Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is ![\alpha =2.538 \ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D2.538%20%5C%20%20rad%2Fs%5E2)
Explanation:
From the question we are told that
The mass of the wheel is m = 6.9 kg
The radius is ![r = 0.69 \ m](https://tex.z-dn.net/?f=r%20%3D%20%200.69%20%5C%20%20m)
The radius of gyration is ![k_G = 0.4\ m](https://tex.z-dn.net/?f=k_G%20%3D%200.4%5C%20%20m)
The angle is ![\theta = 47^o](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2047%5Eo)
The force which the massless bar is subjected to ![F = 22.5 \ N](https://tex.z-dn.net/?f=F%20%3D%2022.5%20%5C%20%20N)
Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that point A is the center of rotation.
Generally the moment of inertia about A is mathematically represented as
![I_a = I_G + M* r^2](https://tex.z-dn.net/?f=I_a%20%3D%20%20I_G%20%2B%20M%2A%20r%5E2)
Here
is the moment of inertia about G with respect to the radius of gyration which is mathematically represented as
![I_G = M * k_G](https://tex.z-dn.net/?f=I_G%20%3D%20%20M%20%2A%20%20k_G)
=>![I_a = k_G* M + M* r^2](https://tex.z-dn.net/?f=I_a%20%3D%20k_G%2A%20%20M%20%2B%20M%2A%20r%5E2)
=>![I_a =0.4 * 6.9 + 6.9 * 0.69^2](https://tex.z-dn.net/?f=I_a%20%3D0.4%20%2A%206.9%20%20%2B%206.9%20%2A%200.69%5E2)
=>![I_a =6.045 \ kg \cdot m^2](https://tex.z-dn.net/?f=I_a%20%3D6.045%20%5C%20%20kg%20%5Ccdot%20m%5E2)
Generally the torque experienced by the wheel is mathematically represented as
![\tau = F * cos (47)](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%20F%20%2A%20%20cos%20%2847%29)
=> ![\tau = 22.5 * cos (47)](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%2022.5%20%2A%20%20cos%20%2847%29)
=> ![\tau = 15.34 \ kg \cdot m^2 \cdot s^{-2}](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%2015.34%20%5C%20kg%20%5Ccdot%20m%5E2%20%5Ccdot%20s%5E%7B-2%7D)
Generally this torque is also mathematically represented as
![\tau = I_a * \alpha](https://tex.z-dn.net/?f=%5Ctau%20%3D%20I_a%20%2A%20%5Calpha)
=> ![15.34 = 6.045 * \alpha](https://tex.z-dn.net/?f=15.34%20%20%3D%20%206.045%20%2A%20%5Calpha)
=> ![\alpha =2.538 \ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D2.538%20%5C%20%20rad%2Fs%5E2)
In nature there are two categories of microorganisms as relating to health. Microorganisms that are considered harmful to humans are called pathogens and these cause disease. Examples include bacteria such as streptococcus which cause sore throat and salmonella which cause typhoid disease.
There are some microorganisms which are helpful to man and they live mostly on the skin of man or in his gut and are mostly bacteria. They are collectively called bacterial normal flora.
In man the normal bacterial flora of the skin include staphylococcus found on dry skin, cornybacteria found in moist skin sites and propionibacteria in the sebaceous sites (head, neck, trunk) of the body. Normal bacterial flora of the gut include Escherichia coli.
One of the major function of bacterial flora is actually to protect our bodies by competing for space with pathogens preventing them from gaining a foothold in our bodies.
Answer:
v=0.94 m/s
Explanation:
Given that
M= 5.67 kg
k= 150 N/m
m=1 kg
μ = 0.45
The maximum acceleration of upper block can be μ g.
a= μ g ( g = 10 m/s²)
The maximum acceleration of system will ω²X.
ω = natural frequency
X=maximum displacement
For top stop slipping
μ g =ω²X
We know for spring mass system natural frequency given as
![\omega=\sqrt{\dfrac{k}{M+m}}](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7B%5Cdfrac%7Bk%7D%7BM%2Bm%7D%7D)
By putting the values
![\omega=\sqrt{\dfrac{150}{5.67+1}}](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7B%5Cdfrac%7B150%7D%7B5.67%2B1%7D%7D)
ω = 4.47 rad/s
μ g =ω²X
By putting the values
0.45 x 10 = 4.47² X
X = 0.2 m
From energy conservation
![\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7DkX%5E2%3D%5Cdfrac%7B1%7D%7B2%7D%28m%2BM%29v%5E2)
![kX^2=(m+M)v^2](https://tex.z-dn.net/?f=kX%5E2%3D%28m%2BM%29v%5E2)
150 x 0.2²=6.67 v²
v=0.94 m/s
This is the maximum speed of system.