Question

A slender uniform rod 100.00 cm long is used as a meter stick. Two parallel axes that are perpendicular to the rod are considered. The first axis passes through the 50-cm mark and the second axis passes through the 30-cm mark. What is the ratio of the moment of inertia through the second axis to the moment of inertia through the first axis?
(A) 1.7
(B) 2.3
(C) 2.1
(D) 1.5
(E) 1.9

Answer 1

Answer:

The correct answer is D    $I_{30}$ /$I_{50}$ =   1.5

Explanation:

In this exercise the moment of inertia equation should be used

    I = ∫ r² dm

In addition to the parallel axis theorem

    I = $I_{cm}$ + M D²

Where  $I_{cm}$ is the moment of the center of mass, M is the total mass of the body and D the distance from this point to the axis of interest

Let's apply these relationships to our problem, the center of mass of a uniform rod coincides with its geometric center, in this case the rod is 1 m long, so the center of mass is in

    L = 100.00 cm (1m / 100 cm) = 1.0000 m

     $x_{cm}$ = 50 cm = 0.50 m

Let's calculate the moment of inertia for this point, suppose the rod is on the x-axis and use the concept of linear density

    λ = M / L = dm / dx

    dm =  λ dx

Let's replace in the moment of inertia equation

    I = ∫ x² ( λ dx)

We integrate

    I =  λ x³ / 3

We evaluate between the lower limits x = -L/2 to the upper limit x = L/2

    I =  λ/3 [(L/2)³ - (-L/2)³] = lam/3  [L³/8 - (-L³ / 8)]

    I =  λ/3  L³/4

    I = 1/12  λ L³

Let's replace the linear density with its value

    I = 1/12  (M/L)  L³

    I = 1/12  M L²

Let's calculate with the given values

   I = 1/12  M 1²

   I = 1/12 M

This point is the center of mass of the rod

    Icm = I = 1/12 M  = 8.333 10-2 M

Now let's use the parallel axis theorem to calculate the moment of resection of the new axis, which is 0.30 m from one end, in this case the distance is

    D = $x_{cm}$ - x

    D = 0.50 - 0.30

    D = 0.20  m

Let's calculate

   $I_{30}$ =$I_{cm}$ + M D²

   $I_{30}$ = 1/12 M + M 0.202

   $I_{30}$ = M (1/12 + 0.04)

   $I_{30}$ = M 0.123

To find the relationship between the two moments of inertia, divide the quantities

  $I_{30}$ / $I_{50}$ = M 0.123 / (M 8.3 10-2)

   $I_{30}$ /$I_{50}$ = 1.48

The correct answer is d 1.5