Question

In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 4.67 kg. They then hang the object on a pivot located 0.180 m from the object's center of mass and set it swinging at a small amplitude. As two of the team members carefully count 129 cycles of oscillation, the third member measures a duration of 249 s. What is the moment of inertia of the object with respect to its center of mass about an axis parallel to the pivot axis

Answer 1

Answer:

The value is  $I_o = 0.6257 \ kgm^2$

Explanation:

From the question we are told that  

   The mass of the object is  $m = 4.67 \ kg$

   The position of the pivot from the center of the mass is  $a = 0.180 \ m$

   The number of cycles of oscillation  is  $n = 129 \ cycles$

     The time taken for the total oscillation is t =249 s

Generally the period of the oscillation is mathematically represented as

         $T = \frac{t}{n}$

=>     $T = \frac{249}{129}$

=>     $T = 1.9302 \ s$

Generally the moment of inertia of the object about the end of the suspension  is mathematically represented as

    $I =[ \frac{T}{2 \pi } ]^2 * [m* g * l ]$

=> $I =[ \frac{1.9302}{2 *3.142 } ]^2 * [4.67* 9.8 * 0.180 ]$

=> $I = 0.777 \ kg m^2$

Generally the moment of inertia of the object with respect to its center of mass about an axis parallel to the pivot axis is mathematically evaluated using the parallel axis theorem as

  =>     $I_o = I- I_p$

Here  $I_p$ is parallel  axis contribution which is mathematically represented as

        $I_p = m * l^2$

=>     $I_p = 4.67 * 0.180^2$

=>     $I_p = 0.1513 \ kg m^2$

So  

     $I_o = I- I_p$

=>   $I_o = 0.777 -0.1513$

=>   $I_o = 0.6257 \ kgm^2$