**The Rockefeller Christmas Tree**
Rotational Dynamics:

What would be the moment of inertia of this pulley as it
raises the star for the Rockefeller Christmas tree? (Pulley rotates about its
center, initially at rest)

Assumptions:

Pulley
radius = .381 meters

Frictional
Torque = t

_{fr}= 2.50 mN
Force
Applied to cord = 2000 N

Angular
speed = 15.0 rad/s

^{2}
St = RF

_{T}- t_{fr}
= (.381 m)(200.0
N) – (5.50 mN) = 756.5 mN

I = St/a

= (73.7 mN) / (15.0
rad/s

^{2}) = 50.43 kg m^{2}
Using the moment of inertia we can now determine the angular
acceleration of the pulley, and finally the linear acceleration of the star

Assumptions:

Mass of
star: 250 kg

a =
(mgR - t

_{fr}) / (I + mR^{2})
=

__(2000 N)(.381 m) – (5.50 mN)__
(50.43 kg m

^{2}) + (250 kg)(.381m^{2})
= 5.19 rad/s

^{2}
Linear Acceleration of star:

a = Ra

= (0.381 m)(5.19
rad/s

^{2})
= 1.98 m/s

^{2}
This answer
makes sense, the set up crew would want to move this star very slowly as to
prevent any swaying or damages to the crystal or its surroundings!

Stress and Fracture

This year’s tree is from Connecticut, with a trunk about 4
feet in diameter and 75 feet high. How much would the tree shorten once the 200
N compressive force of the star is exerted on the tree? (Assuming long uniform
rod)

Length of tree = 75 feet = 22.9 meters

Cross sectional area = p(.61m)

^{2}= 1.17 m^{2}
Compressive Strength of wood parallel to grain = 35 x 10

^{6 }N/m^{2}
F/A = (2000N)/(1.17) = 1709 N/m

^{2 }
(which is less than the compressive
strength so it will not break)

L =

__L___{o}__F__
E A

= [(22.9m) / (10 x 10

^{9}N/m^{2})] x 1709 N/m^{2}
=3.91 x 10

^{-6}m
Therefore the tree will shrink negligibly when the star is
mounted.

Sources:

http://www.nydailynews.com/new-york/rockefeller-center-christmas-tree-lights-city-article-1.1537918

http://www.swarovski.com/Web_US/en/crystal_society?contentid=10007.229549

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