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

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

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

(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

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

thank you so much for your detail work

No worries! I'll be happy to help you solve this problem step by step.

Let's start with part (a):

(a) The question asks for the minimum speed that the bat can have, given its mass of 0.034 kg. To find this, we need to set the vertical component of the lift force equal to the bat's weight, which is given by Fg = mg, where m is the mass of the bat (0.034 kg) and g is the acceleration due to gravity (9.8 m/s^2).

In this case, since the lift force is proportional to the square of the flying speed, we can write the equation as:
FL = mg

Substituting the given values, we get:
(0.020 N·s^2/m^2)v^2 = (0.034 kg)(9.8 m/s^2)

Now we can solve this equation for v. Dividing both sides of the equation by (0.020 N·s^2/m^2) gives us:
v^2 = [(0.034 kg)(9.8 m/s^2)] / (0.020 N·s^2/m^2)

Taking the square root of both sides gives us the value of v:
v = sqrt{ [(0.034 kg)(9.8 m/s^2)] / (0.020 N·s^2/m^2) }

Now, you can plug in the given values and calculate the minimum speed v in meters per second.

Next, let's move on to part (b):

(b) In this part, we are given the maximum speed of the bat, which is 10 m/s, and we need to find the maximum banking angle that allows the bat to stay in a horizontal plane.

To solve this, we need to set the horizontal component of the lift force equal to the centripetal force, which is given by Fc = m*v^2 / r, where m is the mass of the bat (0.034 kg), v is the maximum speed (10 m/s), and r is the radius of the circle.

The horizontal component of the lift force is given by Fh = FL * cos(θ), where θ is the banking angle.

Setting Fh equal to Fc, we get:
FL * cos(θ) = m*v^2 / r

Now, we can substitute the given values and the equation for FL to solve for the maximum banking angle θ.

Moving on to part (c):

(c) Here, we are asked to find the radius of the circle of flight when the bat flies at its maximum speed, which is 10 m/s.

To solve this, we can use the equation we derived in part (b):
FL * cos(θ) = m * v^2 / r

If we rearrange the equation to solve for r, we get:
r = m * v^2 / (FL * cos(θ))

Now, we can substitute the given values to find the radius of the circular flight.

Finally, let's address part (d):

(d) The question asks if the bat can turn with a smaller radius by flying more slowly. To answer this, we need to understand the relationship between the maximum speed (v) and the radius of the flight (r). From our earlier equations, we found that the radius of the flight is inversely proportional to the square of the speed (v^2).

Based on this relationship, we can conclude that as the speed decreases, the radius of the flight will increase. Therefore, the bat cannot turn with a smaller radius by flying more slowly.

I hope this explanation helps you solve the problem! Let me know if you have any further questions.