Centripetal Acceleration grade 12 (forces on a bird pulling out of a dive)

[Physicsstudent2]

Homework Statement

A bird of mass 0.211 kg pulls out of a dive, the bottom, of which can be considered to be a circular arc with the radius of 25.6 m. At the bottom of the arc, the bird’s speed is a constant 21.7 m/s. Determine the magnitude of the upward lift on the bird’s wings at the bottom of the arc.

Relevant Equations

F=me
a= v^2/r

How come the solution requires you to do mv^2/4 when the formula is just v^2 /r?

 

[PeroK]

How come the solution requires you to do mv^2/4 when the formula is just v^2 /r?

What do you say that? What solution?

 

[Physicsstudent2]

this question was in my textbook and i didn't know how to do it so i searched it up online and found that picture explaining how to solve the problem

 

[PeroK]

this question was in my textbook and i didn't know how to do it so i searched it up online and found that picture explaining how to solve the problem

The picture looks good to me. Are you forgetting gravity? Which is odd, given it's clearly included in the solution you posted. They may be able to fly, but birds are still subject to the force of gravity!

 

[Physicsstudent2]

The picture looks good to me. Are you forgetting gravity? Which is odd, given it's clearly included in the solution you posted. They may be able to fly, but birds are still subject to the force of gravity!

no its not about gravity i think i didnt explain myself clearly.
in the photo i sent, i indicated an arrow pointing to "m" of the mv^2/r.
im just confused on why the "m" is there when the standard equation for centripetal acceleration is v^2 /r

 

[PeroK]

no its not about gravity i think i didnt explain myself clearly.
in the photo i sent, i indicated an arrow pointing to "m" of the mv^/r.
im just confused on why the "m" is there when the standard equation for centripetal acceleration is v^2 /r

You're trying to calculate the force. "Lift" is a force.

 

[kuruman]

no its not about gravity i think i didnt explain myself clearly.
in the photo i sent, i indicated an arrow pointing to "m" of the mv^2/r.
im just confused on why the "m" is there when the standard equation for centripetal acceleration is v^2 /r

Newton's law says that the net force is equal to mass times acceleration. Here, the net force is F_L-F_G. That's on the right-hand side of the equation. The left-hand side must be mass times acceleration. If acceleration is v²/r, what is mass times acceleration?

 

[Physicsstudent2]

Fnet = FL - Fg
ma = Fl - Fg
(0.211)(21.7^2/25.6) = Fl - mg
oh, i see it now, i was going too fast. the person who solved it just put the mass to multiple the v^2 directly. thank you for making me realize