moment of inertia help needed??

Question

imagine an object of mass m oscillating, through a small angle, around an axis at point A, which is located a distance D away from the object's centre of mass, cm.
T is the period of oscillation.

how am i meant to show that this object's moment of inertia through the centre of mass, and parallel to the A axis, is given by

I = (MD/4π^2)(gT^2 - 4Dπ^2)

i havent got a clue how to do this, can it be clearly explained please?

thanks.

Answer 1

Let:
t be the angle of inclination to the vertical (small enough to use approximation t = sin(t)),
I1 be the moment of inertia about the axis of suspension.

The equation of motion is:
MgD sin(t) = - I1 t''
approximating to:
MgD t = - I1 t''

Therefore:
t'' = - ( MgD / I1) t

This yields SHM with:
T^2 = 4pi^2 I1 / (MgD) ..(1)

The parallel axis theorem gives:
I1 = I + MD^2

Substituting for I1 in (1):
T^2 = 4pi^2(I + MD^2) / MgD

MgDT^2 = 4pi^2I + 4pi^2 MD^2
I = (MgDT^2 - 4pi^2 MD^2 ) / 4pi^2
= (MD / 4pi^2)(gT^2 - 4D pi^2).

Answer 2

Sorry, not a complete answer.

The period of a pendulum with a small angle of oscillation is usually given by the equation

T = 2 pi sqrt ( I / mgd )

where T is the period, I the moment of inertia, m the mass, g the local acceleration of gravity, and d the distance between the center of mass and the pivot point.

Re-arranging this equation gives you the first part of your answer.

However, the I don't see where second term (-1/2 m d^2) comes from. It doesn't look like it could arise from the second term in the small-angle approximation and you don't have the displacement anyway. Sorry, that's all I can give you - hope it helps.

Answer 3

Frictional torque: PHI Torque utilized via the exterior agent with the clarification for rushing it up: tau Newton's 2d regulation for the time of velocity-up era: tau - PHI = I*alpha_speedup Newton's 2d regulation for the time of decelerate era: PHI = I*alpha_slowdown to locate alpha_speedup, basically use the definition of angular acceleration: alpha_speedup = omega_max/t_speedup And do an identical for alpha_slowdown: alpha_slowdown = omega_max/t_slowdown build device of two equations, 2 unknowns: tau - PHI = I*omega_max/t_speedup PHI = I*omega_max/t_slowdown substitute on an identical time: tau - (I*omega_max/t_slowdown) = I*omega_max/t_speedup deliver at the same time: tau = I*omega_max/t_speedup + I*omega_max/t_slowdown element: tau = I*omega_max*(a million/t_speedup + a million/t_slowdown) clean up for I: I = tau/(omega_max*(a million/t_speedup + a million/t_slowdown)) Simplify: I = tau*t_speedup*t_slowdown/(omega_max*(t_s... Re-substitute: PHI = (tau*t_speedup*t_slowdown / (omega_max*(t_slowdown + t_speedup)) )*omega_max / t_slowdown Simplify: PHI = tau*t_speedup/(t_slowdown + t_speedup) precis of equations offering effects: A) I = tau*t_speedup*t_slowdown / (omega_max*(t_slowdown + t_speedup)) B) PHI = tau*t_speedup/(t_slowdown + t_speedup) suggestion (translate gadgets on omega_max): tau:= 50 N-m; t_speedup := 25 sec; t_slowdown:=a hundred and twenty sec; omega_max:=620*Pi/30 rad/sec; effects: I = 15.ninety 3 kg-m^2 PHI = 8.621 Newton-meters