Solving a Horizontal Arc HW Problem: Centripetal Acceleration

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The discussion focuses on calculating the centripetal acceleration of a hawk flying in a horizontal arc with a radius of 13.5 m at a speed of 3.9 m/s, which is determined to be 1.1267 m/s². The next challenge involves finding the total acceleration when the hawk increases its speed at a rate of 0.54 m/s². Participants clarify that in addition to centripetal acceleration, tangential acceleration must be considered, as it is perpendicular to the centripetal acceleration. The correct approach involves vector summation of both accelerations to find the resultant total acceleration, which is calculated to be 1.2494 m/s². Understanding both types of acceleration is crucial for solving the problem accurately.
exi
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Homework Statement



A hawk flies in a horizontal arc of radius 13.5 m with a speed of 3.9 m/s.

What is its centripetal acceleration? (I correctly found this to be 1.1267 m/s².)

Next question is: If it continues to fly along the same horizontal arc but increases its speed at a rate of 0.54 m/s², what is the magnitude of acceleration under these new conditions?

Homework Equations



Centripetal acceleration = v² / r; possibly another one or two.

The Attempt at a Solution



Finding the first half of this was easy enough, as 3.9² / 13.5 = 1.1267. But I have tried three different ways of incorporating this acceleration and have gotten it wrong each time.

The only other thing I can think of trying is using the circumference of this path as displacement, finding its velocity (10.335 after one "lap"), and using that to find a new acceleration, but I'd rather not blow any more answer submissions on this one question. By the above, I'm coming up with 7.913 m/s². Am I on the right track, or am I making this way too difficult?
 
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exi said:

Homework Statement



A hawk flies in a horizontal arc of radius 13.5 m with a speed of 3.9 m/s.

What is its centripetal acceleration? (I correctly found this to be 1.1267 m/s².)

Next question is: If it continues to fly along the same horizontal arc but increases its speed at a rate of 0.54 m/s², what is the magnitude of acceleration under these new conditions?

Homework Equations



Centripetal acceleration = v² / r; possibly another one or two.

The Attempt at a Solution



Finding the first half of this was easy enough, as 3.9² / 13.5 = 1.1267. But I have tried three different ways of incorporating this acceleration and have gotten it wrong each time.

The only other thing I can think of trying is using the circumference of this path as displacement, finding its velocity (10.335 after one "lap"), and using that to find a new acceleration, but I'd rather not blow any more answer submissions on this one question. By the above, I'm coming up with 7.913 m/s². Am I on the right track, or am I making this way too difficult?


Have you learned about tangential acceleration?

If the speed is constant, there is only a centripetal acceleration directed toward the center of the circle with a magnitude given by a_c = \frac{v^2}{R} as you know.

If the speed is not uniform, there is an addition a tangential acceleration (tangent to the circle) given by a_t = \frac{dv}{dt} i.e. it is equal to the rate of change of speed (notice that this is directly the value provided to you in the question).

Now, notice that the two accelerations are perpendicular to each other. To find the total acceleration you have to do a vector sum of those two accelerations. The magnitude of the total acceleration is easy to find.

Hope this helps.
 
I'm not sure what you calculated there, but you are making things more difficult than necessary. You figured out the radial (centripetal) acceleration; They gave you the tangential acceleration. What's the resultant total acceleration?


Oops... nrqed beat me too it!
 
nrqed said:
Have you learned about tangential acceleration?

If the speed is constant, ... .

No, we haven't really talked about that in class. Thanks for the run-down on it.

And yep, 1.2494 m/s² it is.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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