A uniform spherical shell of mass M = 11.0 kg and radius R = 0.880 m can rotate

Question

A uniform spherical shell of mass M = 11.0 kg and radius R = 0.880 m can rotate 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 = 0.130 kg-m^2 and radius r = 0.0910 m, and is attached to a small object of mass m = 4.90 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 when it has fallen a distance 1.43 m after being released from rest? Use energy considerations. Number 1.0696 Units m/s

Explanation / Answer

change in potential energy = change in kinetic energies of all the objects

v is velocity of block

kinetic energy of shell = 0.5(2/3)(M)r2w2 = 0.5(2/3)(M)(v2) =

kinetic energy of pulley = 0.5Iw2 = 0.5I(v2/r2)

kinetic energy of block = 0.5Mv2

change in potential energy = mgh = (4.9)(9.8)(1.43) = 68.67 J

68.67 = 3.67v2 + 7.85v2 + 2.45v2

v2 = 68.67/13.97 = 4.92

=> v = 2.22m/sec

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