C) Consider a disk used as a pulley, with a very light rope wrapped many times a

Question

C) Consider a disk used as a pulley, with a very light rope wrapped many times around the disk and a box attached to the rope. The box has a mass M and starts at rest at a height h above the ground. The pulley has a mass M_P and radius R and is free to rotate without friction about its center. Its center is at a height H. Important: The rope unwinds off the pulley without slipping, so the velocity of the rope is equal to the velocity of the outer edge of the pulley. Considering the pulley, the rope, the box, and the Earth as your system, write down equations for the total energy of the system in terms of positions and velocities at the following times: i) Initially when the box is at a height h ii) When the box is half-way down, at height h/2 iii) Finally, right before the box hits the ground. D) If M = 0.5 kg, M_P = 2 kg, R = 1.5 m, h = 321.5 m and H = 26 m, what is the velocity of the box when it is at a height h/2? What is the velocity of the box just before it hits the ground?

Explanation / Answer

i)

Ui = initial Total energy = Potential energy of disc + potential energy of box

Ui = Mgh + Mp gH

ii)

v = linear speed of box

I = moment of inertia of disc = (0.5) Mp R2

w = angular speed of the disc = v/R

Uf = Potential energy of disc + potential energy of box + KE of disc + KE of box

Uf = Mgh/2 + Mp gH + (0.5) I w2 + (0.5) M v2

Uf = Mgh/2 + Mp gH + (0.5) ((0.5) Mp R2) (v/R)2 + (0.5) M v2

Uf = Mgh/2 + Mp gH + (0.25) ( Mp v2) + (0.5) M v2

iii)

U'f = Potential energy of disc + potential energy of box + KE of disc + KE of box

U'f = Mg(0) + Mp gH + (0.5) I w2 + (0.5) M v'2

U'f = Mp gH + (0.5) ((0.5) Mp R2) (v'/R)2 + (0.5) M v'2

U'f = Mp gH + (0.25) ( Mp v'2) + (0.5) M v'2

D)

at h1/2 = 21.5/2 m

using conservation of energy

Ui = Uf

Mgh + Mp gH = Mgh/2 + Mp gH + (0.25) ( Mp v2) + (0.5) M v2

Mgh/2 = (0.25) ( Mp v2) + (0.5) M v2

(0.5)(9.8 x 21.5)/2 = (0.25) ( (2) v2) + (0.5) (0.5) v2

v = 8.4 m/s

Just before hitting the ground :

using conservation of energy

Ui = U'f

Mgh + Mp gH = Mp gH + (0.25) ( Mp v'2) + (0.5) M v'2

Mgh = (0.25) ( Mp v'2) + (0.5) M v'2

(0.5)(9.8 x 21.5) = (0.25) ( (2) v'2) + (0.5) (0.5) v'2

v' = 11.85 m/s

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