13. A uniform spherical shell of mass M = 2.0 kg and radius R = 16.0 cm rotates

Question

13.

A uniform spherical shell of mass M = 2.0 kg and radius R = 16.0 cm rotates about a vertical axis on frictionless bearings (see the figure). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 2.32

A uniform spherical shell of mass M = 2.0 kg and radius R = 16.0 cm rotates about a vertical axis on frictionless bearings (see the figure). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 2.32 A 10^-3 kg m^2 and radius r = 4.0 cm, and its attached to a small object of mass m = 4.0 kg. There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object after it has fallen a distance h = 1.5 m from rest: Use work - energy considerations.

Explanation / Answer

?U = m g ?h .

The change in kinetic energy has three terms

kinetic energy of the rotating spherical shell, kinetic energy of the rotating pulley and kinetic energy of the falling mass

?K = 1/2 I_shell (?_sh)^2+ 1/2 I_pulley (?_p)^2 + 1/2 m v^2

Because of no slip (and no extension of the cord) w_sh = v/R_sh and likewise for the pulley

?K = ( 1/2 I_shell / R^2 + 1/2 I_pulley/r^2 + 1/2 m ) v^2

Energy conservation then gives

?K = -?U

v^2 = -mg?h / ( 1/2 I_shell / R^2 + 1/2 I_pulley/r^2 + 1/2 m )

I_shell = 2 / 3 (MR^2)

v^2 = -mg?h / ( 1/3 M_shell + 1/2 I_pulley/r^2 + 1/2 m )

v^2 = -4 * 9.8 * -1.5 / ((1/3) * 2 + (0.5) * 2.32 * 10^-3 / 0.04^2 + 0.5 * 4)

v = 4.164 m/s

speed of the object after it has fallen a distance h = 1.5 m from rest = 4.164 m/s

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