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

In summary: Newton's law says that the net force is equal to mass times acceleration. Here, the net force is FL-FG. 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?You are welcome.
  • #1
Physicsstudent2
7
1
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
1626112987591.png
How come the solution requires you to do mv^2/4 when the formula is just v^2 /r?
 
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  • #2
Physicsstudent2 said:
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?
 
  • #3
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
 
  • #4
Physicsstudent2 said:
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!
 
  • #5
PeroK said:
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
 
  • #6
Physicsstudent2 said:
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.
 
  • #7
Physicsstudent2 said:
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 FL-FG. 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?
 
  • #8
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
 
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FAQ: Centripetal Acceleration grade 12 (forces on a bird pulling out of a dive)

What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle and its magnitude is equal to the square of the object's velocity divided by the radius of the circle.

How does centripetal acceleration affect a bird pulling out of a dive?

When a bird is pulling out of a dive, it experiences a centripetal acceleration towards the center of its circular path. This acceleration helps the bird to change its direction and maintain its circular path.

What forces are acting on a bird pulling out of a dive?

There are two main forces acting on a bird pulling out of a dive: the force of gravity and the force of lift. The force of gravity pulls the bird towards the ground, while the force of lift, generated by the bird's wings, acts in the opposite direction and helps the bird to stay in the air.

How does centripetal acceleration affect the bird's wings during a dive?

As the bird pulls out of a dive, the centripetal acceleration causes a change in the direction of the bird's velocity. This change in direction also affects the direction of the lift force acting on the bird's wings, causing them to tilt upwards and provide the necessary lift to maintain the bird's circular path.

What factors can affect the magnitude of centripetal acceleration on a bird pulling out of a dive?

The magnitude of centripetal acceleration on a bird pulling out of a dive can be affected by the bird's velocity, the radius of its circular path, and the mass of the bird. A higher velocity or a smaller radius will result in a greater centripetal acceleration, while a greater mass will require a larger centripetal acceleration to maintain the circular path.

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