A particle that feels an angular force only

Click For Summary
The discussion revolves around solving a problem involving a particle experiencing only an angular force, leading to the equation r=Ae^theta + Be^-theta. Participants analyze the equations of motion, particularly focusing on the radial and angular forces, and derive relationships between velocity and radius. A key point is the realization that dw/dt = 0 implies angular velocity is constant, simplifying the equations. The challenge lies in integrating the resulting expressions, with one participant noting difficulties in solving a specific integral. The conversation highlights the importance of substitutions and transformations in solving complex differential equations.
retro10x
Messages
66
Reaction score
0

Homework Statement



"Consider a particle that feels an angular force only, of the form F=2mvw(theta direction). Show that r=Ae^theta + Be^-theta, where A and B are constants of integration."
v=dr/dt
w=d(theta)/dt


Homework Equations



F(radial)= 0 = m((dv/dt)-r*w^2)
F(theta)= 2mvw = m(r*(dw/dt)+2vw)

The Attempt at a Solution



So I can solve the F(theta) equation to find that r*(dw/dt)=0, hence dw/dt=0
and I can also solve the F(radial) equation to find that dv/dt=r*w^2
I also know that this question involves a separation of variables and then integrating
However the problem is that I don't know substitutions I can make to put the equations I have into a form that makes that possible. Any hints or ideas are appreciated ^-^
 
Physics news on Phys.org
dw/dt = 0 means w = W = const. So the first equation is then dv/dt = Cr, where C = W^2 = const. Are you saying you don't know how to solve this equation?
 
Oh, thanks for pointing that out
dv/dt can written as v(dv/dr)
So 0.5v^2=0.5c*r^2+D ;where D is the constant of integration
v=sqrt(cr^2 +2D)=dr/dt
I end up with: t+E=intg[dr/sqrt(cr^2 +2D)] ;where E is the constant of integration
Wolfram Alpha can't even solve this integral :O
What am I missing?
 
That integral can be converted to \int \frac {adx} {\sqrt {x^2 + 1}}.
 
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 11 ·
Replies
11
Views
2K
  • · Replies 27 ·
Replies
27
Views
1K
  • · Replies 42 ·
2
Replies
42
Views
6K
Replies
16
Views
2K
  • · Replies 2 ·
Replies
2
Views
526
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 27 ·
Replies
27
Views
5K
Replies
4
Views
3K