PURPOSE: To determine the period of a a solid ring swinging at a pivot and the period of an isosceles triangle about two different pivots.
PART 1:
To find the period of the pendulum of a solid ring, we found angular freqency, ω, by calculating angular acceleration using Torque = Inertia * Angular acceleration. We made angular acceleration look like the equation: α = - (ω^2) θ, rearranged from Newton's Second Law of Motion, F = ma. After substituting for the value for ω^2 and solving for ω, we used T=2π/ω and solved for the period.
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| Image of our solid ring with small radius, r, large radius, R, Ravg (distance from center to pivot) |
We set the pivot of our system to have radius Ravg, which was directly in between the large radius, R, and smaller radius, r. The net torque of the system was only the perpendicular component of its weight, which was mg*sinθ, when the ring was held at angle θ. This force was a distance Ravg from the pivot, giving us Torque = (- mg*sinθ)(Ravg). The inertia of the solid ring at its center of mass was found previously as (1/2M(R^2+r^2), but since we moved the pivot a distance Ravg from the center of mass, the moment of inertia about the ring's new pivot was [1/2M(R^2+r^2 + M(Ravg)^2].
Plugging this into our equation for T=Iα and solving for α:
After solving for α we set up the equation on the right hand side to make it look like α = - (ω^2) θ, and using the Taylor series expansion for sinθ, sinθ is approximated to just θ if θ is small enough. Therefore, we knew that the quantity within the parenthesis was equal to ω^2. We then used this to solve for T using T=2π/ω.
We then measured the solid ring to find R, r, and calculate Ravg:
When we plugged in our data to our equations, our period was 0.718 seconds (see equations above).
We then did the experiment, hanging the ring hung on its pivot, while the other end of the ring moved through a motion sensor that measured the period of the pendulum. Our experimental value for period, T, was 0.72 seconds. Our percent error for found values was 0.28%.
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| Set up used to measure the period of oscillation. |
PART 2:
We then derived for expressions for the period of an isosceles triangle with height, H, and base, B, oscillating about its apex, and then oscillating about the midpoint of its base. After deriving these expressions, we verified them by using a cutout of an isosceles triangle and oscillating it about the two points.
We first derived an expression for a small thin bar, dm, of the entire triangle. Setting it equal to the proportion of the area of this thin bar over total area, we solved for dm. We then found the moment of inertia of this thin bar, treating it as a thin rod rotating about its axis with the equation 1/12ML^2. Each thin bar was a distance, y, away from the pivot. Using this, we added up the moments of inertia of all these thin bars from 0 to H to find the moment of inertia of the entire triangle oscillating about its apex.
We then used dm to solve for the center of mass of the triangle in order to apply this to the parallel axis theorem:
Using the parallel axis theorem, we substituted for our known moment of inertia oscillating about the apex and also substituted distance from apex to center of mass (2/3*H) to find the inertia oscillating about its center of mass.
We then derived for expressions for the period of an isosceles triangle with height, H, and base, B, oscillating about its apex, and then oscillating about the midpoint of its base. After deriving these expressions, we verified them by using a cutout of an isosceles triangle and oscillating it about the two points.
We first derived an expression for a small thin bar, dm, of the entire triangle. Setting it equal to the proportion of the area of this thin bar over total area, we solved for dm. We then found the moment of inertia of this thin bar, treating it as a thin rod rotating about its axis with the equation 1/12ML^2. Each thin bar was a distance, y, away from the pivot. Using this, we added up the moments of inertia of all these thin bars from 0 to H to find the moment of inertia of the entire triangle oscillating about its apex.
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| Moment of inertia of isosceles triangle oscillating about apex. |
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| Center of mass is 2/3 of the height from the apex. |
Once we knew the moment of inertia about the center of mass, we could again apply the parallel axis theorem to find the moment of inertia oscillating about the center of the base of the triangle (a distance 1/3 away from the center of mass).
We used our expressions for moments of inertia in our torque equations, then made each equation resemble α= - ( ω^2 ) θ. By finding ω^2, we were able to solve for the period. We measured our triangle to have a base of 0.14 m and height of 0.147 m. Plugging these values into our equations we solved for theoretical periods of the oscillating triangle.
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| Moment of inertia of triangle oscillating about the midpoint of the base. |
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Theoretical period for triangle oscillating about its apex: 0.69081 seconds.
Theoretical period for triangle oscillating about the midpoint of its base: 0.60238 seconds.
We set up the triangle in both positions, with the other end passing through a photogate:
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| Period for the triangle oscillating about its apex: 0.695362 s. |
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| Period for the triangle oscillating about the midpoint of its base: 0.607886 s. |
Theoretical period for triangle oscillating about its apex: 0.69081 seconds.
Experimental period for triangle oscillating about its apex: 0.695362 seconds.
Error: 0.65%
Theoretical period for triangle oscillating about the midpoint of its base: 0.60238 seconds.
Experimental period for triangle oscillating about midpoint of its base: 0.607886 seconds.
Error: 0.91%
CONCLUSION:
For both parts of our experiment, we were able to use an expression for angular acceleration of a physical pendulum swinging with simple harmonic motion to calculate its angular frequency and then find the period of the system. For part 1, we used a net torque equation, and for part 2, we used moments of inertia of the shape oscillating about 2 different pivots, and we also used the parallel axis theorem.
In part 1, our theoretical time for period was 0.718 seconds, while our experimental time was 0.72 seconds, giving us an error of 0.28%, which shows that our calculated data for the period of a pendulum was very precise to the experimental result.
In part 2, our percent erros were 0.65% for the triangle oscillating about its apex and 0.91% for the triangle oscillating about the midpoint of its base. In all 3 cases, the error was less than 1%, which just shows us that there is very high precision in the periods of pendulums that oscillate with simple harmonic motion.
In part 2, our percent erros were 0.65% for the triangle oscillating about its apex and 0.91% for the triangle oscillating about the midpoint of its base. In all 3 cases, the error was less than 1%, which just shows us that there is very high precision in the periods of pendulums that oscillate with simple harmonic motion.
