A physical pendulum consists of a disc of radius R and mass m fixed at the end of a rod of mass m and length l .
(a) Find the period of the pendulum for small angles of oscillation. Express your answer in terms of m, R, l and acceleration due to gravity g as needed (enter m for m, R for R, l for l, g for g and pi for π).
Tfixed=
(b) For small angles of oscillation, what is the new period of oscillation if the disk is mounted to the rod by a frictionless bearing so that it is perfectly free to spin? Express your answer in terms of m, R, l and acceleration due to gravity g as needed (enter m for m, R for R, l for l, g for g and pi for π).
Tfree=
http://phys.thu.edu.tw/~ycchen/physics/ebook-8.01x/chapter24_physical_pendulum_v03.pdf
(a) To find the period of the pendulum for small angles of oscillation, we can use the formula for the period of a physical pendulum:
T = 2π * √(I / (mgh))
where T is the period, I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and h is the distance from the center of mass to the axis of rotation.
In this case, the moment of inertia I of the pendulum can be calculated as the sum of the moment of inertia of the rod and the moment of inertia of the disk:
I = I_rod + I_disk
The moment of inertia of a rod about its end is given by:
I_rod = (1/3) * m * l^2
The moment of inertia of a disk about its center of mass is given by:
I_disk = (1/2) * m * R^2
Since the disk is fixed at the end of the rod, the center of mass of the disk coincides with the end of the rod. Therefore, the distance from the center of mass to the axis of rotation is equal to the length of the rod, l.
Substituting these values into the formula for the period, we have:
Tfixed = 2π * √((1/3) * m * l^2 + (1/2) * m * R^2) / (m * g * l)
Simplifying the expression further, we get:
Tfixed = 2π * √((l^2 + (3/2) * R^2) / (3 * g * l))
So, the period of the pendulum for small angles of oscillation is Tfixed = 2π * √((l^2 + (3/2) * R^2) / (3 * g * l)).
(b) When the disk is mounted to the rod by a frictionless bearing, it is perfectly free to spin. In this case, the moment of inertia of the system changes. The moment of inertia of the rod remains the same, but the moment of inertia of the disk changes because it can freely rotate.
The moment of inertia of a freely rotating disk about its center of mass is given by:
I_disk = (1/2) * m * R^2
Since the disk is now free to rotate, the center of mass of the disk is no longer fixed at the end of the rod. Therefore, the distance from the center of mass to the axis of rotation is the radius of the disk, R.
Substituting these values into the formula for the period, we have:
Tfree = 2π * √((1/3) * m * l^2 + (1/2) * m * R^2) / (m * g * R)
Simplifying the expression further, we get:
Tfree = 2π * √((2 * l^2 + 3 * R^2) / (6 * g * R))
So, the new period of oscillation when the disk is free to spin is Tfree = 2π * √((2 * l^2 + 3 * R^2) / (6 * g * R)).