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I have an Aluminium square cross sectional rod of $20mm$ X $20mm$. Length 700mm. At the end of this rod there is a mass of $660$ $g$ attached. My question is what torque do i need to lift this mass up assuming my pivot point is at the other extreme of the rod? And how do i consider the mass of the rod in the calculations!

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  • $\begingroup$ Do you mean to lift the mass like what cranes do ? I mean pure vertical or over an angle ? $\endgroup$ – Sam Farjamirad Oct 6 '18 at 9:48
  • $\begingroup$ Not exactly . The motor is lifting the rod as well as the mass(attached at the end) together over an angle unlike Cranes. Lets assume the the rod and the mass are initially horizontal to a plane. $\endgroup$ – Mojiz Oct 6 '18 at 9:57
  • $\begingroup$ You can consider the mass of the beam as a weightless beam and a point load, at the midpoint of beam $\endgroup$ – Jonathan R Swift Oct 6 '18 at 10:12
  • $\begingroup$ @JonathanRSwift Point load as in taking the mass of beam plus the additional mass at the end as one acting at midpoint? $\endgroup$ – Mojiz Oct 6 '18 at 10:20
  • $\begingroup$ No, You have to superpose the 660 g and the mass of the beam, if you want to work very accurate you can integrate over the mass of beam $\endgroup$ – Sam Farjamirad Oct 6 '18 at 10:34
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The angular momentum of a rod or beam rotating about one end is its angular momentum about the center of mass of the rod plus mass times (l/2)^2, in this case, parallel axis momentum theory. $$L = m\frac{l^2}{12} +m\times (l/2)^2 = ml^2/3 $$ Where m is the mass and l the length of the aluminum rod, 700mm.

So the torque needed is

$T= 1/3(ml^2) + 660l$

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