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Mechanics Assignment
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QUESTION 1:
Calculate the moment of inertia I for the following objects:
a.
A rod of length L with density P(X) = KX2 and axis of rotation in the center of the rod. X = 0
is define to be at the center of the rod.
Irod = 1/12 *mass * (rod length) 2 = 1/12ML2
= 1/12 * KX2 * L2
I= kx2L2/12
b.
A solid disc of radius R with axis of rotation at its center and density p(r) = ke-r
Idisc = 2/5 * mass * (radius) 2
= 2/5 * ke-r * R2
= 2 ke-r R2/5
QUESTION 2:
Find the period of a physical pendulum with the following shapes: (Hints: The moment of inertia for
two objects attached to each other can be calculated by individually calculating their moments of
inertia and adding them together. You will need to use the parallel axis theorem in this problem.)
a. A thin rod of length L and mass m with a solid disc of radius L/4 and mass m centered at the end.
The axis of rotation is at the end of the rod opposite the solid disc.
Calculating the moment of inertia of the objects attached:
I (rod) = ML2/3 = mL2/3
I (disc) = 2/5mr2 = 2/5 * m * (L/4)2 = mL2/40
I (rod &disc) = mL2/3 + mL2/40 = 43mL2/120
Using the parallel axis theorem to find moment of inertia at pivot point:
I (pivot point) = I (center mass) + MD2
I (pivot point) = 43mL2/120 + mL2
I (pivot point) = 163 mL2/120
Calculating for period:
T = 2 π √(163 mL2/120)/mgL
b. A thin rod of length L and mass m with a hoop of radius L/4 and mass m centered at the end. The
axis of rotation is at the end of the rod opposite the hoop. c. Which shape has the larger period?
Explain why your answer makes sense conceptually.
Calculating the moment of inertia of the objects attached:
I (rod) = ML2/3 = mL2/3
I (hoop) = mr2/2= m * (L/4)2/2 = mL2/8
I (rod &hoop) = mL2/3 + mL2/8 = 11mL2/24
Using the parallel axis theorem to find moment of inertia at pivot point:
I (pivot po...