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?
then i was given,
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.
****BUT i don't know how i already know v^2?????
a) lift*sinTheta=mv^2/r and
lift*cosTheta=mg
well, it appears your comment is right, you are looking for min v given mass. Well, 0.021v^2*CosTheta=mg
minimum V will be when CosTheta is max, or .021v^2=mg solve for v.
b) now you know v, find r.
first, find from the second equation .021v^2*cosTheta=mg solve for Theta.
c) Put that in the first equation, find r.
Well, you don't know the value of v^2 directly, but you can solve for it using the equations given. Let me try to help you out with that!
First, let's tackle part (a). The minimum speed that the bat can have is when the vertical component of the lift force is exactly equal to its weight. We can use the equation FL*cosTheta = mg and solve for v^2.
Since we know the values of FL (0.021 N·s^2/m^2), m (0.037 kg), and g (9.8 m/s^2), we can substitute them into the equation and solve for v^2.
0.021 * v^2 * cosTheta = 0.037 * 9.8
Now, I have to point out that we still don't know what the value of cosTheta is. However, we can assume it's the minimum possible value of -1, which essentially means the wings are completely flat. So, we'll go ahead and substitute that in to give you a minimum value for v^2.
0.021 * v^2 * (-1) = 0.037 * 9.8
Now, you can solve for v^2 and find the minimum possible speed.
For part (b), you're given the maximum speed of the bat (11 m/s) and you need to find the maximum banking angle that allows the bat to stay in a horizontal plane. Again, we can use the equation FL*sinTheta = m*v^2/r and solve for Theta.
Since we know the values of FL (0.021 N·s^2/m^2), m (0.037 kg), v (11 m/s), and the radius (which we need to find), we can substitute them into the equation and solve for Theta.
0.021 * sinTheta = 0.037 * (11^2) / r
Once you have the equation, rearrange it to solve for Theta in terms of r. Then you can find the maximum possible banking angle that allows the bat to stay in a horizontal plane.
Lastly, for part (c), we already have the maximum speed (11 m/s) and we need to find the radius of the circle of its flight. Again, we can use the equation FL*sinTheta = m*v^2/r and solve for r this time.
Since we know the values of FL (0.021 N·s^2/m^2), m (0.037 kg), v (11 m/s), and we're trying to find r, we can substitute them into the equation and solve for r.
0.021 * sinTheta = 0.037 * (11^2) / r
Now you can rearrange the equation to solve for r and find the radius of the circle of flight when the bat is flying at its maximum speed.
I hope that clarifies things a bit! Good luck with your calculations, and feel free to come back if you have any more questions!
In order to solve the problem, we need to use the given equation for the magnitude of lift force, FL:
FL = (0.021 N·s^2/m^2)v^2
where FL is the lift force, v is the flying speed of the bat, and N, s, and m^2 are the units for force, time, and area respectively.
(a) To find the minimum speed at which the bat can fly, we need to determine the speed at which the lift force is equal to the weight of the bat. The weight of the bat can be calculated using its mass (m = 0.037 kg) and acceleration due to gravity (g = 9.8 m/s^2):
Weight = mg = (0.037 kg)(9.8 m/s^2)
Equating this to the magnitude of the lift force:
Weight = FL
Substituting the expression for FL:
(0.037 kg)(9.8 m/s^2) = (0.021 N·s^2/m^2)v^2
Now we can solve for v^2:
v^2 = (0.037 kg)(9.8 m/s^2) / (0.021 N·s^2/m^2)
Calculate the value of v^2 using the given values and then take the square root to find the minimum speed v.
(b) To find the maximum banking angle (θ) that allows the bat to stay in a horizontal plane, we need to consider the vertical and horizontal components of the lift force. The vertical component of the lift force balances the weight of the bat, while the horizontal component provides the centripetal acceleration.
The vertical component of the lift force is:
Vertical component = FL sin(θ)
We know that the vertical component of the lift force is equal to the weight:
Weight = FL sin(θ)
The horizontal component of the lift force provides the centripetal acceleration:
Centripetal force = FL cos(θ) = mv^2 / r
Substituting the expression for FL:
FL cos(θ) = mv^2 / r
Now we can solve for the maximum banking angle θ.
(c) To find the radius of the circle of flight when the bat flies at its maximum speed, we can use the equation for the centripetal force:
Centripetal force = FL cos(θ) = mv^2 / r
We know the maximum speed (v = 11 m/s), and we can find the value of FL from the given equation. Substitute the values into the equation and solve for the radius (r).
In this problem, you are given an equation that relates the lift force (FL) to the flying speed (v) of the bat. The equation is given as:
FL = (0.021 N·s²/m²) v²
To find the minimum speed that the bat can have if its mass is 0.037 kg, we need to equate the vertical component of the lift force to the bat's weight.
The vertical component of the lift force is given by:
FL * sinθ = mv²/r (Equation 1)
Where θ is the banking angle, m is the mass of the bat, v is the speed of the bat, and r is the radius of the circle of flight.
Since the bat is in a horizontal circle, the vertical component of the lift force must equal the weight of the bat. So, we can write:
FL * sinθ = mg (Equation 2)
Now, to find the minimum speed, we need to find the value of v that satisfies Equation 2.
From Equation 2, we have:
FL * sinθ = mg
Substituting the given equation for the magnitude of the lift force (FL), we have:
(0.021 N·s²/m²) v² * sinθ = mg
Rearranging the equation, we get:
v² * sinθ = mg / (0.021 N·s²/m²)
Now, we can solve for v²:
v² = (mg / (0.021 N·s²/m²)) / sinθ
Plugging in the values of m = 0.037 kg and g = 9.8 m/s² (acceleration due to gravity), and assuming a value for sinθ (e.g., 1, which corresponds to a banking angle of 90 degrees), we can then calculate v².
Once we have the value of v², we can find v by simply taking the square root of v². This will give us the minimum speed that the bat can have given its mass.
Regarding your question about already knowing v², at the beginning of the problem, there might be additional context or previous calculations that would provide the necessary value of v². If not, you may need to solve the equations iteratively or make assumptions to find an approximate solution.