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July 30, 2014

July 30, 2014

Posted by **Ana** on Saturday, March 13, 2010 at 7:28pm.

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

- Physics -
**Damon**, Saturday, March 13, 2010 at 7:54pmI 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

- Physics -
**Damon**, Saturday, March 13, 2010 at 8:03pmb. )

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

- Physics -
**Damon**, Saturday, March 13, 2010 at 8:10pm(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

- Physics -
**Damon**, Saturday, March 13, 2010 at 8:16pmcos 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

- Physics -
**Ana**, Saturday, March 13, 2010 at 8:29pmthank you so much for your detail work

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