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3

I am not sure what you actually want to ask. So I would recommend that you put more effort into your question. Assuming that this is the setup that you have, a scissor in gray and the cross section of the rope in orange: Say the toque applied to the joint of the scissors is $\tau$. What is this force $F$, then? The distance of the rope contact point to ...


2

The angular velocity is related to the tangential speed $v_t$ by: $$ \omega = \frac{v_t}{r} $$ However the velocity $v$ being used here is not the tangential speed. It is the linear velocity $v_l$ of the centre of mass of the disk relative to the floor. The question has chosen a value for the angular velocity of the disk that appears similar to the ...


2

If a body of mass m hanged on a string is moving, let uniformly, on a circle fixed relatively to the ground, then an observer G on the ground uses the 2nd Newton Law : $$ \mathbf{F}=m\cdot \mathbf{a} \tag{01} $$ and finds the relation between the force $\mathbf{F}$ and the acceleration $\mathbf{a}$. For observer G there exists a "real" force, the tension ...


1

So the definition of angular momentum should be this: "Angular momentum is the product of the angular velocity of the body or system and its moment of inertia with respect to the rotation axis, and that is directed along the rotation axis". That's not a useful definition at all, because (i) it does not specify what this "moment of inertia" thing is, and ...


1

Note that you have to swing the pail with a certain minimum speed for the water to stay in. That minimum speed is such that when the pail is at the top of the arc, the rope accelerates the pail downward faster than gravity accelerates the water downward. Otherwise, the water falls out.


1

$\omega=\frac{v}{r}$ is used for rotating without slipping. Your disc is rotating and slipping. So, it is possible that $\omega=\frac{v}{3r}$.


1

I think sections 4.1.2 and 4.1.3 of this lecture on dynamics explains it by looking at each component separately. Since $\frac{{\rm d}}{{\rm d}t} \sin \theta = \dot{\theta} \cos\theta$ and $\frac{{\rm d}}{{\rm d}t} \cos \theta = -\dot{\theta} \sin\theta$ the components of ${\rm d}u$ are perpendicular to $u$. Our first step is to choose cartesian axes, ...


1

But why must $d\hat{u}$ be orthogonal to $\vec{\Omega}$ too (i.e. be tangential to a circle orthogonal to $\vec{\Omega})$? To get such a precession there must be a clockwise torque in the plane of the screen acting on the system which means that the torque vector must be pointing into the screen. That torque produces a change in the angular momentum ...


1

The proof is very simple and comes right form the definition. let $R(t)$ be a rotation matrix as a function of time. $R$ is an orthogonal matrix so its inverse is equal to its transpose: $I = R(t)R^T(t).$ ($I$ is the identity matrix) The time derivative of the above equation is $0 = \frac{d[R(t)R^T(t)]}{dt} = ...


1

While you are correct saying disk 2 has rotational kinetic energy, you are missing that no matter the situation, since ground is frictionless, the work done be external force(F, in this case), is same in both cases. Thus by work energy theorem, $$Work=change in KE$$.Thus since work in case 1 equals work in case 2 thus both disk have same kinetic energies.


1

"As we can see from the picture, both disks have the same force being applied to them and they also go the same distance d→" This is the erroneous assumption - the 2 discs do not go the same distance. Some of the distance that the rope is pulled will rotate disc 2 as it unravels. As a result the liner distance is less and the balance of work goes into ...



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