A pendulum built from a steel sphere with radius r cm 5 and density stl kg m S 3 7800 is attached to an aluminum bar with length l m 1 thickness t cm 0 8. and width w cm 4 and density . al kg m S 3 2820 a. Calculate the mass moment of inertia of the pendulum about its center of mass, . cm I b. Calculate the mass moment of inertia of the pendulum about its pivot point, o I .

a)  I = 0.0198 kg m² ,  b)    I = 21.85 kg m²

Explanation:

For this exercise we will use the definition of moment of inertia

        I = ∫ r² dm

For body with high symmetry they are tabulated

sphere  I = 2/5 m r²

bar with respect to  center of mass I = 1/12 m L²

let's calculate the mass of each body

bar

        ρ = m / V

        m = ρ V

        m = ρ l w h

where we are given the density of the bar rho = 32840 kg / m³ and its dimensions 1 m, 0.8 cm and 4 cm

        m = 32820 1 0.008 0.04

        m = 10.5 kg

Sphere

       M = ρ V

       V = 4/3 pi r³

       M = rgo 4/3 π r³

give us the density 37800 kg / m³ and the radius of 5 cm

       M = 37800 4/3 π 0.05³

       M = 19.8 kg

a) asks us for the moment of inertia of the sphere with respect to its center of mass

        I = 2/5 M r²

        I = 2/5 19.8 0.05²

        I = 0.0198 kg m²

b) the moment of inertia with respect to the turning point, for this we will use the theorem of parallel axes

        I = I_cm + M d2

where d is the distance from the body to the point of interest

        I_cm = 0.0198 kg m²

the distance to the pivot point is

        l = length of the bar + radius of the sphere

        l = 1 + 0.05 = 1.005 m

        I = 0.0198 + 19.8 1.05²

        I = 21.85 kg m²

6/12=1/2

step-by-step explanation: