I am not sure I understand your lift function.
Normally
Lift = Cl (1/2) rho v^2 * area
so I guess you are saying that
Cl*(1/2)rho*area = 0.020
And lift force = L = 0.02 v^2
That will be the force up in the frame of reference of the bat.
Say the bat flies at angle A from horizontal while traveling horizontal
The vertical component of lift is then
L cos A = .02 v^2 cos A
This must equal the weight if traveling horizontal
0.02 v^2 cos A = m g = 0.034 (9.8)
or V^2 cos A = 16.7
cos A = 16.7/v^2
cos may not be greater than 1
v^2 = 16.7
v at least sqrt(16.7)
then
horizontal component = 0.02 v^2 sin A
= m v^2/r = .034 v^2/r
so.02 sin A = .034/r
sin A = 1.7/r
The maximum lift force on a bat is proportional to the square of its flying speed v. For the hoary bat (Lasiurus cinereus), the magnitude of the lift force is given by
FL (0.020 N·s2/m2)v2
The bat can fly in a horizontal circle by "banking" its wings at an angle θ, as shown in the figure below. In this situation, the magnitude of the vertical component of the lift force must equal the bat's weight. The horizontal component of the force provides the centripetal acceleration.
(a) What is the minimum speed that the bat can have if its mass is 0.034 kg?
answer in m/s
(b) If the maximum speed of the bat is 10 m/s, what is the maximum banking angle that allows the bat to stay in a horizontal plane?
answer in °
(c) What is the radius of the circle of its flight when the bat flies at its maximum speed? answer in m
(d) Can the bat turn with a smaller radius by flying more slowly?
Yes or No
I'm lost in this problem
5 answers
b. )
max is 10 m/s
remember V^2 cos A = 16.7 to have vertical component equal to weight
so
cos A = 16.7/v^2 = 16.7/v^2 = 0.167 at v^2=100
then A = 80.4 degrees
after that, gravity wins
max is 10 m/s
remember V^2 cos A = 16.7 to have vertical component equal to weight
so
cos A = 16.7/v^2 = 16.7/v^2 = 0.167 at v^2=100
then A = 80.4 degrees
after that, gravity wins
(c) What is the radius of the circle of its flight when the bat flies at its maximum speed? answer in m
Remember horizontal component = m v^2/r and therefore
sin A = 1.7/r
That Angle for minimum radius = 80.4 at 10m/s
so
r = 1.7 / sin 80.4
r = 1.72 meters is minimum turn radius
Remember horizontal component = m v^2/r and therefore
sin A = 1.7/r
That Angle for minimum radius = 80.4 at 10m/s
so
r = 1.7 / sin 80.4
r = 1.72 meters is minimum turn radius
cos A = 16.7/v^2
as v gets smaller, the cos gets bigger, therefore smaller angle A for least turn radius (flies more level)
and
sin A = 1.7/r
or r = 1.7/sinA
as A gets smaller, r gets BIGGER,
so
min turn radius gets bigger, not smaller with lower speed
thus no
as v gets smaller, the cos gets bigger, therefore smaller angle A for least turn radius (flies more level)
and
sin A = 1.7/r
or r = 1.7/sinA
as A gets smaller, r gets BIGGER,
so
min turn radius gets bigger, not smaller with lower speed
thus no
thank you so much for your detail work
1 answer