Respuesta :
Answer:
the correct result is L = 0.319 m
Explanation:
This system is a physical pendulum whose angular velocity is
w² = [tex]\frac{M \ g \ d}{I}[/tex]
where d is the distance from the center of mass to the point of rotation and I is the moment of inertia of the system
The Moment of Inertia is a scalar, therefore an additive quantity
I = I_bar + I_disk
the moment of inertia of each element with respect to the pivot point can be found with the parallel axes theorem
let's use M for the mass of the bar and m for the mass of the disk
Bar
I_bar = I_{cm} + Md²
the moment of inertia of the center of mass is
I_{cm} = [tex]\frac{1}{12}[/tex] M L²
the distance from the center of mass
d = L / 2
we substitute
I_bar = [tex]\frac{1}{12}[/tex] M L² + M ([tex]\frac{L^2}{4}[/tex])
Disk
I_disk = I_{cm} + m d²
moment of inertia of the center of mass
I_{cm} = ½ m R²
the distance d is
d = L
we substitute
I_disk = 1/2 m R² + m L²
the total moment of inertia is
I = [tex]\frac{1}{12}[/tex] M L² +[tex]\frac{1}{4}[/tex] M L² + [tex]\frac{1}{2}[/tex] m r² + m L²
I = [tex]\frac{1}{4}[/tex] M L² + m L² + ½ m r²
I = L² (m + [tex]\frac{1}{4}[/tex] M) + ½ m r²
The position of the center of mass of the system can be found with the expressions
d_{cm} = [tex]\frac{1}{M} \sum r_i m_i[/tex]
d_{cm} = [tex]\frac{1}{m+M} \ ( M \frac{L}{2} + m L)[/tex]
d_{cm} = [tex]L \frac{m + M/2}{m +M }[/tex]
now we can substitute in the expression for the angular velocity
w² = (m + M) g L [tex]\frac{m + \frac{M}{2} }{m+M}[/tex] [tex]\frac{1}{L^2 (m+ \frac{M}{4} ) + \frac{1}{2} m r^2 }[/tex]
w² = g (m + [tex]\frac{1}{2}[/tex] M) [tex]\frac{L}{ L^2 ( m +\frac{1}{4} M ) + \frac{1}{2} m r^2}[/tex]
angular velocity and period are related
w = 2π/T
sustitute
4π²/T² = g (m + [tex]\frac{1}{2}[/tex] M) [tex]\frac{L}{ L^2 ( m +\frac{1}{4} M ) + \frac{1}{2} m r^2}[/tex]
L² (m + [tex]\frac{1}{4}[/tex] M) + ½ m r² = [tex]\frac{T^2}{4 \pi ^2 } \ g ( m + \frac{1}{2} M ) \ \ L[/tex]
we substitute the values and solve the second grade equation
L² (0.1 + [tex]\frac{1}{4}[/tex] 0.3) - [[tex]\frac{1^2}{4\pi ^2}[/tex] 9.8 (0.1 + 0.3/2) ] L + ½ 0.1 0.2² = 0
L² 0.175 - 0.06206 L + 0.002 = 0
the equation remains after simplifying
L² - 0.3546 La + 0.01143 = 0
solve us
L = [tex]\frac{0.3546 \ \pm \sqrt{ 0.3546^2 - 4 \ 0.01143 }}{2}[/tex]
L = [tex]\frac{0.3546 \ \pm \ 0.28288 }{2}[/tex]
L₁ = 0.319 m
L₂ = 0.036m
the correct result must have a value greater than the radius of the disk. The correct result is L = 0.319 m
