Physics quiz question: falling box, rotating disk

Physics 8A Quiz 9, spring semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Young and Freeman, University Physics, 11/e, Problem 9.86

A 3.0 kg solid disk of radius 0.80 m that is mounted on a horizontal frictionless axis through its center. A light string is wrapped around its rim, and a 1.0 kg box is attached to the other end of the string. The box is released from rest from an unknown height above the floor, and has a speed of 4.2 m/s just as it strikes the floor. Neglect friction and drag. The initial height of the box above the floor is:
(A) 0.45 m.
(B) 0.90 m.
(C) 1.4 m.
(D) 2.3 m.

Correct answer: (D)

No external work is done on this system, so the decrease in gravitational potential energy of the box is equal to the sum of the increases of the translational kinetic energy of the box, and the rotational kinetic energy of the disk. The translational speed of the box v is related to the angular speed of the disk via v = r·ω. With initial conditions v_i = 0, and ω_i = 0; final conditions v_f = r·ω_f, and y_f = 0; and the moment of inertia of the disk is I_disk = (1/2)·m_disk·r², the energy conservation transfer-balance equation is:

W_nc = ∆K_tr + ∆K_rot + ∆U_grav + ∆U_elas,

0 = (1/2)·m_box·(v_f² - 0²) + (1/2)·I_disk·(ω_f² - 0²) + m_box·g·(0 - y_i) + 0,

0 = (1/2)·m_box·v_f² + (1/2)·((1/2)·m_disk·r²)·(v_f/r)² - m_box·g·y_i,

0 = (1/2)·m_box·v_f² + (1/4)·m_disk·v_f² - m_box·g·y_i,

Solving for y_i:

y_i = ((1/2)·m_box + (1/4)·m_disk)·v_f²/m_box·g = 2.25 m,

or to two significant figures, 2.3 m.

Choice (A) stems from a sign error in the kinetic energy terms. Neglecting the rotational kinetic energy of the disk results in (B); neglecting the translational kinetic energy of the box results in (C).

Student responses:
(A) : 6 students
(B) : 12 students
(C) : 10 students
(D) : 2 students