Learn how to calculate the moment of inertia of a uniform rod about the axis perpendicular to the rod and passing through one of its ends. Practice mass moment of inertia of a rod, ring, and disk.
\(\int{dI}\) = \(\int {r^2dm}\)
⇒ \(I=\int {r^2dm}\)
Examples :
\(I=\int\limits_{–L/2}^{L/2}r^2dm\)
where, dm is mass of the element.
Integrating equation (1) on both sides within proper limits,
\(\int{dI}\) = \(\int {r^2dm}\)
\(I=\int {r^2dm}\)
\(I=\int\limits_{–L/2}^{L/2}r^2dm\)
where dm is mass of the element.
therefore, \(dm=\dfrac{M}{L}.{dr}\) ...(1)
\(I=\int\limits_{-L/2}^{L/2}r^2dm\)
\(I=\dfrac{M}{L}\)\(\int\limits_{-L/2}^{L/2}r^2dr\) ...[from (1)]
\(I=\dfrac{M}{L}\)\(\left[\dfrac{r^3}{3}\right]_{-L/2}^{L/2}\)
\(I=\dfrac{ML^2}{12}\)
A 144 kgm2
B 140 kgm2
C 130 kgm2
D 150 kgm2
\(\int{dI}\) = \(\int {r^2dm}\)
⇒ \(I=\int {r^2dm}\)
\(I=\int\limits_{–L/2}^{L/2}r^2dm\)
where, dm is mass of the element.
A \(\dfrac {ML^2}{9}\)
B \(\dfrac {ML^2}{6}\)
C \(\dfrac {ML^2}{3}\)
D \(\dfrac {ML^2}{5}\)
Area of section of disk =\(\dfrac{\theta_0}{2}R^2\)
Mass per unit area = \(\dfrac{M}{\dfrac{\theta_0}{2}R^2}\)
Area of the element chosen, \(dA=r\theta_0.{dr}\)
Mass of element, \(dm=\dfrac{M}{\dfrac{\theta_0}{2}R^2}×r\theta_0.{dr}\)
\(I=\int{dmr^2}\)
\(I=\dfrac{2M}{R^2}\)\(\int\limits_{0}^{R} r^3dr\)
\(I=\dfrac{2M}{R^2}\) ×\(\dfrac{R^4}{4}\)
\(I=\dfrac{MR^2}{2}\)
A 40 kg-m2
B 25 kg-m2
C 16 kg-m2
D 20 kg-m2
\(I=\int{r^2}dm\)
\(I=\int{R^2}dm\)
\(I=R^2\int{dm}\)
\(= MR^2\)
A 1 kg-m2
B 2 kg-m2
C 3 kg-m2
D 5 kg-m2
Area of ring, dA = Length × Breadth
⇒ \(dA=2\pi{x}\;dx\)
\(dm=\dfrac{M}{\pi{R^2}}{(dA)}\)
\(dm=\dfrac{M}{\pi{R^2}}\) × \(2\pi{x}dx\)
\(dm=\dfrac{2Mxdx}{{R^2}}\)
\(dI=x^2dm\)
= \(x^2\dfrac{2Mxdx}{{R^2}}\)
Integrating both sides within proper limits -
\(\int\limits_O^IdI\) = \(\int\limits_O^Rx^2\dfrac{2Mxdx}{R^2}\)
\(I=\dfrac{2M}{R^2}\)\(\int\limits_O^Rx^3dx\)
\(I=\dfrac{MR^2}{2}\)
A 2 kg-m2
B 1 kg-m2
C 3 kg-m2
D 4 kg-m2
\(I_1=mR^2\)
Moment of inertia of all the rings,
\(I=R^2(m_1+m_2+m_3+...)\)
\(I=R^2M\)
\(I=MR^2\)
Moment of inertia of all the rods,
\(I=\dfrac{L^2}{3}(m_1+m_2+m_3+...)\)
\(=\dfrac{ML^2}{3}\)
A \(\dfrac{ML^2}{3}\)
B \(\dfrac{ML^2}{12}\)
C \(\dfrac{ML^2}{6}\)
D \(\dfrac{ML^2}{9}\)
Thus, \(I_X=I_Y\)
A 1 kg-m2
B 3 kg-m2
C 2 kg-m2
D 4 kg-m2