Solving a Horizontal Arc HW Problem: Centripetal Acceleration

In summary, the centripetal acceleration of a hawk flying in a horizontal arc of radius 13.5 m with a speed of 3.9 m/s is 1.1267 m/s². If the hawk increases its speed at a rate of 0.54 m/s² while continuing to fly along the same arc, the magnitude of its acceleration under these new conditions is 1.2494 m/s². This is calculated by adding the tangential acceleration (0.54 m/s²) to the centripetal acceleration (1.1267 m/s²) using vector addition.
  • #1
exi
85
0

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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  • #2
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 [itex] a_c = \frac{v^2}{R} [/itex] as you know.

If the speed is not uniform, there is an addition a tangential acceleration (tangent to the circle) given by [itex] a_t = \frac{dv}{dt} [/itex] 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.
 
  • #3
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!
 
  • #4
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.
 

1. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It always points towards the center of the circle and is necessary to keep an object moving along a curved path.

2. How do you calculate centripetal acceleration?

The formula for calculating centripetal acceleration is a = v^2/r, where a is the centripetal acceleration, v is the velocity, and r is the radius of the circle.

3. What is the difference between centripetal acceleration and tangential acceleration?

Centripetal acceleration is the acceleration towards the center of a circle, while tangential acceleration is the acceleration along the tangent of the circle. They are both components of the total acceleration of an object moving in a circular path.

4. How does mass affect centripetal acceleration?

Mass does not directly affect centripetal acceleration. However, a larger mass will require a greater centripetal force to keep it moving in a circle at a constant speed.

5. How is centripetal acceleration used in real life?

Centripetal acceleration plays a role in many real-life situations, such as the motion of planets around the sun, the rotation of a car around a curve, and the spinning of a ball on a string. It is also important in amusement park rides, such as roller coasters and carousels.

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