinder with speed $V$, the velocity of the centre of the cylinder relative to him is $V--V=2V$, as you have asserted. However, the velocity of the top of the cylinder relative to the man is $2V--V=3V$ not $4V$. So if the cylinder travelled a distance of $l$, a length of $3l$ of rope would have passed through the man's ...

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Thanks to the OP for posting this great question. The figures are of great help in explaining this interesting aspect of friction related phenomena. In what follows, this answer provides a detailed analysis of the rolling without slipping phenomenon which underlies the physics addressed in the OP. $$\underline{\textit{Qualitative analysis}}$$ It is easily ...

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the particle P has a tendency to slip to the left as a wheel rotates in a clockwise manner. That is true only if you assume information that is not given in the diagram (like something producing a torque on the axle). It could be that the wheel is just rolling along. In which case, there is no tendency to slip. On a slope, the direction of tendency to ...

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Starting from first principles, we can write the position vector of an object as $\vec r = R \hat r$ where $\hat r$ is a unit radial vector. So the velocity vector is $\displaystyle \vec v = \frac {d \vec r}{dt} = \frac {dR}{dt} \hat r + R \frac {d \hat r}{dt}$ If the object is moving in a circle about the origin then $\frac{dR}{dt}=0$ so $\displaystyle \vec ... 1 the Energy is: $$E=\frac 12\,m{v}^{2}+\frac 12\,I_{{D}}{\omega}^{2}+m\,g\,h=0$$ at$~t=0$$$E_i=E\left(v=0~,\omega=0~,h=h_0\right)$$ and at the final position $$E_f=E\left(h=\frac{L}{2}\,\sin(\alpha)\right)$$ with$~E_i=E_f~$you obtain the velocity of the center of mass at the final position $$v_{\text{CM}}^2=-{\frac {I_{{D}}{\omega}^{2}+m\,g\,L\sin \left( \... 1 Assume the rod to be of mass m, uniform and completely rigid. The floor is also perfectly rigid and there's no friction between floor and rod, which prevents any sideways motion on or after impact. The collision is therefore perfectly elastic. The translational kinetic energy is completely converted to rotational kinetic energy, for rotation about the CoG: ... 1 The acceleration of the center of mass of the rod is the net force on the rod divided by the mass of the rod. In reality the rod will bounce as it impacts the ground, but let's assume it does not to address your question. The net force downward at the instant of impact is the force of gravity minus the normal force and has no sideways component, so the ... 2 Vertical forces. The reaction to the normal force from the ground on the rod is the normal force from the rod on the ground. The reaction to the gravitational force from the Earth on the rod is the gravitational force from the rod on the Earth. While the normal force and gravitational force are opposite in direction here, there's no reason they must be equal ... 0 Collision of two ellipses \mathbf w_i~ start velocities \mathbf v_i~ final velocities \omega_i=\frac{d\phi_i}{dt}~ angular velocities S_i~ center of masses \mathbf t~ tangential direction \mathbf n~ normal direction I_i~ moment of inertia dp~ linear momentum with$$\mathbf w_i=w_{in}\,\mathbf n+w_{it}\,\mathbf t\\ \mathbf v_i=v_{in}\,\... 2 Add forces$A'$,$A''$and$B'$,$B''$acting at the centre of mass$G$of the rod. Forces$A$and$A'$act as a couple of magnitude$Aa$in a counter-clockwise direction. Forces$B$and$B'$act as a couple of magnitude$Bb$in a counter-clockwise direction. So the net couple on the rod is$Aa+Bb$counter-clockwise and the rod will rotate about the centre ... 0 If your pulley has mass, you have to consider the rotational inertia of it. Inertia is the tendency to resist movement. Your rope does not have equal tensions. Consider$T_1$and$T_2$where$T_1$is the tension on the side of$M_1$and$T_2$is the tension on the side of$M_2$. Assuming$M_1$is greater than$M_2$, we have$M_1g-T_1 = M_1a$,$T_2-M_2g=M_2a$,... 1 See my 2012 answer for an explanation of gyroscopic precession, and the forces involved. This exposition of the physics of gyroscopic precession is not in terms of the angular momentum vector. Instead this exposition capitalizes on symmetry; a wheel is very symmetric. Specific to the device demonstrated in that video: The suspension of the gyro wheel of ... 0 The moment of inertia of the object in the drawing is equal to the moment of inertia of a disk of radius "b" minus the moment of inertia of a disk of radius "a". 1 Interesting question. The cross product does generalize to higher dimensions. Especially, for 1+3d spacetime, i.e., Minkowski, we can either consider$(A^\mu,B^\nu)\mapsto u^\mu\varepsilon_{\mu\nu\rho\sigma}A^\rho B^\sigma$with$u^\mu$a timelike unit vector (e.g., the observer’s 4-velocity) or$(F^{\mu\nu},G^{\mu\nu})\mapsto\Delta^{\kappa\lambda}{}_{\mu\nu\...

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The sense (direction) of the fiction force $F_f$ (in my diagram) is determined by the relative motion of the cylinder's surface and the ground. The moment you start applying force on the string the cylinder 'wants' to start rotating clockwise and its surface wants to move left wrt the floor. Friction of course always opposes motion, that's just what it does!...

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