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armis
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Homework Statement
An 8m and 200kg telephone post is laying on the ground. What work one needs to do to reposition the post vertically?
Homework Equations
The Attempt at a Solution
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Yep, we actually already did that. Read the previous postscalef said:Why even bother with torque?
If you lift this log the 4 requisite meters and put in a magical frictionless bearing in the center, every orientation of the log has equal potential energy. So isn't the solution the gravitational potential of the center of mass at 4 meters?
Any solution involving torque would include the inertia of the pole about some hinge, which, although not unnecessarily complicated, is far more complicated than the gravitational potential formula.
armis said:Thanks dx
Any book which explains torgue in detail? I imagine all on classical mechanics must have an explanation, just tell me
dx said:Now consider some particle on the plane with a force [tex] (F_x, F_y) [/tex] on it. It you now rotate the particle about the origin by and angle [tex] \Delta \theta [/tex], the work done on it can be shown to be
[tex](xF_{y} - yF_{x})\Delta \theta[/tex].
dx said:In the case of your problem, the torque is [tex] - \frac{l}{2} mg cos\theta [/tex] where [tex] \theta [/tex] is the angle between the ground and the rod. If you integrate minus this from 0 to pi/2, you get the answer.
Displace the particle by a small angle [tex] \Delta \theta [/tex], and the find the small displacements in the x and y coordinates in terms of this. If these are [tex] \Delta x [/tex] and [tex] \Delta y [/tex], then the work will be [tex] F_{x} \Delta{x} + F_{y} \Delta{y} [/tex].armis said:How can it shown to be [tex](xF_{y} - yF_{x})\Delta \theta[/tex] ?
armis said:why is it l/2 if the force is acting at the end of the pole?Is it because of the mass distribution?If it wouldn't be symmetric, then what?
thanks
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