- #1
Daniiel
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For part a
We know Icm = m/12 (L^2 + W^2) where W is width and L is length
using parallel axis theorem to find the I with the pivot L/4 down the ruler
I = Icm + m(L^2/16)
so then with the I we find the torque
T = IO O = theta double dot (d^2 theta / dtheta ^2)
then the torque would be = L/4 (-mg sin(theta))
so IO = L/4 (-mg sin(theta)) for small theta sin theta > theta
then the sin term is omega squared
so
T = 2pi/omega and yea that's the period
is that correct?
im not sure if the parallel axis theorem is correct.
and for B, how do you find the new inertia with the mass on the top of the ruler? is it the same kind of thing?
Uploaded with ImageShack.us
For part a
We know Icm = m/12 (L^2 + W^2) where W is width and L is length
using parallel axis theorem to find the I with the pivot L/4 down the ruler
I = Icm + m(L^2/16)
so then with the I we find the torque
T = IO O = theta double dot (d^2 theta / dtheta ^2)
then the torque would be = L/4 (-mg sin(theta))
so IO = L/4 (-mg sin(theta)) for small theta sin theta > theta
then the sin term is omega squared
so
T = 2pi/omega and yea that's the period
is that correct?
im not sure if the parallel axis theorem is correct.
and for B, how do you find the new inertia with the mass on the top of the ruler? is it the same kind of thing?
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