Calculation of Oscillation Period- Short Problem

[BDM333]

Homework Statement

You drill a hole at the end of a solid thin rod of length 1.3 m, and then attach the rod through this hole onto an approximately frictionless pivot. You then release the rod from rest at a reasonably small angle from vertical, so that it executes simple harmonic motion.

L=1.3m

a.) What will be its oscillation period? _____ seconds

b.) If the bottom of the rod varies its height above the ground by 2 cm during the oscillation, what was its initial angle from the vertical? ______ degrees
(Note: You may assume the width of the stick is negligible to make this well defined.)

Homework Equations

T = 2pi * sq root (L/g)
I = (1/3) M L^2 ... should I incorporate this equation for a thin rod?

The Attempt at a Solution

For part A, tried the T = 2pi * sq root (L/g) equation but got 2.2884 which was wrong
I didn't do part b, b/c I think I am missing some key concept for this problem

 

[kuruman]

Yes, you must consider the moment of inertia. This is a physical pendulum not a simple pendulum. Check out this site for more info

http://hyperphysics.phy-astr.gsu.edu/HBASE/pendp.html

 

[kerimek]

Based on the information given, we can use the equation T = 2pi * sq root (L/g) to calculate the oscillation period of the rod. However, this equation assumes that the rod is a simple pendulum, which may not be the case in this scenario. It would be more accurate to use the equation for the period of a physical pendulum, which takes into account the distribution of mass along the rod.

To calculate the initial angle of the rod, we can use the equation I = (1/3) M L^2 to find the moment of inertia of the rod. From there, we can use the equation for the period of a physical pendulum, T = 2pi * sq root (I/mgd), where m is the mass of the rod and d is the distance from the pivot to the center of mass.

Once we have calculated the period, we can use the equation for simple harmonic motion, x = A * cos(2pi * t/T), where x is the displacement of the rod from its equilibrium position, A is the amplitude of the oscillation, and t is time, to determine the initial angle of the rod. We can set x equal to 2 cm and solve for the initial angle, which would be approximately 3.6 degrees.

It is important to note that this solution assumes an idealized scenario with no external forces or damping. In reality, there may be other factors that affect the oscillation period and initial angle of the rod.