Posted by **Rhi** on Saturday, June 2, 2012 at 9:45am.

A uniform spherical shell of mass M = 19.7 kg and radius R = 1.62 m rotates about a vertical axis on frictionless bearings. A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 1.02 kg m2 and radius r = 0.63 m, and is attached to a small object of mass m = 6.6 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 falls a distance h = 4.63 m from rest? Use energy considerations.

- Physics -
**Elena**, Saturday, June 2, 2012 at 11:20am
Assume that I1=2MR²/3 and ω1 are the momemt of inertia and

angular velocity of the sphere of radius R;

I and ω2 are the momemt of inertia and

angular velocity of the disc of radius r;

m and v are the mass and linear velocity of the load,

and ω1•R= ω2•r = v.

From the energy conservation law

m•g•h = m•v²/2 + I1•ω1²/2 + I2•ω2²/2 = m•v²/2 +{(1/2)•(2MR²/3) •(v/R)² } + 1/2•I•(v/r)² =

= m•v²/2 +Mv²/3 +I•v² / 2• r².

v =sqrt[m•g•h/(m/2 +M/3 + I/2r²)] = 5.18 m/s.

(check my calculations)

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