A uniform spherical shell of mass M = 17.0 kg and radius R = 0.890 m can rotate

Question

A uniform spherical shell of mass M = 17.0 kg and radius R = 0.890 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 = 0.0840 kg middot m^2 and radius r = 0.140 m, and is attached to a small object of mass m = 3.70 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 0.990 m after being released from rest? Use energy considerations.

Explanation / Answer

Here ,

let the speed of the object is v

as the string does not slip

angular speed of pulley = v/r

angular speed of sphere = v/R

Now , Using conservation of energy

0.5 *Isphere * wsphere^2 + 0.5 * I * wpulley + 0.5 * m * v^2 = m * g * h

0.5 * (2/3) * 17 * 0.890^2 * (v/0.890)^2 + 0.5 * 0.0840 * (v/0.140)^2 + 0.5 * 3.70 * v^2 = 3.70 * 9.8 * 0.990

solving for v

v = 1.93 m/s

the speed of the block after falling 0.990 m is 1.93 m/s

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