Pendulum, Rotational Inertia and Center of mass

[bieon]

Homework Statement

A physical pendulum which is a mass of 0.25 kg and length 1.0 m with an attached mass (0.25 kg) represented by the white circle which can oscillate about the black circle that goes through one end of the rod.

The mass can be moved to different positions on the rod. The distance from the rotation axis to the mass is x.

Use the period formula below to derive the formula for the period T of the physical pendulum as a function
of the distance x.

Relevant Equations

For small-amplitude oscillation, the period of the physical pendulum is theoretically given by

$T=2\pi \sqrt\frac {I}{0.5gd}$

where
$0.5$ is the total mass of the physical pendulum (which is composed of the rod and particle), in $kg$

$g$ is the acceleration due to gravity, $9.8 m/s^2$
Rotation axis

$I$ is the rotational inertia of the physical pendulum relative to the rotation axis, in $kg.m^2$

$d$ is the distance from the rotation axis to the centre of mass of the physical pendulum, in $m$

This is the figure given.

My Attempt

$T=2\pi \sqrt\frac {I}{0.5gd}$

$\frac {m_r} {l}$
$dm= \frac {m} {l}dx$

$dI = dm_r x^2$
$dI=(\frac {m_r} {l}dx)x^2$
$I= \int_l^0 (\frac {m_r} {l}dx)x^2 \, dx$

$I_c.m=\frac {m_r l^2}{3}$

$I_,, = \frac {m_r l^2}{3}+m_p x^2$

Given
$m_r=0.25$
$m+p=0.25$
$l=1$

$I_,, = \frac {(0.25 (1)^2}{3}+(0.25) x^2$

$I_,, = \frac {0.25+0.75x^2}{12}$

From here on I got stuck because I have no idea on how to find the d along with the centre of the mass.

 

[Doc Al]

From here on I got stuck because I have no idea on how to find the d along with the centre of the mass.

Where's the center of mass of the rod alone? And of the attached mass? Then find the composite center of mass of the system using the definition of center of mass. (Measure the center of mass from the axis of rotation.)

 

[bieon]

Where's the center of mass of the rod alone? And of the attached mass? Then find the composite center of mass of the system using the definition of center of mass. (Measure the center of mass from the axis of rotation.)

Is this correct?

Centre of Rod:

$l_1= \frac {\int_0^x l dm}{M}$

Where,

$\lambda = \frac {dm} {dl} = \frac {M}{x} ; dm=\lambda dl$

$l_1= \frac {\int_0^x l \lambda dl}{M}$

$l_1= \frac {\int_0^x l \frac {M}{x} dl}{M}$

$l_1= \frac{1}{x} \int_0^x l dl$

$l_1=\frac {1}{x} [\frac {l^2}{2}]_0^x$

$l_1=\frac {x}{2}$

Centre of attached mass:

$\frac {\sum xm}{M}$

$l_2=\frac {m_r \frac{2}{3} + m_p \frac{1}{3}}{M}$

$l_2=\frac {0.25 \frac{2}{3} + 0.25\frac{1}{3}}{0.5}$

$i_2=\frac{1}{4}$

$d=\frac {x}{2} +\frac{1}{4}$

 

[Doc Al]

Is this correct?

Centre of Rod:

$l_1= \frac {\int_0^x l dm}{M}$

Where,

$\lambda = \frac {dm} {dl} = \frac {M}{x} ; dm=\lambda dl$

$l_1= \frac {\int_0^x l \lambda dl}{M}$

$l_1= \frac {\int_0^x l \frac {M}{x} dl}{M}$

$l_1= \frac{1}{x} \int_0^x l dl$

$l_1=\frac {1}{x} [\frac {l^2}{2}]_0^x$

$l_1=\frac {x}{2}$

The center of mass of the rod is right in the middle; its length is given as 1.0 m. (x is the position of the attached mass.)

Centre of attached mass:

$\frac {\sum xm}{M}$

$l_2=\frac {m_r \frac{2}{3} + m_p \frac{1}{3}}{M}$

$l_2=\frac {0.25 \frac{2}{3} + 0.25\frac{1}{3}}{0.5}$

$i_2=\frac{1}{4}$

Since the mass is just a point, its "center" is a distance x from the axis of rotation.

 

[bieon]

The center of mass of the rod is right in the middle; its length is given as 1.0 m. (x is the position of the attached mass.)Since the mass is just a point, its "center" is a distance x from the axis of rotation.

Sorry I didn't quite catch that, can you elaborate?

 

[Doc Al]

Sorry I didn't quite catch that, can you elaborate?

To find the center of mass of a system of two bodies, use: $d = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$. The two bodies are the rod and attached mass, of course.

For the purpose of finding the center of mass, treat the rod as having its mass at its center, thus $x_1 = 0.5$ m. $x_2$ is just the position of the attached mass, thus $x_2 = x$.

 

[bieon]

To find the center of mass of a system of two bodies, use: $d = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$. The two bodies are the rod and attached mass, of course.

For the purpose of finding the center of mass, treat the rod as having its mass at its center, thus $x_1 = 0.5$ m. $x_2$ is just the position of the attached mass, thus $x_2 = x$.

So,

$d=\frac {0.25(0.5)+0.25x} {0.5}$

$d=\frac{1}{4}+\frac {x}{2}$

Ahhh I complicated myself with the integrations and others! Thank you!

 

[Doc Al]

So,

$d=\frac {0.25(0.5)+0.25x} {0.5}$

$d=\frac{1}{4}+\frac {x}{2}$

Exactly.

Ahhh I complicated myself with the integrations and others!

Yes, you were making things more difficult than necessary.

 

[bieon]

Exactly.Yes, you were making things more difficult than necessary.

Thank you! I really appreciate the help!