A uniform spherical shell of mass M = 4.5 kg and radius R = 8.5 cm can rotate ab

Question

A uniform spherical shell of mass M = 4.5 kg and radius R = 8.5 cm can rotate about a vertical axis on frictionless bearings. A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 3.0 × 10-3 kgm2 and radius r = 5.0 cm, and is attached to a small object of mass m = 0.60 kg. There is no friction on the pulley’s axle; the cord does not slip on the pulley.

a. What is the speed of the object when it has fallen 82 cm after being released from rest? Use energy considerations.

b. What is the angular acceleration of the sphere?(adapted from Problem 66, Chapter 10, Fundamentals of Physics 8th ed. by Jearl Walker, Wiley)

Explanation / Answer

  Initially, the only energy in the system is the potential energy of the hanging block.
PE = m × g × h
PE = (0.6 kg)×(9.8 m/s)×(0.82 m)
PE = 4.8216 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 = (0.6 kg) × V² / 2
KE = (0.3 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 = (0.003 kgm²) × V² / 2(0.05 m)²
RE = (0.6 kg) × V²

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

so
I = 2 × (4.5 kg) × (0.085 m)² / 3
I = 0.021675 kgm²
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 = ( 0.021675 kgm²) × V² / 2(0.085 m)²
RE = (1.5 kg) × V²

Applying conservation of energy, you get:
PE = KE + RE + RE
(4.8216 J) = (0.3 kg)×V² + (0.6 kg)×V² + (1.5 kg)×V²
(4.8216 J) = (2.4 kg)×V²
V² = 2.009
V = 1.417 m/s
V = 1.42 m/s

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