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=
Tfixed=2*pi*((8*l^2+3*R^2)/(9*g*l))^(1/2)
but i haven't yet checked my answer for Tfree.
Tfree=2*pi*((8*l^2)/(9*g*l))^(1/2)
To find the period of a pendulum, we can use the equation:
T = 2π√(I/mgd)
where T is the period, I is the moment of inertia, m is the mass, g is acceleration due to gravity, and d is the distance of the center of mass from the pivot point.
For a physical pendulum, the moment of inertia is given by:
I = Icm + mR^2
where Icm is the moment of inertia of the disc about its center of mass, and mR^2 is the moment of inertia of the rod about its pivot point.
(a) For small angles of oscillation, we can assume that the center of mass of the pendulum stays at a fixed distance from the pivot point. Therefore, the distance d is equal to the length of the rod, l. The moment of inertia of the disc about its center of mass and the moment of inertia of the rod about its pivot point can be calculated using the known formulas.
Icm = (1/2)mR^2
mR^2 = (1/3)mR^2 (for a rod rotating about one of its ends)
Substituting these values into the equation for the period, we get:
Tfixed = 2π√((1/2)mR^2 + (1/3)mR^2)/(mgl)
Simplifying this expression, we have:
Tfixed = 2π√((5/6)mR^2)/(mgl)
Tfixed = 2π√((5/6)R^2)/(gl)
So, the period of the pendulum for small angles of oscillation is Tfixed = 2π√((5/6)R^2)/(gl).
(b) If the disc is mounted to the rod by a frictionless bearing so that it is perfectly free to spin, the moment of inertia changes. In this case, the moment of inertia of the disc about its center of mass and the moment of inertia of the rod about its pivot point are added as they are no longer fixed together.
Icm = (1/2)mR^2
mR^2 = (1/3)mR^2
Substituting these values into the equation for the period, we get:
Tfree = 2π√((1/2)mR^2 + (1/3)mR^2)/(mgl)
Simplifying this expression, we have:
Tfree = 2π√((5/6)mR^2)/(mgl)
Tfree = 2π√((5/6)R^2)/(gl)
So, the new period of oscillation when the disk is mounted to the rod by a frictionless bearing is Tfree = 2π√((5/6)R^2)/(gl).