Speed boat question: calculating centripetal acceleration

In summary, a speedboat with a length of 10.2m and a mass of 247 kg is negotiating a circular turn (radius = 33 m) around a buoy. The engine causes a net tangential force of magnitude 521 N to be applied to the boat, with an initial speed of 4 m/sec. The tangential acceleration of the boat 2 seconds into the turn is 2.1 m/sec^2, while the tangential velocity is 8.2 m/sec. The correct formula for centripetal acceleration is a_c = v^2/r, and the total acceleration is found using the formula a = (aT^2 + aC^2)^1/2.
  • #1
zooboodoo11
4
0
1.Speedboat
A speedboat with a length of 10.2m and a mass of 247 kg is negotiaging a circular turn (radius = 33 m) around a buoy. During the turn, the engine causes a net tangential force of magnitude 521 N to be applied to the boat. The intial speed of the boat going into the turn is 4 m/sec.

a) What is the tangential acceleration of the boat, 2 sec into the turn?
aT = m/sec2 *

b) What is the tangential velocity of the boat, 2 sec into the turn?
vT = m/sec *
8.2 OK

c) What is the centripetal acceleration, 2 sec into the turn?
acen = m/sec2
528 NO

d) What is the total acceleration, 2 sec into the turn?
a = m/sec2






2. I got parts A and B, but for part C I tried using the formula: a(cp)= rw^2, i tried using the intitial velocity, 4 m/s and also tried using the calculated velocity from part b, 8.2 and neither are being accepted, am i making a conceptual mistake here in trying to apply this formula here?



3. Solution attempt: a=(33)(4)^2 / 33 and also, a=(33)(8.2)^2 / 33
I think I understand that the total will just be (At^2 + Acp^2) ^ .5 This is my first post so I hope this was in a clear enough format and not misplaced, thanks
 
Physics news on Phys.org
  • #2
sorry, the answer from part A was 2.1m/s^2
 
  • #3
hmm I've tried using different equations and not seeming to get them to work either, could it just be a problem with the computer system taking my answers?
 
  • #4
Hi zooboodoo11,

For part c, in your formula for centripetal acceleration:

[itex]
a_c [/itex]= r w^2


w is the angular velocity, but it looks to me like you are plugging in the linear velocities. There is another formula for centripetal acceleration that uses the linear velocity (which is what you found in part b). What is that formula? Once you find that, I think you'll get the correct answer.

(Or equivalently you could convert the linear velocity from part b into an angular velocity and use your original formula.)
 
  • #5
A(t)= linear velocity/time
A(centripetal)= velocity(linear)^2/radius

A(total)= [A(t)^2+A(c)^2]^1/2
 
  • #6
Hi physixguru,

physixguru said:
A(t)= linear velocity/time
A(centripetal)= velocity(linear)^2/radius

A(total)= [A(t)^2+A(c)^2]^1/2

I don't believe your first equation is quite right. In this problem the tangential acceleration magnitude would be

[tex]
a_T = \frac{\mbox{change in linear speed}}{\mbox{time interval}}
[/tex]

(because we know the tangential part is constant). More generally it is the derivative of the linear speed with respect to time.
 
  • #7
That was not said with respect to the problem in any case.It was a generalization from my side.It was not question specific at all.
 

1. What is the formula for calculating centripetal acceleration?

The formula for calculating centripetal acceleration is a = v^2/r, where a is the acceleration in meters per second squared (m/s^2), v is the velocity in meters per second (m/s), and r is the radius of the circular path in meters (m).

2. How does the speed of a boat affect the centripetal acceleration?

The speed of a boat directly affects the centripetal acceleration. As the speed of the boat increases, the centripetal acceleration also increases, as seen in the formula a = v^2/r. This means that the boat is traveling at a higher velocity and is able to make sharper turns at a given radius.

3. What is the role of the radius in calculating centripetal acceleration?

The radius plays a significant role in calculating centripetal acceleration. As the radius decreases, the centripetal acceleration increases. This is because the boat has to travel a shorter distance to complete a full circle, which requires a higher acceleration.

4. Can you use this formula for objects other than a speed boat?

Yes, this formula can be used for any object that is moving in a circular path, regardless of its size or speed. It is commonly used in physics and engineering to calculate the acceleration of objects in circular motion, such as satellites orbiting around a planet or cars making turns on a racetrack.

5. What other factors can affect centripetal acceleration?

Apart from speed and radius, other factors that can affect centripetal acceleration include the mass of the object and the force acting on it. A heavier boat will require a larger force to maintain a certain centripetal acceleration, while a stronger force will result in a higher acceleration. Friction can also affect centripetal acceleration by slowing down the object and reducing its speed and acceleration.

Similar threads

Replies
12
Views
565
  • Introductory Physics Homework Help
Replies
4
Views
929
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
760
  • Introductory Physics Homework Help
Replies
24
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
958
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top