Thank you for the respond!! I appreciate it. But it says on my paper

360 rev/min = 360 (2pi/60) rad/s = 12pi rad/s and so
D= 2r = 2(s/theta) = 2(40/12pi) ft = 20/3pi ft = 2.12 ft should be the answer :(

I have 2 more if that's alright.

A point on the rim of a turbine wheel of diameter 10 ft moves with a linear speed 45 ft/s. Find the rate at which wheel turn (angular speed) in rad/s and in rev/s.

Circumference = pi*D = 3.14 * 10 = 31.4 ft.

Va = 1rev/31.4ft. * 6.28rad/rev * 45ft./s = 9 rad/s.

Va = 9rad/s * 1rev/6.28rad =

To find the rate at which the wheel turns (angular speed) in rad/s and in rev/s, you can use the formula:

Angular speed (ω) = Linear speed (v) / Radius (r)

Given that the diameter of the turbine wheel is 10 ft, the radius (r) would be half of that, which is 5 ft. The linear speed (v) is given as 45 ft/s.

1. Finding the angular speed in rad/s:
Angular speed (ω) = 45 ft/s / 5 ft
= 9 rad/s

Therefore, the angular speed of the wheel is 9 rad/s.

2. Finding the angular speed in rev/s:
To convert the angular speed from rad/s to rev/s, you need to use the conversion factor: 1 revolution = 2π radians.

So, angular speed in rev/s = 9 rad/s / (2π rad/revolution)
= 4.54 rev/s

Therefore, the angular speed of the wheel is approximately 4.54 rev/s.