Respuesta :
Answer:
The linear speed is 1.33 m/s.
Explanation:
Given that,
Length of rod = 90.0 cm
Mass of slender rod = 0.120 kg
Mass of small sphere = 0.0200 kg
Mass of another small sphere = 0.0400 kg
Suppose, we need to find the linear speed of the 0.0500-kg sphere as it passes through its lowest point?
We need to calculate the the change in potential of the complete system
Using formula of change in potential
m₂ and m₃ are the masses at the rod ends.
The rod center of mass neither gains nor loses potential
[tex]\Delta U=m_{2}gy_{2}+m_{3}gy_{3}[/tex]
Put the value into the formula
[tex]\Delta U=0.0400\times9.8\times(-45\times10^{-2})+0.0200\times9.8\times45\times10^{-2}[/tex]
[tex]\Delta U=-0.0882\ N-m[/tex]
We need to calculate the moment of inertia of the rod
Using formula of moment of inertia of the rod
[tex]I_{1}=2\rho \int_{0}^{r}(r^2 dr)[/tex]
Put the value into the formula
[tex]I_{1}=2\times\dfrac{0.120}{90\times10^{-2}} \int_{0}^{0.45}(r^2 dr)[/tex]
[tex]I_{1}=2\times\dfrac{0.120}{90\times10^{-2}}\times(\dfrac{(0.45)^3}{3}-0)[/tex]
[tex]I_{1}=0.0081\ kg-m^2[/tex]
We need to calculate the moment of inertia of the end masses
Using formula of moment of inertia
[tex]I_{2+3}=\sum mr^2[/tex]
Put the value into the formula
[tex]I_{2+3}=(0.0400+0.0200)\times0.45^2[/tex]
[tex]I_{2+3}=0.01215\ kg-m^2[/tex]
We need to calculate the change in potential energy to the system kinetic energy
Using formula of kinetic energy
[tex]\Delta U=K.E[/tex]
[tex]\Delta U=\dfrac{1}{2}(I_{1}+I_{2+3})\omega^2[/tex]
Put the value into the formula
[tex]0.0882=\dfrac{1}{2}(0.0081+0.01215)\omega^2[/tex]
[tex]\omega^2=\dfrac{2\times0.0882}{0.02025}[/tex]
[tex]\omega=\sqrt{\dfrac{2\times0.0882}{0.02025}}[/tex]
[tex]\omega=2.95\ rad/s[/tex]
We need to calculate the linear speed
Using formula of linear speed
[tex]v = r\omega[/tex]
Put the value into the formula
[tex]v=0.45\times2.95[/tex]
[tex]v=1.33\ m/s[/tex]
Hence, The linear speed is 1.33 m/s.
