A very light rigid rod with a length of 1.03 m extends straight out from one end
ID: 2039662 • Letter: A
Question
A very light rigid rod with a length of 1.03 m extends straight out from one end of a meter stick. The stick is suspended from a pivot at the far end of the rod and is set into oscillation.
Ip = ICM + MD2
(a) Determine the period of oscillation. [Suggestion: Use the parallel-axis theorem equation given above. Where D is the distance from the center-of-mass axis to the parallel axis and M is the total mass of the object.]
_______________ s
(b) By what percentage does the period differ from the period of a simple pendulum 1 m long?
Explanation / Answer
IP=ML2/12 +MD2 =M*12/12 +M*(1.03+(1/2))2
IP=2.424233M
Period of oscillation
T=2pisqrt(I/MgD] =2pi*sqimrt[2.424233M/M*9.81*(1.03+0.5)]
T=2.525 s
b)
Period With L=1 m
T=2pi*sqrt[1/9.81] =2.006 s
% difference =[(2.525-2.006)/2.006]*100
=25.87 %