Learn how to calculate moment of inertia of ring and disk about axis passing through their circumference. Practice parallel axis theorem example.
\( I = I_{CM} + Md^2\)
where, d is the perpendicular distance between AB and CZ and M is the mass of the body.
Note :
\(I_0=\dfrac {ML^2}{12}\) ....(1)
The distance between the two parallel axes considered is,
\( d =\) \(\dfrac {L}{2}\) ... (2)
\(I=\dfrac {ML^2}{12}\)+ \(M\)\(\left(\dfrac {L}{2}\right)^2\) ... [from (1) & (2)]
\(I\) = \(\dfrac {ML^2}{3}\)
? \(I_0=MR^2\) ...(1)
using parallel-axis theorem,
\(I=I_0+Md^2\)
Thus, \( I = MR^2 + M(R)^2 \) [ from (1) and (2)]
\( I = 2MR^2\)
\(I_0=\dfrac {MR^2}{2}\) ...(3)
using parallel-axis theorem,
\(I=I_0+Md^2\)
Thus, \(I=\dfrac{MR^2}{2}+M(R)^2\)... [from equation (3) and (4)]
\(I=\dfrac {3MR^2}{2}\)
A 1 kg m2
B 2 kg m2
C 3 kg m2
D 4 kg m2
\(dm=\dfrac {Mass \;of \;body} {Area \;of \;body}{(Area\; of \;element)}\)
\(dm=\dfrac{M}{L^2}(Ldx)\)
\(dm=\dfrac{M}{L}(dx)\) ...(1)
\(dI_0=\dfrac{dmL^2}{12}\)
Thus, \(I_z=dI_0+Md^2\)^{ }
^{ }⇒ \(dI=\dfrac{dmL^2}{12}+dmx^2\)
\(I=\int dI\)
=\(\int \left(\dfrac {L^2}{12}+ x^2\right)dm\)
= \(\dfrac{M}{L}\) \(\int\limits_{–L/2}^{L/2}\)\(\left(\dfrac {L^2}{12}+ x^2\right)dx\) ...[from(1)]
\(I\) = \(\dfrac{M}{L}\) \(\left[\dfrac {L^2x}{12}+ \dfrac{x^3}{3}\right]_{-L/2}^{L/2}\)
\(I=\dfrac{ML^2}{6}\)
A 16 kg m2
B 8 kg m2
C 6 kg m2
D 10 kg m2
\(I_{o_1}=\dfrac{M_1R_1^2}{2}\)
\(I=I_{o_1}+Md^2\)
\(I_1=\dfrac{M_1R_1^2}{2}\) + \(M_1R_1^2\)
\(I_1=\dfrac {3}{2}M_1R_1^2\) ...(1)
\(I_{o_2}=M_2R_2^2\)
\(I_2=I_{o_2}+Md^2\)
\(I_2=M_2R_2^2\) + \(M_2R_2^2\)
\(I_2=2M_2R_2^2\) ... (2)
Mathematically,
\(I=I_1+I_2\)
\(I=\dfrac {3}{2}M_1R_1^2\) + \(2M_2R_2^2\)
mass (M) = Density \(\left(\rho\right)\) × volume (V)
\(M=\rho\times(A\times\,t)\)
\(M=\rho\) \(\times\,\pi{R}^2\,t\) ... (1)
Mass , m = Density × Volume
\(m=\rho\times(A_1\times\,t)\)
\(m=\rho\) \(\times\,\pi{r}^2×t\)
\(\rho\) and t for smaller disk are same as that of large disk
\(\therefore\) \(m=\dfrac{M}{\pi{R^2×t}} ×\pi r^2×t\)...[from(1)]
\(m=\dfrac{Mr^2}{R^2}\) or \(\dfrac {m}{M}\)= \(\left(\dfrac{r}{R}\right)^2\)
then,
Moment of inertia of the whole disk about the axis CB is
\(I_0=I+I' \)
\(\because\,I_0=\dfrac{MR^2}{2}\), \(I'=\dfrac{mr^2}{2}+md^2\)
\(\therefore\, I=I_0-I'\)
\(\Rightarrow I=\dfrac{MR^2}{2}\)– \(\dfrac{mr^2}{2}-md^2\)
Substituting the value of m
\(I=\dfrac{MR^2}{2}\) – \(\dfrac{Mr^2}{R^2}\)\(\left(\dfrac {r^2}{2}+d^2\right)\)
A \(\dfrac{43}{2} \,M\)
B \(\dfrac{45}{2} \,M\)
C \(\dfrac{42}{2} \,M\)
D \(\dfrac{41}{2} \,M\)