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.

Maybe the problem lies in the fact that you used cm...instead of meters. So it should be 0.32^2 and not just 32^2.

The problem lies in the intertia equation. You found the intertia of a disk around its center, not around its rim, as the problem calls for.

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.

To find the correct answer for part (a), you need to consider the rotational inertia of the disk when it's rotating about a point on its rim, not its center. The correct formula for the rotational inertia of a uniform disk rotating about an axis through a point on its rim is:

I = 1/2 * MR^2 + MR^2

where M is the mass of the disk and R is its radius. Notice that the first term represents the rotational inertia of the disk about its center (which you calculated correctly), and the second term represents the additional rotational inertia when rotating about the rim. This second term is obtained by using the parallel-axis theorem, which states that the rotational inertia about any parallel axis is equal to the rotational inertia about the center plus the product of the mass and the square of the distance between the axis and the center.

So, using this corrected formula for the rotational inertia, let's calculate the period of oscillation for part (a). Remember to convert the radius to meters for consistency in units:

R = 32.0 cm = 0.32 m

I = 1/2 * (1 kg) * (0.32 m)^2 + (1 kg) * (0.32 m)^2
= 0.1088 kg*m^2 + 0.1088 kg*m^2
= 0.2176 kg*m^2

Now we can substitute the values into the period equation:

T = 2π * sqrt(I / mgh)
= 2π * sqrt(0.2176 kg*m^2 / (1 kg * 9.8 m/s^2 * 0.32 m))
≈ 2.810 seconds

So, the correct answer for part (a) is approximately 2.810 seconds.

Now, moving on to part (b), you need to find the radial distance r from the center of the disk at which you can suspend it to achieve the same period as in part (a). To do this, we'll use the formula mentioned in the hint:

T = 2π * sqrt(Icom + mr^2) / mgr

Where Icom is the rotational inertia for rotation about an axis through the center of mass. Since we have a uniform disk, the rotational inertia about the center of mass is given by:

Icom = 1/2 * MR^2

Substituting the values:

T = 2π * sqrt(1/2 * (1 kg) * (0.32 m)^2 + (1 kg) * r^2) / (1 kg * 9.8 m/s^2 * r)

Simplifying the equation, we have:

2.810 = sqrt(0.1088 + r^2) / (9.8 * r)

Squaring both sides and rearranging, we get:

(2.810 * 9.8 * r)^2 = 0.1088 + r^2

Taking the square of 2.810 * 9.8 * r on the left side:

(2.810 * 9.8 * r)^2 - r^2 = 0.1088

Simplifying further:

(79.924r)^2 - r^2 = 0.1088

6301.638r^2 - r^2 = 0.1088

Combining like terms:

6300.638r^2 = 0.1088

Now, divide both sides by 6300.638:

r^2 = 0.1088 / 6300.638

Taking the square root of both sides:

r ≈ √(0.1088 / 6300.638)

r ≈ 0.0053 m

Finally, convert the radial distance r to centimeters:

r ≈ 0.0053 m * 100 cm/m ≈ 0.53 cm

So, the correct answer for part (b) is approximately 0.53 cm.