Answer:
It's may help you to answer
Answer:
-775 rad/s^2
Explanation:
Knowing the initial and final angular speed is
and
, respectively. We can use the following formula for equation of motion to calculate the average angular acceleration in t = 0.569 seconds
![\Delta \omega = \alpha t](https://tex.z-dn.net/?f=%5CDelta%20%5Comega%20%3D%20%5Calpha%20t)
![\omega - \omega_0 = \alpha t](https://tex.z-dn.net/?f=%5Comega%20-%20%5Comega_0%20%3D%20%5Calpha%20t)
![0 - 440.9 = \alpha 0.569](https://tex.z-dn.net/?f=0%20-%20440.9%20%3D%20%5Calpha%200.569)
![\alpha = \frac{-440.9}{0.569} = -775 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B-440.9%7D%7B0.569%7D%20%3D%20-775%20rad%2Fs%5E2)
That's a reasonable request, but right now, there's no way to do it yet. First, we need all the information that's in the REST of the question . . . the whole part BEFORE where you started copying.
Answer:
![\omega_1 = 6.06 rad/s](https://tex.z-dn.net/?f=%5Comega_1%20%3D%206.06%20rad%2Fs)
Explanation:
Initially rod is bent into L shaped at mid point
So we will have rotational inertia of the rod given as
![I = \frac{mL^2}{3} + (\frac{mL^2}{12} + m(L^2 + \frac{L^2}{4}))](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BmL%5E2%7D%7B3%7D%20%2B%20%28%5Cfrac%7BmL%5E2%7D%7B12%7D%20%2B%20m%28L%5E2%20%2B%20%5Cfrac%7BL%5E2%7D%7B4%7D%29%29)
![I = \frac{5mL^2}{3}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B5mL%5E2%7D%7B3%7D)
Now when rod becomes straight during the rotation
then we will have
![I_2= \frac{(2m)(2L)^2}{3} = \frac{8mL^2}{3}](https://tex.z-dn.net/?f=I_2%3D%20%5Cfrac%7B%282m%29%282L%29%5E2%7D%7B3%7D%20%3D%20%5Cfrac%7B8mL%5E2%7D%7B3%7D)
now from angular momentum conservation we have
![I \omega_o = I_1 \omega_1](https://tex.z-dn.net/?f=I%20%5Comega_o%20%3D%20I_1%20%5Comega_1)
![(\frac{5mL^2}{3}) (9.7) = \frac{8mL^2}{3}\omega_1](https://tex.z-dn.net/?f=%28%5Cfrac%7B5mL%5E2%7D%7B3%7D%29%20%289.7%29%20%3D%20%5Cfrac%7B8mL%5E2%7D%7B3%7D%5Comega_1)
![\omega_1 = 6.06 rad/s](https://tex.z-dn.net/?f=%5Comega_1%20%3D%206.06%20rad%2Fs)