Asked by Geee
Find the diameter of a pulley which is driven at 360 rpm by a belt moving at 40 ft/s.
Then in 1 s the pulley turns through an angle theta measuring 12 pi radians and a point on the rim travels a distance s= 40 ft.
Then in 1 s the pulley turns through an angle theta measuring 12 pi radians and a point on the rim travels a distance s= 40 ft.
Answers
Answered by
Reiny
let the diameter be D ft
in one rotation the belt moves Dπ ft.
The wheel rotates at 360 rpm, or at 60 rotations per second
In that 1 second the belt covers 40 ft.
so 60(Dπ) = 40
Dπ = 2/3
D = 2/(3π) ft = appr .2122 ft or appr 2.546 inches
in one rotation the belt moves Dπ ft.
The wheel rotates at 360 rpm, or at 60 rotations per second
In that 1 second the belt covers 40 ft.
so 60(Dπ) = 40
Dπ = 2/3
D = 2/(3π) ft = appr .2122 ft or appr 2.546 inches
Answered by
Henry
Circumference = pi*D = 3.14D ft.
Vp = Vb, Vp is velocity of the pulley, and Vb is velocity of the belt.
3.14D ft/rev * 360rev/60s = 40.
D = 2.12 Ft. = Diameter of the pulley.
Vp = Vb, Vp is velocity of the pulley, and Vb is velocity of the belt.
3.14D ft/rev * 360rev/60s = 40.
D = 2.12 Ft. = Diameter of the pulley.
Answered by
Anonymous
360 rpm=6 rps
(360rev./min.)(1min./60sec.)=6rps
(360rev./min.)(1min./60sec.)=6rps
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