Question

A light meterstick is loaded with masses of5.0 kg and 3.0 kg

Answer 1

Answer:

$I=4.47\ kg.m^{2}$

Explanation:

Given:

A light meter stick (assuming mass-less) is loaded with point masses at the following positions.

• mass of first object, $m_1=5\ kg$

• mass of second object, $m_2=3\ kg$

• position of the first mass, $x_1=36\ cm=0.36\ m$

• position of the second mass, $x_2=89\ cm=0.89\ m$

As we know that moment of inertia for point mass is given by:

$I=m.R^2$

where:

$m=$ mass of the object

$R=$ radial distance from the axis of rotation

Now the total moment of inertia:

$I=m_1.x_1^2+m_2.x_2^2$

$I=5\times 0.36+3\times 0.89$

$I=4.47\ kg.m^{2}$

Answer 2

Answer:$I=3.024\ kg-m^2$

Explanation:

Given

mass of first $m_1=5 kg$

mass of second $m_2=3 kg$

Distance of $m_1$ from origin $r_1=36 cm$

Distance of $m_2$ from origin $r_2=89 cm$

Moment of inertia is given by multiplication of mass and square of distance

$I=\sum mr^2$

$I=m_1r_1^2+m_2r_2^2$

$I=5\times 0.36^2+3\times 0.89^2$

$I=0.648+2.376$

$I=3.024\ kg-m^2$