A uniform circular disk whose radius R is 32.0 cm is suspended as a physical pendulum from a point on its rim.
(a) What is its period of oscillation? __ s
(b) At what radial distance r < R is there a point of suspension that gives the same period? __ cm
in the book.. it gives me a hint:
..... the period of oscillation is given by T = 2pi sqrt(I/mgh), where I is the rotational inertia of the disk, 'm' is its mass, and 'h' is the distance from the center of mass to the pivot point. In this case 'h' is a radius of the disk.
Use the rotational inertia for rotation about an axis through the disk center and use the parallel-axis theorem to find the rotational inertia for rotation about a parallel axis that is a distance 'R' away.
To find the position of another pivot point for which the period is the same, solve T = 2pi sqrt(Icom + mr^2)/mgr) for 'r'. Here 'Icom' is the rotational inertia for rotation about an axis through the center of mass.
.. so i tried this and i got it wrong:
I = 1/2MR^2
512 = 1/2(1)(32cm)^2
..(in this case, would i even need to use a mass??)
and then..
T = 2pi sqrt(I/mgh)
T = 2pi sqrt((512)/(1)(9.8m/s^2)(32cm)
T = 8.03
..and that's wrong for (a).
i still havent gotten to (b) yet. may i get some assistance on how to work this problem please? thanks..
IN the period equation, you used I of the disk when rotating about the center. You did not convert that to a new piviot point with the parallel axis thm. Mass will divide out in the period equation, it does not matter the value.
2 answers
The equation for inertia should be:
I = 0.5mr^2 + mr^2
Therefore the coefficient when you solve should be 1.5 and not 0.5. Marianne was right when she said to convert centimeters to meters, and you're right that you don't need the mass.
It cancels out because the equation for a physical pendulum reads:
T = 2*pi * sqrt(I/mgh), and I has masses in both terms, cancelling with the m in mgh.
1 answer