Centripital Acceleration Question

  • Thread starter Thread starter pinky2468
  • Start date Start date
  • Tags Tags
    Acceleration
Click For Summary
SUMMARY

The discussion focuses on calculating tangential and centripetal acceleration for a 220 kg boat negotiating a circular turn with a radius of 32 meters. The net tangential force applied to the boat is 550 N, resulting in a tangential acceleration of 2.5 m/s². After 2 seconds into the turn, the centripetal acceleration is determined to be 0.781 m/s², using the formula a = v²/r, where the constant tangential speed is 5.0 m/s.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with circular motion concepts
  • Knowledge of tangential and centripetal acceleration formulas
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of centripetal acceleration formulas
  • Explore the relationship between tangential and angular acceleration
  • Learn about forces acting on objects in circular motion
  • Investigate real-world applications of centripetal acceleration in engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for practical examples of acceleration calculations.

pinky2468
Messages
86
Reaction score
0
I can get the first part, but I am getting tripped up on the second part.

A 220 kg boat is negotiating a circular turn (radius=32m) around a buoy. During the turn, the engine causes a net tangetntial force of magnitude 550N to be applied to the boat. The initial tangential speed of the boat going around the turn is 5.0 m/s a)find the tangential acceleration b)after the boat is 2.0 s into the turn find the centripetal acceleration.

So for part a)
F=mr(alpha) (alpha=angular acceleration, I can't make the symbol) alpha= .078 rad/s^2

Tang. acceleration= r(alpha)= 2.5 m/s^2

I am stuck on part b, I think the fact that I am given the initial tangential speed. So do I find the final and take the average?
 
Physics news on Phys.org
part b

Since you are given the force of the engine, and the intial speed, you know the acceleration of the boat. From that you can deduce the final speed of the boat after 2 seconds from v2 = v1 + at(t = 2,a=550/massofboat,v1=5m/s). Then the acceleration you seek is simply
a = v^2/R m/s^2
 


For part b), you do not need to find the final tangential speed. You can use the formula for centripetal acceleration: a = v^2/r, where v is the tangential speed and r is the radius of the circular turn. Since the tangential speed remains constant at 5.0 m/s, you can plug that in for v and the given radius of 32m to find the centripetal acceleration.

a = (5.0 m/s)^2 / 32m = 0.781 m/s^2

So the centripetal acceleration of the boat after 2.0 seconds into the turn is 0.781 m/s^2.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
Replies
12
Views
2K
Replies
15
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
3
Views
3K
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K