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Angular Momentum?
The hour and minute hands of Big Ben, the famous Parliament Building tower clock in London, are 2.70 m and 4.60 m long and have masses of 61.0 kg and 101 kg, respectively. Calculate the total angular momentum of these hands about the center point. Treat the hands as long thin rods
Alright, so sorting out the data:
The Big Hand
Length=4.60 m
m=101 kg
The Small Hand
Length=2.70 m
m=61.0 kg
I assume I can also calculate the average angualr speed:
angular speed (large hand) = (pi)/21600;
angular speed (small hand) = (pi)/1800
L = Iw
Since it's a long thin rod with the axis of rotation through the end, I used I = (1/3)*M*(L^2) for each
I (large) = (1/3)*(101)*(4.60^) = 712.3866....
I (small) = (1/3)*(61)*(2.70^2) = 148.23
Iw (large) = .1036124407
Iw (small) = .258710155
Adding them gives me .3623225955 which is not the correct answer. Where am I going wrong?
Any long hand of a clock like Big Ben (or of a two-handed watch on your wrist) must be 2pi radians per 1 hour, i.e. W=2*3.1415926/3600 (rad/s).
Any short hand covers the same 2pi in 12 hours. That means that its angular speed is 12 times lower than that of the long hand, i.e. w=W/12.
If the hand is L meters long and its mass = M kg, then its I=integral[from 0 to L] of ((M/L)*x^2*dx)=m*L^2/3, thus A1=I1*W is angular momentum of long hand, A2=I2*w is angular momentum of short hand, A=A1+A2 is total angular momentum.
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