Finding angular acceleration from revolutions and velocity

Click For Summary
To find the angular acceleration of an object starting from rest with a final angular velocity of 6 rad/s after completing 2 revolutions, the revolutions are treated as the distance traveled in angular terms. Using the equation wf^2 = wi^2 + 2αd, where d is the angular distance in radians (2 revolutions equals 4π radians), the calculation leads to α = 1.43 rad/s². The user expresses uncertainty about the correctness of this solution due to the lack of a provided answer. The discussion emphasizes the importance of correctly interpreting revolutions as angular distance in such calculations.
hpthgpjo
Messages
2
Reaction score
0

Homework Statement


an object starts from rest and has a final angular velocity of 6 rad/s. the object makes 2 complete revolutions. find the object's angular acceleration.

Homework Equations


wf^2=wi^2+2αd

The Attempt at a Solution


Not sure what to do with the revolutions, would it take act as the distance traveled? I am not given the radius, so the most I can do is:
6^2=0+2α(2rev*2π)
α=1.43rad/s^2
I'm not given the answer for this question so I am not sure if I am right or extremely wrong
 
Physics news on Phys.org
hpthgpjo said:

Homework Statement


an object starts from rest and has a final angular velocity of 6 rad/s. the object makes 2 complete revolutions. find the object's angular acceleration.

Homework Equations


wf^2=wi^2+2αd

The Attempt at a Solution


Not sure what to do with the revolutions, would it take act as the distance traveled? I am not given the radius, so the most I can do is:
6^2=0+2α(2rev*2π)
α=1.43rad/s^2
I'm not given the answer for this question so I am not sure if I am right or extremely wrong
Looks good.
 
alright thank you!
 
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?

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
918
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
67
Views
4K