Question

What is the moment of inertia i of this assembly about the axis through which it is pivoted? Express the moment of inertia in terms of mr

Answer 1

Answer:

I = x^2(m1+m2+1/3mr)

Explanation:

Answer 2

Answer:

The moment of inertia of the assembly is $= (\frac{m_r}{3} +m_1+m_2)x^2$

Explanation:

Moment of inertia of a rod of mass M and length L about its center is given by

$I = \frac{ML^2}{12}$

Substitute${m_r}$ for M and 2x for L

$I_r = \frac{m_r (2x)^2}{12}$

$I_r = \frac{m_r x^2}{3}----(1)$

Moment of inertia of mass ${m_1}$, distance x from the axis:

$I_1 = m_1x^2----(2)$

Moment of inertia of mass ${m_2}$, distance x from the axis:

$I_2 = m_2x^2----(3)$

The moment of inertia of each of the components is found out. Moment of inertia of point particle with mass MM is given by the product of its mass and the square of its distance from the axis

The total moment of inertia of the system is given by:

$I_{total}= I_r+I_1+I_2$

Substitute from $\left( 1 \right),\left( 2 \right),\left( 3 \right)$ to the above relation:

$I_{total} = \frac{m_rx^2}{3} + m_1x^2+m_2x^2$

$= (\frac{m_r}{3} +m_1+m_2)x^2$

The moment of inertia of the assembly is $= (\frac{m_r}{3} +m_1+m_2)x^2$