Answer:
it's possible to calculate an object's velocity at any moment along its path. This is called instantaneous velocity and it is defined by the equation v = (ds)/(dt),in other words, the derivative of the object's average velocity equation.
Customer satisfaction is considered to be the "driving force" in order to achieve an efficient supply chain. An efficient supply chain takes place when the organization, or the company itself, meets with the demands of the consumers to improve and provides services that satisfies the people.
Answer:
the active region is bound by cutoff region and saturation or power dissipation region.
Explanation:
Explanation:
It is given that,
Mass of person, m = 70 kg
Radius of merry go round, r = 2.9 m
The moment of inertia, ![I_1=900\ kg.m^2](https://tex.z-dn.net/?f=I_1%3D900%5C%20kg.m%5E2)
Initial angular velocity of the platform, ![\omega=0.95\ rad/s](https://tex.z-dn.net/?f=%5Comega%3D0.95%5C%20rad%2Fs)
Part A,
Let
is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :
![I_1\omega_1=I_2\omega_2](https://tex.z-dn.net/?f=I_1%5Comega_1%3DI_2%5Comega_2)
Here, ![I_2=I_1+mr^2](https://tex.z-dn.net/?f=I_2%3DI_1%2Bmr%5E2)
![I_1\omega_1=(I_1+mr^2)\omega_2](https://tex.z-dn.net/?f=I_1%5Comega_1%3D%28I_1%2Bmr%5E2%29%5Comega_2)
![900\times 0.95=(900+70\times (2.9)^2)\omega_2](https://tex.z-dn.net/?f=900%5Ctimes%200.95%3D%28900%2B70%5Ctimes%20%282.9%29%5E2%29%5Comega_2)
Solving the above equation, we get the value as :
![\omega_2=0.574\ rad/s](https://tex.z-dn.net/?f=%5Comega_2%3D0.574%5C%20rad%2Fs)
Part B,
The initial rotational kinetic energy is given by :
![k_i=\dfrac{1}{2}I_1\omega_1^2](https://tex.z-dn.net/?f=k_i%3D%5Cdfrac%7B1%7D%7B2%7DI_1%5Comega_1%5E2)
![k_i=\dfrac{1}{2}\times 900\times (0.95)^2](https://tex.z-dn.net/?f=k_i%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20900%5Ctimes%20%280.95%29%5E2)
![k_i=406.12\ rad/s](https://tex.z-dn.net/?f=k_i%3D406.12%5C%20rad%2Fs)
The final rotational kinetic energy is given by :
![k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2](https://tex.z-dn.net/?f=k_f%3D%5Cdfrac%7B1%7D%7B2%7D%28I_1%2Bmr%5E2%29%5Comega_1%5E2)
![k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2](https://tex.z-dn.net/?f=k_f%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%28900%2B70%5Ctimes%20%282.9%29%5E2%29%280.574%29%5E2)
![k_f=245.24\ rad/s](https://tex.z-dn.net/?f=k_f%3D245.24%5C%20rad%2Fs)
Hence, this is the required solution.