Answer to Question #109474 in Mechanics | Relativity for happy45

Question

[img]https://upload.cc/i1/2020/04/13/EQ6TnM.jpg[/img]


question is in picture

d8=4

Answer 1

Problem: $R=48cm=0.48m$ ; $r_h=20cm=0.2m$ ; $L=80cm=0.8m$ ; $M_{rod}=1.2 kg$ ; $\rho_d=4kgm^{-2}$ ; $d_0=20cm=0.2m$ ;

Solution: (a) We will use the information [1] . The moment of inertia of a thin rod about its center of mass is $I_{rod}=\frac{M_{rod}\cdot L^2}{12}$ . The moment of inertia of a  thin full disc  about an axis through its center and perpendicular to its face is $I_R=M_d\frac{R^2}{2}$   where $M_d=\rho_d\cdot \pi R^2=4kgm^{-2}\cdot 3.14\cdot (0.48m)^2=2.9kg$

For further determination, we also determine the moment of inertia of a thin disk with a radius of the hole $I_h=M_h\frac{r_h^2}{2}$ , $M_h=\rho_d\cdot \pi r_h^2=4kgm^{-2}\cdot 3.14\cdot (0.2m)^2=0.5kg$

We determined all the moments of inertia of the constituent parts of our pendulum. Now we must take into account that their axis of rotation is shifted to the pivot point O. We should use the parallel axis theorem for planar case (also known as Huygens–Steiner theorem, or just as Steiner's theorem):

$I _R=I_C+Md^2$ , where $d$  is the distance from the center of mass $C$ to the reference point $R$ .

The center of rotation a thin rod is shifted on $d_r=L/2-d_0=0.4-0.2=0.2m$ thus its moment of inertia about pivot point O is

(1) $I_1=I_{rod}+M_{rod}\cdot d_r^2=M_{rod}\cdot(\frac{L^2}{12}+d_r^2)=\\=1.2\cdot (0.8^2/12+0.2^2)=0.112 kgm^2$

The center of rotation a full disk is shifted on $d_d=d_0+R=0.2+0.48=0.68m$ thus its moment of inertia about pivot point O is

(2) $I_2=I_R+M_d\cdot d_d^2=M_d\frac{R^2}{2}+M_d\cdot d_d^2=M_d\cdot(\frac{R^2}{2}+ d_d^2)=\\=2.9\cdot(\frac{0.48^2}{2}+ 0.68^2)=1.675kgm^2$

The center of rotation a holl disk is shifted on

$d_h=20cm+30cm+20cm=70cm=0.7m$ (look on the figure of problem) thus its moment of inertia about pivot point O is

(3) $I_3=I_h+M_h\cdot d_h^2=M_h\frac{r_h^2}{2}+M_h\cdot d_h^2=M_h\cdot(\frac{r_h^2}{2}+ d_h^2)=\\=0.5 \cdot(\frac{0.2^2}{2}+0.7^2)=0.255 kgm^2$

The moment of inertia of the compound pendulum is

$I=I_1+I_2 - I_3=0.112+1.675-0.255=1.53kgm^2$

(b) The distance of the center of mass of the compound pendulum from pivot O is

$h_{cm}=\frac{ M_d\cdot d_d-M_h\cdot d_h-M_{rod}\cdot d_r}{M_d-M_h+M_{rod}}=\frac{ 2.9\cdot 0.68-0.5\cdot 0.7-1.2\cdot 0.2}{2.9-0.5+1.2}=\frac{1,382}{3.6}=0.38m$

We use negative sign for $M_h$ due to this is holl and its mass was already add in $M_d$ and negative sign for $d_r$ as the center of thin rod mass is distance upward from pivot O.

(c) The angular frequency $w=\frac{2\pi}{T}=\frac{2\cdot 3.14}{2.1}=3radian\cdot s^{-1}=171\degree s^{-1}$ The period of pendulum you can get from (d)

(d) The period of physical pendulum is $T=2\pi\sqrt{\frac{I}{mgl}}$ . In our case $l=h_{cm}$ , $m=M_d-M_h+M_{rod}$ =3.6kg if we get $g=9.8ms^{-2}$ then

$T=2\cdot 3.14\cdot \sqrt{\frac{1.53}{3.6\cdot 9.8\cdot 0.38}}=2.1s$

Answer: (a) The moment of inertia of the compound pendulum at a pivot O is $1.53kgm^2$

(b) The distance of the center of mass of the compound pendulum from pivot O is 0.38m

(c) The angular frequency of small oscillations is 3 rad/s

(d) The period of small oscillations of pendulum is 2.1s

[1] https://en.wikipedia.org/wiki/Moment_of_inertia