Question

A thin, round disk made of acrylic plastic with a uniform density of 1.1g/cm3 is 20.0 cm in diameter and 1.0 cm thick. A very small hole is drilled through the disk at a point =8.0 cm from the center. The disk is hung from the hole on a nail and set into simple harmonic motion on the surface of Earth with a maximum angular displacement (measured from vertical) of theta=7.0∘.

Calculate the period T of the motion

Answer 1

The period of the simple harmonic motion of the acrylic disk is approximately 0.80 seconds.

Step-by-step explanation:

To calculate the period T of the simple harmonic motion of the acrylic disk, we can use the formula:

T = 2π√(I/mg)

Where I is the moment of inertia, m is the mass, and g is the acceleration due to gravity. The moment of inertia for a thin disk rotating about an axis through its center is given by the formula:

I = 0.5mr2

Substituting the values given in the question, we have:

T = 2π√((0.5mr2)/(mg))

Simplifying further:

T = 2π√(0.5r/g)

Plug in the values: r = 10.0 cm = 0.10 m, and g = 9.8 m/s2:

T = 2π√(0.5(0.10)/(9.8))

T = 2π√(0.0051)

T ≈ 0.80 seconds

Answer 2

The period of a physical pendulum can be calculated by finding the moment of inertia about the pivot point, using the mass of the disk, the radius, and the distance from the pivot to the disk's center in the equation for the period of SHM.

Step-by-step explanation:

The student is dealing with a simple harmonic motion (SHM) problem in physics where the period of the oscillation is being sought. Based on the information provided, and assuming small angular displacements, the physical pendulum model can be applied to solve for the period T. The period of a physical pendulum can be calculated using the formula T = 2π√(I/mgh), where I is the moment of inertia of the disk about the pivot point, m is the mass of the disk, g is the acceleration due to gravity, and h is the distance from the pivot to the center of mass of the disk.

For a uniform disk, the moment of inertia about an axis through its center is I_center = 0.5 * m * r^2, where r is the radius of the disk. To find the moment of inertia about the pivot point, we use the parallel axis theorem, I = I_center + m * d^2, with d being the distance from the center to the pivot point, which is given here as 8.0 cm or 0.08 m. The radius of the disk is half its diameter, so r = 10 cm or 0.1 m.

After calculating the mass of the disk using its volume (V = π * r^2 * thickness) and density (ρ = mass/V), and thus obtaining the moment of inertia about the pivot point, one can substitute all values into the period formula to find T. The surface area of the disk and its round shape play an essential role in calculating the moment of inertia, which is directly related to T.