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The frame is only instantaneously aligned with the body frame. The measuring frame is not moving, but the body frame is. So the motion and momentum measure non zero because is it only the alignment that is used and not the motion for measuring. The equations of motion are still on an inertial frame, just not aligned with the world coordinate system. The way ...

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When initially you exert a force $F_i$ to get things going, you're actually exerting a torque $T$ about the centre point of the circle: $$T=F_i R,$$ with $R$ the radius of the circle. According to Newtonian physics, this torque causes an angular acceleration $\dot{\omega}$ as follows: $$F_i R=I\dot{\omega},$$ where $I$ is the Moment of Inertia of the ...

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There has been an apprehension about whether the cyclist can take a turn without steering. In my opinion he can. There are two principles he can use. These are By shifting the line of normal reaction sideways to the line of force acting upon the center of gravity. By trying to rotate the cycle sideways. For simplicity consider the case of a standing man ...

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Its not that tough. You can work it out by using just two equations. But the one thing you should keep in mind is that when the comet is at the minimum distance from the sun, its velocity must be perpendicular to the radial vector (sun to comet). So the minimum distance is itself the minimum perpendicular distance used in the angular momentum formula at ...

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It's like any balancing problem. You constantly move your point of support to cause yourself to fall one way or the other. If you don't like the way you are falling, you move your point of support to stop that fall, and then start falling the other way. Your point of support is never stationary. If it is, you fall over. On a bike, you move your point of ...

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Your derivation is correct, although your assumption about $v$ (it's constant) must be made before evaluating the relevant integral. Physically speaking, make the transformation to the moving frame: $y' = y$ $x' = x - vt$, and the implicit form becomes $y'^2 + x'^2 = R^2$. So, this is indeed a cycloid, because we see a circular path in the moving frame. ...

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Center or rotation isn't located where those two gears make contact, it's in a point closer to the smaller gear's center

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Your calculations are wrong, Hint: $\frac{1}{2}I\omega^2 + \frac{1}{2}Mv^2 = Mgd\sin(t)$ $wr=v$

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It would take the same time as it took with A, so $v_b=\frac{2\pi R}{t}$ $t=\frac{2\pi r}{v}$ $v= c/2$ Note that if $R$ is big enough, it may seem as if the point $B$ is travelling faster than light, it is. But the fact is that no physical object is really travelling faster than light.

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When you are changing gears you are trading speed for torque (or vice versa). The overall power transmitted maintains the same so $P=\omega_I T_I = \omega_O T_O$. The way this works is by the chain forcing the same tangential velocity on the two sprockets (input and output sprocket) from which their angular velocity is found $\omega_I = \frac{v}{r_I}$ and ...

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Check out this chart as very rough baseline. Solar energy, even with 88 days of Mercury level sunshine, Wouldn't reach nearly as far into Mercury as you suggest. A few KM, perhaps 10 or 20, but not 4,000. I could back that up with a thermal energy calculation of Mercury's mantle and compare it to annual solar energy it gets hit by, but I'm quite sure ...

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Firstly, a more general advice(something that was told to my class by my professor): The "opposite to the motion" direction of friction is not the best way to see it. In fact, nothing in physics should be viewed as being an absolute rule except from the very basic foundations of physics, which are its laws. One case in which friction is not opposite of ...

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If it doesn't slip, you can model it as rotating around the end of the rod (where it touches the ground). When it strikes the ground, this restriction means that you know both the angular speed of rotation and the speed of the center of mass. If there is no friction, then the rod will rotate around the center of mass. This changes the moment of inertia ...

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