Well, they tell you what the period is
45.8/20 = 2.29 seconds
However that is not the point
d = pivot to center (CG) (I think it is 30 cm to the center of the meter stick but they do not say that)
force down at CG = weight = Mg
torque about pivot = Mg d sin theta
for small angle sin theta = theta in radians
so
torque = -M g d theta
torque = I alpha
if theta = A sin w t
then alpha =-Aw^2 sin wt
then
Torque = -I A w^2 sin w t
where w = 2 pi/T
M g d A sin w t = I w^2 A sin w t
M G d = I w^2
w = sqrt(M g d/I)
2 pi/T = sqrt (M g d/I)
T = 2 pi sqrt (I/Mgd)
I am calling your I20 simply I
a meter rule is suspended from a horizontal pin through the 20 cm mark and allowed to oscillate (with simple harmonic motion). In this position, the ruler completes 20 oscillations in 45.8 seconds.
derive an expression for the period T in terms of I20, M, g and d where I20 is the moment of inertia about an axis through 2o cm mark, g is the acceleration due to gravity and d is the distance between the point of suspension and the center of mass...
I AM AT A LOSS as even where to start
2 answers
By the way, you know d = 30 from the geometry of the meter stick
You can compute I20 assuming the meter stick is uniform in mass distribution along its length.
That I20 will have M in it, but it cancels the M in the period equation
Use g = 9.8
Then you can actually compute T in seconds.
You can compute I20 assuming the meter stick is uniform in mass distribution along its length.
That I20 will have M in it, but it cancels the M in the period equation
Use g = 9.8
Then you can actually compute T in seconds.