Posted by Taylor on Thursday, November 4, 2010 at 8:53pm.
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?????

PHYSICS HELP  bobpursley, Thursday, November 4, 2010 at 9:04pm
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.
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