Question

A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I1 is the moment of inertia of this object with respect to an axis passing through the center of the rod and perpendicular to it and I2 is the moment of inertia with respect to an axis passing through one of the masses, it follows that:______
a. I1 > I2
b. I2 > I1.
c. I1 = I2.

Answer 1

Answer:

The correct answer to the following question will be Option A (I1 > I2).

Explanation:

Method for moment of inertia because of it's viewpoint including object at a mean distance "r" from the axis is,

⇒ mr²

For Case 1:

Let the length of a rod be "r".

The axis passes via the middle of that same rod so that the range from either the axis within each dumbbell becomes "$\frac{r}{2}$".

Now,

Now total moment of inertia = sum of inertial moment due to all of the dumbbell

⇒  $l1=m(\frac{r}{2})^2+m(\frac{r}{2})^2$

⇒  $\frac{mr^2}{2}$

For Case 2:

Axis moves via one dumbbell because its range from either the axis becomes zero (0) and its impact is zero only at inertia as well as other dumbbell seems to be at a range "r" from either the axis

Now,

Total moment of inertia = moment of inertia of dumbbell at distance "r".

$l2=mr^2$

And now we can infer from this one,

⇒ $mr^2>\frac{mr^2}{2}$

So that "I1 > I2" is the right answer.