Hot answers tagged rotational-dynamics
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You can always decompose a motion like this into two parts: (1) rolling without slipping and (2) slipping without rolling.
What is slipping without rolling? It means the object moves uniformly in one direction along the surface, with no angular velocity about the object's own center of mass. For instance, a box that is pushed along the ground can easily ...
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There's are many things wrong in your concepts .
Let's attend them one by one ,
A body without a force acting on it always can never rotate as its velocity is changing at each hence momentum is changing at each instant . But to maintain constant velocity , the force must be such that it never does any work , so as to maintain the constancy of Kinetic ...
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How can we apply angular momentum conservation when friction is present?
Why not? If we have a closed system, momentum and angular momentum are conserved. In this case, the full system is disk A and disk B, and there are no external forces, so the system is closed. There are internal forces, namely in this case, friction, but that doesn't matter.
You ...
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It depends on the friction of the contact. With a frictionless plane the top would precess around its center of gravity and the contact point will prescribe a circle.
Add friction, and the friction force translates the center of gravity the same way tire traction translates a car. Here you have the cases of a) pure rolling, or b) rolling with slipping.
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If the wheel is rolling without slipping, what's the velocity of the point at the base of the wheel?? It is... zero! Convince yourself that the velocity must be zero. Since if it wasn't zero, the wheel wouldn't be rolling without slipping.
So far the explanation is correct. "No slipping" refers really to some non-zero interval of time, and to the state of ...
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Basically , it means at each instant the bottom most point has $0 $ velocity , it doesn't mean that the point has no acceleration . But at an instant it has $0$ velocity . And because of that at each instant $v_{cm}=\omega r$ for the bottom most point , and if this doesn't happen , then static friction acts to make it $0$ .
Its like suppose you are walking ...
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The relative speed of the point of contact of the rolling body w.r.t. the surface on which it rolls is zero.
If the surface is at rest then the velocity of the point of contact of rolling body and surface is zero.
Mathematically:
$$v_1 -\omega R=v_2$$
Also we can get the reltation in accelerations ..... Differentiate the above eq.
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You don't need to apply Steiner's theorem onto the point mass. The point mass finds itself at a distance (apparently) $R$ of the x-axis. Since the moment of inertia is an extensive value, you can simply add all moments of inertia.
There's the moment of inertia of the solid disk with respect to it's diameter. You have to 'Steiner' that away from a distance ...
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Firstly , definition of torque is $\vec{r}\times \vec{F}$ and angular momentum $\vec{r}\times \vec{p}$.
And now w.r.t. your frame $\vec{F}$ & $\vec{p}$ & $\vec{r}$ are all relative . but newton's second law of rotation holds for all frames.
.Because all points are just frames and to maintain the distances in frame , you've to move with that frame , ...
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I'll expand my comment into an answer.
I would take $\mathbf{T}=d\mathbf{L}/dt$ as the definition of torque, but it sounds like the OP takes $\mathbf{T}=\mathbf{r}\times\mathbf{F}$ as the definition. Either way, we need to prove that the two expressions are equivalent for a system of particles.
The total angular momentum is
$$\mathbf{L}_{tot}=\sum ...
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