Two physical pendulums (not simple pendulums) are made from meter sticks that are suspended from the ceiling at one end. The sticks are uniform and are identical in all respects, except that one is made of wood (mass = 0.17 kg) and the other of metal (mass = 0.85). They are set into oscillation and execute simple harmonic motion. Determine the period of (a) the wood pendulum and (b) the metal pendulum.
The answer is 1.64 s for both. Can someone show the steps to see how to get this?
To determine the period of a physical pendulum, you need to use the equation for the period of a physical pendulum:
T = 2π√(I / mgd)
where T is the period of the pendulum, I is the moment of inertia of the pendulum about its pivot point, m is the mass of the pendulum, g is the acceleration due to gravity, and d is the distance from the pivot point to the center of mass of the pendulum.
In this case, both pendulums are made from meter sticks, so the length (d) of each pendulum is 1 meter. Furthermore, since both pendulums are uniform and identical in all respects except for their masses, we can assume that they have the same shape and distribution of mass. Therefore, for simplicity, we can assume that the pivot point is at the center of mass for both pendulums.
Now, let's calculate the moment of inertia for each pendulum.
For a physical pendulum with a small oscillation angle, the moment of inertia about the pivot point is given by:
I = (1/3)ml^2,
where m is the mass of the pendulum and l is the length of the pendulum.
(a) For the wood pendulum:
m = 0.17 kg
l = 1 m
Substituting these values into the equation for the moment of inertia, we get:
I_wood = (1/3)(0.17 kg)(1 m)^2 = 0.0567 kg.m^2
Now, substituting the values of I_wood, m, g, and d into the equation for the period, we have:
T_wood = 2π√(0.0567 kg.m^2 / (0.17 kg)(9.8 m/s^2)(1 m)) = 1.64 s
(b) For the metal pendulum:
m = 0.85 kg
l = 1 m
Similarly, we can calculate the moment of inertia for the metal pendulum:
I_metal = (1/3)(0.85 kg)(1 m)^2 = 0.2833 kg.m^2
Substituting the values of I_metal, m, g, and d into the equation for the period, we get:
T_metal = 2π√(0.2833 kg.m^2 / (0.85 kg)(9.8 m/s^2)(1 m)) = 1.64 s
Therefore, the period of both the wood and metal pendulums is 1.64 seconds.