Answer:
a) 3.94 rad/s
b) 11.84 rad/s
Step-by-step explanation:
Given:
Distance, s = 78 cm = 0.78 m
Now the time taken, t
we know
![s = ut +(1)/(2)gt^2](https://img.qammunity.org/2020/formulas/physics/high-school/ulcmebdgewam4tocqgkkn5qx9xsua4482n.png)
where,
s = distance
u = initial speed
g = acceleration due to gravity
since it is a free fall. thus, u = 0
thus, we get
![0.78 = 0* t +(1)/(2)9.8* t^2](https://img.qammunity.org/2020/formulas/physics/high-school/uirwzjb0ophspwtyxcbmam2mt0s2w09y1i.png)
or
![t=\sqrt{(2* 0.78)/(9.8)}=0.398 s](https://img.qammunity.org/2020/formulas/physics/high-school/vspyk6k3ilew6fuscs5dgopqh47tediudu.png)
a) now, the smallest angle will be 1/4 of the revolution so as to fall on the butter side i.e 90° or (
)
also,
![angular\ speed\ (\omega) = (Change\ in\ angle)/(time)](https://img.qammunity.org/2020/formulas/physics/high-school/xzakalak99wnq0x4sa2gl2ro80f6zsdq4n.png)
thus, we have
![angular\ speed\ (\omega) = ((\pi)/(2))/(0.398) = 3.94rad/s](https://img.qammunity.org/2020/formulas/physics/high-school/98c3k03kydcjmbmzxjzd469pk9hdfdj736.png)
b)now, the largest angle will be 3/4 of the revolution so as to fall on the butter side i.e 270° or (
) contributing to the largest angular speed
![angular\ speed\ (\omega) = (Change\ in\ angle)/(time)](https://img.qammunity.org/2020/formulas/physics/high-school/xzakalak99wnq0x4sa2gl2ro80f6zsdq4n.png)
thus, we have
![angular\ speed\ (\omega) = ((3\pi)/(2))/(0.398) =11.84rad/s](https://img.qammunity.org/2020/formulas/physics/high-school/n18duo9b7h223fqqejavmyuo4yxbjn7j3b.png)