# Tag Info

## Hot answers tagged rotational-kinematics

3

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 ...

1

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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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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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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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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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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