Question

Greek engineers had the unenviable task of moving large columns from the quarries to the city. One engineer, Chersiphron, tried several different techniques to do this. One method was to cut pivot holes into the ends of the stone and then use oxen to pull the column. The 4-ft diameter column weighs 13200 lbs, and the team of oxen generates a constant pull force of 1500 lbs on the center of the cylinder G. Knowing that the column starts from rest and rolls without slipping, determine the velocity of its center G after it has moved 5 ft and the minimum static coefficient of friction that will keep it from slipping.

Answer 1

I believe this should be everything you need:)

Answer 2

The velocity is a vector quantity. The velocity of center G of the column after it has moved 5 ft is 5.18 ft.

From work Energy equation:

$U_{12} = T_2$

First, the potential energy of column,

$U_{12} = \int\limits^5_0 {F} \, dx$

Here, the column moves from zero and displaces 5 ft.

$F$ - force = 1500 lbs

Thus,

$U_{12 } = 7500\rm \ ft\ lbs$

Now, total kinetic energy at position 2,

$T_ 2 = \dfrac 12 mv_2^{-2} + \dfrac 12 \vec I \omega_2^2$

Where,

m  = 372.67 slugs

r = 2 ft

Since,

$\omega_2 = \dfrac {\vec V}{r}$

So,

$7500 = \dfrac {12000}{2g}v_2^{-2}+\dfrac {12000}{4g}r^2 \times \dfrac {\vec V}{r}\\\\\vec v_2 = 5.18 \rm \ ft$

Therefore, the velocity of center G of the column after it has moved 5 ft is 5.18 ft.

Learn more about the work Energy equation:

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