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The internal forces don't cancel each other they are just equal and in opposite directions. This results in an internal momentum change of zero. The same is true for angular momentum. As one object in the isolated system begins to rotate another object must rotate with exactly opposite angular momentum so there is a net change of zero.


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Faced by the same question and a background that includes courses in vector calculus, I have sought a simpler answer. My answer is much that same as to why one can easily balance on a typical bicycle. Bicycles are constructed so that the point where the extension of the front fork pivot would hit the ground is in front of the point where the front tire ...


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The reason why a gyroscope does behave in this strange way is that if you try to rotate it's axis in some direction, the "endpoints" of this axis have to be pushed perpendicular to what our first intuition would say. In order to verify why the axis starts rotating in this strange way, let's make some simplifications: the gyroscope consists of two identical ...


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No tricks needed. You can make it basically without a special design, as long as it follows the general guidelines: piece with hole tied with string to piece with rod. Of course the hole should be a fit loosely the rod. As for string length, it should allow movement of the dangling piece, and the shorter the harder it will be to master. As for the physics, ...


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Say the projectile was thrown with a velocity $v$ at an angle $\theta$ with respect to the horizontal. We ignore all friction effects (air drag, side winds). Define a coordinate system with a vertical $y$-axis, a horizontal $x$-axis and the point of origin $O$ the point from which the projectile starts its trajectory. The trajectory can now be decomposed in ...


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In theory there is no torque from friction once the cylinder is rolling. Imagine that there is no supporting surface but no gravity - then the cylinder will just continue to "roll" forever. So all the supporting surface needs to provide to the cylinder is a force through its centre of mass to counteract the gravitational force. This is a force perpendicular ...


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Does a body always rotate purely about its center of mass? Well, that depends. The first assumption you need is that the body is rigid. Violate this assumption and all bets are off the table because you can't even necessarily classify all motions as "rotations": for example if a long thin board starts twisting sinusoidally into/out-of a helix shape, ...


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A body in free motion does not necessarily rotate about the center of mass. The center of mass might have straight linear motion in addition any rotation. The general motion is a screw motion with a rotation about some instantaneous axis and parallel translation at the same time. Consider an arbitrary body rotating by $\vec{\omega}$ and at some instant the ...


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Here's how I see it: Assume the friction force is proportional to surface area $S$, via (all other things being constant): $F_\text{friction} = \mu S$ with $\mu$ the coefficient of friction. Then for an infinitesimal ring with thickness $dr$ at $r$: $dF_\text{friction} = 2\mu \pi rdr$. And $dT=rdF_\text{friction}=2\mu \pi r^2dr$ Integrating between ...


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Energy is conserved, but if you ignore some kinds of energy then it will look like it isn't conserved. You can imagine a really big disk with some radial pointing two by fours attached at the one o'clock, two o'clock etcetera positions then attach springs to each two by four with the spring pointing in the clockwise/counter-clockwise directions. Add a nice ...


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As one of the comments mentions, it is simpler to consider a linear case. Dropping a body of mass $m$ on one moving with mass $M$ and velocity $v$ is essentially considered the instantaneous transformation $M \to M + m$. Momentum must be conserved in the collision, but the mass of the system effectively increases, producing a smaller kinetic energy: $$ ...


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It seems to me what you're asking is pretty simple. You say you can control the angular velocity of each wheel. That, times the wheel radius, give you the forward velocity of each wheel on the ground. That tells you the robot's forward speed (the average of the forward speeds of the wheels), and it tells you the rate at which the robot is turning (the ...


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1- Is it right to say "the motion of the robot can be described as a transitional motion of center of mass plus a rotational motion about that point?" Pick a point on (or off) your robot; pick any point. The motion can always be described in terms of the translational motion of that point plus a rotational motion about that point In general, the ...


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Any rigid body in motion can be described as rotating about in instantaneous axis of rotation (IAR) and translating along the same axis at the same time. Example/Proof A rigid body in moving and at time instant a point A riding on the rigid body has position vector $\vec{r}_A$ and instantaneous linear velocity $\vec{v}_A$ at A. The whole body is rotating ...


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No difficulties at all. If the net force applied on a body is zero then the center of mass is not going to move accelerate. This leads to the conclusion that the only motion allowed is a rotation about the center of mass. For more details refer to: http://physics.stackexchange.com/a/81078/392 The relevant equations are: $$ \mathbf{F} = m \,\mathbf{a}_C ...


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In the case of sphere when it roles you have to analyse in which direction is the force acting. Consider a force acting along a line passing through its center of mass, now this force is not going to roll the sphere it's going to slide the sphere. For rolling a sphere an external torque must be given to it and as we know friction always opposes relative ...


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First, let's grasp the idea that there is no force that is "the centripetal force." There are forces like gravity, Lorentz (EM), tension, and spring which can have components directed toward a center of curvature. A net force toward a center of curvature is called a centripetal acceleration, and the product of mass times this centripetal acceleration is ...



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