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 (0.021 N·s2/m2)v2
The bat can fly in a horizontal circle by "banking" its wings at an angle è, 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 0.037 kg?

(b) If the maximum speed of the bat is 11 m/s, what is the maximum banking angle that allows the bat to stay in a horizontal plane?

(c) What is the radius of the circle of its flight when the bat flies at its maximum speed?

lift*cosTheta=mg

lift*sinTheta=mv^2/r

on the first equation
.021v^2*cosTheta=mg
you know m, g, and v^2, solve for Theta

Then on the second, solve for r.

How do I know v^2? for a)

To solve these problems, we need to apply the given information and use the principles of circular motion and forces. Let's go step by step:

(a) What is the minimum speed that the bat can have if its mass is 0.037 kg?

In this situation, the magnitude of the vertical component of the lift force must equal the bat's weight. We can equate the lift force to the weight:

FL = mg

Given: m = 0.037 kg, g = 9.8 m/s^2

0.021 N·s^2/m^2 * v^2 = 0.037 kg * 9.8 m/s^2

Simplifying the equation:

v^2 = (0.037 kg * 9.8 m/s^2) / 0.021 N·s^2/m^2

v^2 = 0.1717 kg·m^2/s^2

Taking the square root:

v = √(0.1717 kg·m^2/s^2)

Using a calculator, we find:

v ≈ 0.414 m/s

Therefore, the minimum speed that the bat can have is approximately 0.414 m/s.

(b) If the maximum speed of the bat is 11 m/s, what is the maximum banking angle that allows the bat to stay in a horizontal plane?

To find the maximum banking angle, we need to consider the vertical component of the lift force balancing the bat's weight. At the maximum speed, the lift force must be equal to the weight, given by:

FL = mg

Given: m = 0.037 kg, g = 9.8 m/s^2, and v = 11 m/s

0.021 N·s^2/m^2 * (11 m/s)^2 = 0.037 kg * 9.8 m/s^2

0.021 N·s^2/m^2 * 121 m^2/s^2 = 0.037 kg * 9.8 m/s^2

Simplifying the equation:

2.541 N = 0.3626 N

Since the equation doesn't hold, it means that the lift force is not enough to balance the weight at the maximum speed. Therefore, the bat cannot maintain a horizontal plane at its maximum speed. Hence, there is no maximum banking angle.

(c) What is the radius of the circle of its flight when the bat flies at its maximum speed?

To find the radius of the flight circle, we need to use the centripetal force equation:

Fc = mv^2 / r

Where Fc is the centripetal force, m is the mass, v is the velocity, and r is the radius of the circle.

Given: m = 0.037 kg, v = 11 m/s

We need to calculate the centripetal force Fc using the magnitude of the lift force FL:

Fc = FL * sin(θ)

FL = 0.021 N·s^2/m^2 * (11 m/s)^2

Fc = 0.021 N·s^2/m^2 * 121 m^2/s^2 * sin(θ)

Since sin(θ) can vary between -1 and +1, the centripetal force Fc can be calculated with the maximum and minimum values:

Fc_min = 0.021 N·s^2/m^2 * 121 m^2/s^2 * sin(-1)
Fc_max = 0.021 N·s^2/m^2 * 121 m^2/s^2 * sin(1)

Using these values, we can calculate the range of possible radii for the flight circle.