It’s important because if you were let’s say training to see how far you go in 10sec you need to record your data to see how much you are improving in that skill or something
Answer:
a) 5.197rev/s
b) Kf/Ki =2.28
Explanation:
a) Angular momentum of the system L = Iw
ButLi=Lf
Kiwi =Ifwf
wf = (Ii/If)will = (4.65/3.4)×3.8=5.197rev/s
b)Kinetic energy KE= 0.5Iw^2
Ki = 0.5Iiwi^2
Kf=0.5Ifwf^2
Kf/Ki = Ifwf/Iiwi
Kf/Ki = (4.65/3.4))(5.197/3.8)
Kf/Ki = 1.22(1.368)^2
Kf/Ki = 2.28
Answer:

Explanation:
Given that :
mass of the SUV is = 2140 kg
moment of inertia about G , i.e
= 875 kg.m²
We know from the conservation of angular momentum that:

![mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=mv_1%20%2A0.765%20%3D%20%5BI%2Bm%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)
![2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=2140v_1%2A0.765%20%3D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)



From the conservation of energy as well;we have :

^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%280.4262%20%5C%20v_1%29%5E2%20-2140%289.81%29%5B%5Csqrt%7B0.76%5E2%2B0.895%5E2%7D%20-0.765%5D%5D%20%3D0)





