Motion of a Particle: Solutions & Examples

Click For Summary
The discussion focuses on understanding the trajectory of a particle given its acceleration in a specific format. The user is exploring the condition where a constant vector 'a' satisfies the equation a•r=constant, which would imply conical motion. However, there is skepticism regarding the validity of this conclusion, as the proposed equation suggests spherical symmetry rather than a defined cone orientation. The conversation highlights the need for further examples or resources to clarify the relationship between acceleration and trajectory in this context. Ultimately, the inquiry emphasizes the complexity of deriving motion patterns from acceleration data.
Einstenio
Messages
2
Reaction score
1
Homework Statement
Show that a point with acceleration given by:
a=c*((dr/dt)×r)/|r|3
where c is a constant, moves on the surface of a cone.
Relevant Equations
v=dr/dt
This is jut an example to illustrate my doubt. I don't know how to obtain the tracjectory given only the acceleration in this format. I realized that if i can show that there is an constat vector 'a' that satisfy a•r=constant, than the motion would be on the surface of a cone. So i tried to make use of some vectorial identity multiplying by cross product on both sides and try to use the 'BAC-CAB' rule, but that didnt lead to anywhere.

Is there any example similar to this case or anywhere i can study to have a better understanding?
 
Physics news on Phys.org
Einstenio said:
if i can show that there is an constat vector 'a' that satisfy a•r=constant, than the motion would be on the surface of a cone.
Seems to me that would be motion in a plane normal to ##\vec a##.
 
##\ddot{\vec r}=c\frac{\dot{\vec r}\times\vec r}{|r|^3}##?
Seems most unlikely that would give a cone. A cone's axis has an orientation in space, whereas that equation appears to have spherical symmetry.
 
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 1 ·
Replies
1
Views
1K
  • · Replies 62 ·
3
Replies
62
Views
11K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
177
Views
30K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
21
Views
7K
  • · Replies 11 ·
Replies
11
Views
3K