Physics: Rotational Variables Problem Set?
A propeller is rotating about an axis perpendicular to its center, as the drawing shows. The axis is parallel to the ground. An arrow is fired at the propeller, travels parallel to the axis, and passes through one of the open spaces between the propeller blades. The angular open spaces between the three propeller blades are each π/3 rad (60.0˚). The vertical drop of the arrow may be ignored. What is a maximum value ω for the angular speed of the propeller, beyond which the arrow cannot pass through an open space without being struck by one of the blades. Find this maximum value when the arrow has length L and speed v shown in the following table.
- SlowfingerLv 61 week ago
Open spaces between the three propeller blades are each
θ = π/3 rad.
Time this space is opened for an arrow to pass through is
t = θ / ω ...... (1)
An arrow must pass with all of its length so the corresponding time for which an arrow is passing through the propeller plane is
t = L / v ..... (2)
Equalizing with (1) yields
θ / ω = L / v
ω = θ v / L
For each case as follows
(a) ω = (π/3) * 73 / 0.67 = 114.1 rad/s
(b) ω = (π/3) * 96.6 / 0.67 = 151.0 rad/s
(c) ω = (π/3) * 96.6 / 0.96 = 105.4 rad/s
- billrussell42Lv 71 week ago
depends on where in that open space the arrow passes, and on the diameter of the arrow. The space near the hub is not well defined, but that is the space where the window is smallest.
You have not explained A or B or a or b or c, so this cannot be answered. But I'll do some calculations.
You want a max for ω. There is no max, as it could rotate at a huge rate and not allow the arrow to pass. Perhaps you want the minimum speed? but that depends on where in the open space the arrow passes. If I assume it passes thru at the outside, and the arrow diameter is small...
1/ω is time for 1 rad.
open space is (1/6)2π = π/3 rad
thus (1/ω) time/rad x π/3 rad/opening = π/3ω time/opening
take L as length of arrow, and V speed
time for it to pass it's length = L/V
so set the two equal
L/V = π/3ω