A crate, A , with weight WA = 385 lb is hanging from a rope wound around a unifo

Question

A crate, A , with weight WA = 385 lb is hanging from a rope wound around a uniform drum, D, of radius r = 1.1 ft, weight WD = 112 lb, and center C. The systems is initially at rest when the restraining system holding the drum stationary fails, thus causing the drum to rotate, the rope to unwind, and. consequently. the crate to fall. Assuming that the rope does not stretch or slip relative to the drum and neglecting the inertia of the rope, determine the speed of the crate 1.5 s after the system starts to move, After drawing FBDs for the crate and drum, write a linear impulse/momentum equation for the crate and an angular impulse/momentum equation for the drum. Note that, if the rope is unwinding without slip, the relationship between the speed of the crate and the angular speed of the drum is v = r omega.

Explanation / Answer

Let tension in rope be T.

Torque = T*r = I

Moment of inertia of drum I = 1/2*mdrumr2

Thus, T*r = 1/2*mdrumr2

Thus, angular acceleration of drum, = 2T/(mdrumr)

Linear acceleration of crate a = r

Thus a = 2T/mdrum

For equilibrium of crate, mcrateg-T = mcratea

Thus, T = mcrate(g-a)

Thus, a = 2mcrate(g-a)/mdrum

Thus a = 2mcrateg/(mdrum+2mcrate)

Putting values, a = 2*385*32.2/(112+2*385) = 28.11 ft/s2

Now using, v = u + at

v = 0 + 28.11*1.5 = 42.17 ft/s

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