A uniform spherical shell of mass M = 14.0 kg and radius R = 0.480 m can rotate

Question

A uniform spherical shell of mass M = 14.0 kg and radius R = 0.480 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 = 0.240 kg·m2 and radius r = 0.130 m, and is attached to a small object of mass m = 3.00 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 1.26 m after being released from rest? Use energy considerations.

Explanation / Answer

Initially, the only energy in the system is the potential energy of the hanging block.
PE = m × g × h
PE = (3 kg)×(9.8 m/s)×(1.26 m)
PE = 37.044 J

After falling a distance h, that energy is converted to
* kinetic energy of the block
* rotational energy of the pulley
* rotational energy of the sphere

Let V be the speed of the block at that time.

The kinetic energy of the block is:
KE = m × V² / 2
KE = (3 kg) × V² / 2
KE = (1.5 kg) × V²

The rotational energy of the pulley is:
RE = I × ² / 2
where
= V / r
so
RE = I × (V/r)² / 2
RE = I × V² / 2r²
RE = (0.240 kgm²) × V² / 2(0.130 m)²
RE = (14.20 kg) × V²

The moment of inertia of a uniform spherical shell is:
2 × M × R² / 3

so
I = 2 × (14kg) × (0.480 m)² / 3
I = 2.1504kgm²
The rotational energy of the sphere is:
RE = I × ² / 2
where
= V / R
so
RE = I × (V/R)² / 2
RE = I × V² / 2R²
RE = (2.1504 kgm²) × V² / 2(0.480 m)²
RE = (4.666 kg) × V²

Applying conservation of energy, you get:
PE = KE + RE + RE
( 37.044 J J) = (1.5 kg)×V² + (14.20 kg)×V² + (4.666 kg)×V²
( 37.044 J J) = (1.9667 kg)×V²
V² = 1.81
V = 1.345 m/s
V = 1.3 (to 2 sf) m/s < - - - - - - - - - - - - - - - answer

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