Respuesta :
Answer:
I=[tex]\frac{4}{3}ML^2+2MR^2+2MRL[/tex]
Explanation:
We are given that
Mass of rod=M
Length of rod=L
Mass of hoop=M
Radius of hoop=R
We have to find the moment of inertia I of the pendulum about pivot depicted at the left end of the slid rod.
Moment of inertia of rod about center of mass=[tex]\frac{1}{12}ML^2[/tex]
Moment of inertia of hoop about center of mass=[tex]MR^2[/tex]
Moment of inertia of the pendulum about the pivot left end,I=[tex]\frac{1}{12}ML^2+M(\frac{L}{2})^2+MR^2+M(L+R)^2[/tex]
Moment of inertia of the pendulum about the pivot left end,I=[tex]\frac{1}{12}ML^2+\frac{1}{4}ML^2+MR^2+MR^2+ML^2+2MRL[/tex]
Moment of inertia of the pendulum about the pivot left end,I=[tex]\frac{1+3+12}{12}ML^2+2MR^2+2MLR[/tex]
Moment of inertia of the pendulum about the pivot left end,I=[tex]\frac{16}{12}ML^2+2MR^2+2MRL[/tex]
Moment of inertia of the pendulum about the pivot left end,I=[tex]\frac{4}{3}ML^2+2MR^2+2MRL[/tex]
Answer:
Explanation:
Rod of pendulum of mass M and length L.
Mass of hoop is M and radius of hoop is R.
Moment of inertia of the rod about its centre is
[tex]I_{R}=\frac{ML^{2}}{12}[/tex]
Moment of inertia of the rod about the pivot point
Use the parallel axis theorem
[tex]I=I_{R}+M\left (\frac{L}{2} \right )^{2}[/tex]
[tex]I=\frac{ML^{2}}{12}+\frac{ML^{2}}{4}=\frac{ML^{2}}{3}[/tex] .... (1)
Moment of inertia of the hoop about its centre is
[tex]I_{H}=MR^{2}[/tex]
Moment of inertia of the hoop about the pivot point
Use the parallel axis theorem
[tex]I'=I_{H}+M(R+L)^{2}[/tex]
[tex]I'=M(2R^{2}+L^{2}+2RL)[/tex] .... (2)
Total moment of inertia of the pendulum about the pivot is
[tex]I=\frac{ML^{2}}{3}+M\left ( 2R^{2}+L^{2}+2RL \right )[/tex] .... from (1) and (2)
[tex]I=M\left ( 2R^{2}+\frac{4L^{2}}{3}+2RL \right )[/tex]
