7

Okay so let me give you the intuition. Now you would've heard of cases of a pendulum made of a rod instead of a string. (You can checkout this: http://hyperphysics.phy-astr.gsu.edu/hbase/penrod.html) Okay, so in this case you can clearly see that the torque is taken about the hinge. And you apply gravitational force at the centre of mass and take its torque ...


4

It has to do with angular force balance [torque = R * force(R)] around the center of mass of a rigid body. Those objects that have wider bases have, by definition, larger R at the base to compensate for external torque applied. Those objects that have narrower bases have smaller R at the base to compensate for torque applied, even torque applied by the body'...


4

Why a body does not rotate if force is applied on the centre of mass? The question is: Why does it rotate if it is not applied to the center of a mass? First we look at two rules that students of mechanical engineering (I don't know about physics) learn at university: Multiple forces applied to the same point of a solid body have the same effect as a ...


3

This is easily explained by Newton's second law. If there is no net force applied to a body then the center of mass will not accelerate. it will either be stationary or move in a straight line. The only allowed motion is a rotation about the center of mass. Changes in entropy, arise from the exchange of energy at some temperature and has nothing to do with ...


3

Based on your free body diagram, yes, the center of mass would not experience any acceleration, but the object would rotate about the center due to the torque from force $f$. However, based on what you have drawn, perhaps you have misunderstood how the forces are at work here. It looks like to me that you have a ball, (or disk, etc.) that is on an incline. ...


3

The 3D motion of a rigid body requires some familiarity with vector and matrix operations. The rotation about the center of mass does not have an analytical solution. Suppose at $t=0$ the body has orientation matrix $\mathbf{R}_0$ as well as rotational velocity $\boldsymbol{\omega}_0$. This means the 3×3 mass moment of inertia tensor initially is calculated ...


2

In this particular type of problem, the mechanical energy, E, is constant because no external work is done on the system. What you seem to quote ("They write ...") is not an accurate copy. Rather, they write $$E = \frac{1}{2}m\left(\dot{r}^2+r^2\dot{\phi}^2\right)-\frac{GmM}{r}.$$ That makes all the difference in the world because, when you change ...


2

In this answer I am referring to top position only.Similar arguments can be made for any position. Well think it like this.You know that radius of curvature of circular motion is given as: R= F/(m ω^2) where F is net force. Notice if we keep angular speed decreases, radius decreases. However you know that such a thing is not happening here.The reason of this ...


2

Basically, if we know the concrete is traveling in a circle with frequency $\omega$, we know its trajectory - that is, its position vector as a function of time. So, by taking the second derivative of that function, we know its acceleration. For circular motion, that works out to be a vector of magnitude $R\omega^2$, always pointing toward the center of ...


2

One solution is the above. The other one is from the rotation at the point when the spool touching the ground. For simplicity, let $\alpha = 0$: kinetic relation: $a_{||} = R \frac{d \omega}{dt}$ Newton law in rotation: $F(R-r) = (\gamma + 1) R^2\frac{d \omega}{dt}$ So that: $a_{||} = \frac{1- \frac{r}{R}}{\gamma + 1} \frac{F}{m}$


2

You appear to be missing the concept of static friction and thinking only of kinetic friction. Kinetic friction is the friction between two objects in relative motion, also called slipping. This friction force has a magnitude $F_{kinetic}=\mu_k N$ where $N$ is the magnitude of the normal force and $\mu_k$ is the coefficient of kinetic friction. The direction ...


2

Your statements do seem to apply to a rolling object. If there is acceleration, the friction provides the torque required for the corresponding angular acceleration.


2

The short answer is that when you write $$T = \frac{1}{2}M v_{CM}^2 + \frac{1}{2}I\omega^2$$ you are assuming that the moment of inertia is being calculated about the center of mass. It's true that the moment of inertia can be calculated about whatever axis you'd like; however, if you don't choose the center of mass, then the expression for the total ...


2

The rotation of a body depends upon the point at which it is hinged. So if a force applied at the center of mass (COM) then it may cause rotation about the axis (if it's not the one through COM) at which it is hinged by causing a torque to act. But what about the case when it isn't hinged at any point with the force still acting on the COM? A torque still ...


2

Although the quaternion has 4 parameters, it really has 3 degrees of freedom, since it must obey the unity condition $\sqrt{x^2+y^2+z^2+w^2}=1$. Also note that $\boldsymbol{\omega} \neq \boldsymbol{\dot q}$ and the same applied for their derivatives also. Baseline Method So the process goes like this Given known orientation $\boldsymbol{q}$ calculate the 3×...


1

With a large base, the force from the supporting surface can shift further away from a point directly below the center of gravity when necessary to oppose any external tipping torque. Any good physics text has a section on “Statics”.


1

Because force is the time derivative of momentum, and momentum is linked to the motion of the center of mass. If you consider a rigid body as a collection of particles glued together and their position split into the position of the center of mass $\boldsymbol{r}_{\rm COM}$ plus some other relative position $\boldsymbol{d}_i$, then $$ \boldsymbol{r}_i = \...


1

Let $\vec r_1$ denote the position of the centre of mass of an object of mass $m$, given by the formula below. $$\vec r_1 = \frac{1}{m}\int \rho \vec r^\prime \mathrm{d^3} \vec r^\prime$$ If an object is not rotating, then all of its points must have the same acceleration. Therefore, if a single force applied to the object does not cause rotation, it must be ...


1

They should get displaced by same angle, as the torque equations would be same in both cases.


1

After sliding down through an angle θ without friction, the kinetic energy of each bead is: (1/2)m$v^2$ = mgR(1 - cos θ). To move along the circular ring requires a centripetal force: m$v^2$/R = mg cos(θ) + N. The reaction force on the ring is N = 2mg(1 – cos θ) – mg cos(θ). The lift provided by the vertical components of N from both beads (while θ is ...


1

I don't think you can avoid this singularity by using Quaternions because they are good for rotation . but for numerical simulation you can use this concept: your geodetic equations are: $$\ddot{\theta}=-2\,{\frac {\dot{\theta}\dot{\phi}\cos \left( \phi \right) }{\sin \left( \phi \right) }} \tag 1$$ and $$\ddot{\phi}={\frac {\sin \left( \phi \right) \left( ...


1

It's not just pendulums. It's a major issue for spacecraft navigation, especially if, for instance, your trajectory involves the deployment of a supersonic parachute on Mars. Hence: quaternions are the standard. They are also computationally faster, which can be a factor with space-qualified CPUs, which are not fast.


1

Consider the bicycle and cyclist as a system. If the bike is on a flat road, there are two relevant external forces: Friction with the ground pushes the bicycle forward. Because the velocity of the wheel's contact point with the ground is zero, the power supplied by friction is exactly zero. Sources of dissipation, such as air resistance, do negative work ...


1

It's a matter of identifying your systems. As you mentioned, power is derived from force; hence, just like force, power is delivered from one system to another system. You can talk about the power from the person or the ground or both together, and to the wheels or the bicycle as a whole. Since the wheels are rotating, you apply the torque formula to them, ...


1

we cannot apply it about centre of mass as external torque about centre of mass is not 0. Is my reasoning correct? Once the ball enters pure rolling friction force stops acting hence you can apply conservation of angular momentum but before that you can't. Now after starting rolling, it collides with a wall inelastically so that linear velocity becomes say ...


1

In the formula $$ \vec{L} = \vec{L'}+M\vec{r}_{cm}\times\vec{v}_{cm}$$ Which represents the total angular momentum we have two parts/components: First let's look at the second term, which is $M\vec{r}_{cm}\times\vec{v}_{cm}$: This represents only the linear motion of the center of mass. Take the vector from the origin to the COM and call it $r_{cm}$ and take ...


1

The formula $$\mathbf L=\mathbf L' +M\mathbf r\times\mathbf v_{\rm com}$$ is used only for finding the value of angular momentum with respect from different locations. It is not a dynamical formula explaining any dynamics of the system. So it can't give you any relation between the torque applied and the subsequent angular momentum. So the answer of How ...


1

The motion of the body can be modeled by the system of first-order ordinary differential equations in vector/matrix: \begin{align} J&\frac{d\vec{\Omega}}{dt} \, = \, J\,\vec{\Omega} \times \vec{\Omega} \, + \, \vec{X}_F\times \vec{F}\\ &\\ &\frac{d}{dt} U \, = \, U\, \big(\vec{\Omega} \times \cdot\big)\\ &\\ &\frac{d\vec{x}_G}{dt} \, = \,...


1

When it will be in equilibrium, it will have some angular velocity and due to inertia it will continue to rotate till the work done is completely converted to potential energy.


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