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I do not believe the pin has any influence in the motion of the gears. At least not if the pin is light enough. The reason is that in such a case internal forces will be zero. Let us look at the two extreme cases: 1) Minimum $x$: $x=0$, then the pin (and the arm) will simply rotate parallel to the bearing at $\omega_c$ 2) Maximum $x$: $x=r_0$, then the ...


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I think the problem is that when you drive one wheel the other free wheels and thus turning the robot in circles. You have to consider carefully the relationship between the torque characteristics of the motors with any gearing to the wheels to establish reasonable control of the robot. In you case is it really possible to fix a motor (not moving) even ...


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You haven't given enough information: What you are asking for is that missing bit. I think I can guess just what that is: If $\vec{A}$ is a position vector (e.g. points from the origin to a point that is represented in both the inertial and the non-inertial frame of reference) and your non-inertial frame of reference rotates at a constant angular velocity ...


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First he spins the stick and then he lets it rotate in air. If you watch carefully, you can see that he stops it with his left hand before kicking and then walks away: he didn't kick it during rotation!.


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This is a problem that involves only calculation of velocities from other velocities, no influence (forces) needs to be considered. Imagine the system as it appears in the inertial system where the point of contact of the two wheels is at rest. Since the contact point is at rest, the mass points of both wheels that touch each other are at rest. Since any ...


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OK, I'm assuming you want the formal proof of this well known kinematics formula! So here goes: Let the particle rotate about the axis OO' ... Within time interval dt let its motion be represented by the vector dφ whose direction is along axis obeying right-hand-corkscrew rule, and whose magnitude is equal to the angle dφ. Now, if elementary displacement ...


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In a way centripetal force causes circular motion, centripetal force that accelerates a body to undergo circular motion. If you imagine a ball on a piece of string: when the ball is spun around the centripetal force will be tension, the tension of the string pulls the ball as it spins to cause it to move in a circle. Without the string, the ball will just ...


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Physics does not have a proper, rigorous concept of causation. There are the terms locality and causality, but they are technical terms with precise meanings that do not occur in Newtonian physics. Nevertheless, Newton's law, $\vec F = m \ddot{\vec x}$ is often seen to embody causation in a certain sense: You are given, as external circumstance, the total ...


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If no force acts on a body it moves in a straight line. To make the body deviate from a straight line you have to apply a force to it. Therefore applying the centripetal force to the body is what makes it move in a circle. If you apply a constant force at right angles to the direction of motion then your object will indeed move in a circle.


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The rotational kinetic energy of a (uniform) solid sphere rotating about an axis passing through the center of mass is given by $\frac{1}{2}I\omega^{2}$, where $I=\frac{2}{5}MR^{2}$. So $K=\frac{1}{5}MR^{2}\omega^{2}$. Using $M=6\times10^{24}\,\mbox{kg}$, $R=6400\,\mbox{km}$, and $\omega=\frac{2\pi}{T}$, with $T=24\,\mbox{hrs}$, we get ...


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A general planar body that translates on its center of mass and rotates is going to have some point (either on the body, or somewhere in space) with zero linear velocity. This is called the instant center of rotation as the body can be said that it instantaneously rotates about this point. If the center of rotation happens to be on a contact point then it ...


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A simple example of this is a wheel that rolls without slipping. I'll try to make the example as illustrative as I can. Imagine you're stuck to the side of a giant runaway wheel that is rolling without slipping along the ground. In this scenario, right as the wheel touches the ground you're not moving. This is just the condition for a wheel to be considered ...


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However as seen by A , B remains at fixed distance and also doesn't rotate (relative angular velocity is zero). But it does rotate, if your reference frame does not rotate, but just gets centered on your point of interest. If you want to consider a rotating reference frame, then all points (that are fixed to the disc, or to the frame) are obviously ...


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A rotating reference frame is an accelerated reference frame so $A$ and $B$ are at rest in an accelerated reference frame. Assume an inertial reference frame $S_0$ and another reference frame $S$, with a common origin and rotating with respect to $S_0$. Let the (constant) angular velocity vector of $S$ be $\mathbf \Omega$. Then, the time rate of change of ...


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If I understand your question correctly you are saying that: $$ v = r\omega $$ and therefore: $$\begin{align} v_A &= r\omega \\ v_B &= \tfrac{1}{2}r\omega \\ v_A &= 2v_B \end{align} $$ but how can $A$ and $B$ have different velocities when they are both attached to the disk so the separation between is fixed? The answer is that $A$ and ...


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Yes, $B$ does rotate when seen from a static frame of coordinates outside the disk: As to velocities and accelerations, see the article in Wikipedia. It says, $$\vec {v_s} = \vec {v_r} + \vec {\Omega} \times \vec r,$$ where $v_s$ is the velocity in the static frame and $v_r$ in the rotating. If you apply this formula for both points $A$ and $B$, their ...



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