A uniform spherical shell of mass M = 5.50 kg and radius R = 0.920 m can rotate

Question

A uniform spherical shell of mass M = 5.50 kg and radius R = 0.920 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 = 0.170 kg

A uniform spherical shell of mass M = 5.50 kg and radius R = 0.920 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 = 0.170 kg·m2 and radius r = 0.0700 m, and is attached to a small object of mass m = 3.40 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 0.600 m after being released from rest? Use energy considerations

Explanation / Answer

use the law of conservation of energy as

the sum of KE's   in two rotation bodies   = loss in PE - KE of the mass

so Moment of Inertia Is = 2/3 mR^2   = 2/3 * 5.5 * 0.92*0.92

Is = 3.103 Kgm^2

Moment of inertial Ip = 0.170 kgm^2

so Eenrgy

0.5 Is Ws^2   + 0.5 Ip Wp^2    = mgh - 0.5 mv^2

Ws = V/R     and Wp = V/r

so

0.5 * 3.103 * (V/0.92)^2    + 0.5 * 0.170 * (v/0.07)^2   = (3.4 * 9.8 * 0.6) -(0.5 * 3.4*v^2)

1.833v^2 + 17.34 v^2    = 19.99 -(1.7 v^2)

20.873 v^2 = 19.99

V^2 = 0.957

V = 0.978 m/s ---------------------<<<<<<<<<<<<<<<Answer

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