Dynamics Problem (Find Angular Velocity)

  • Thread starter parislad
  • Start date
  • #1
19
0
Hope you can offer me some help with this question - I feel close to getting it myself but have spent ages looking at it now.

Question and working so far is attached.

I think what I need to do to enable me to perform the integral is to get theta as a function of time, or instead somehow convert the integral so I am integrating with respect to theta (with limits 0 and pi/2 ). But I am stuck at this point.

n.b. Angular velocity omega is written as theta_dot in my notation.

Thanks for any help in advance.
 

Attachments

Answers and Replies

  • #2
121
0
Try using an energy method: Potential energy of the cocked spring is equal to gravitational potential energy of the upright rod plus kinetic energy (translational & rotational) of the moving rod.
 
  • #3
19
0
obafgkmrns, thanks.

I've solved the question using your recommended approach , and I've attached the working just for closure.
 

Attachments

  • #4
19
0
Although if anyone does know how they would solve it via something like my original route, it would still be useful.
 
  • #5
121
0
You can do it using your original approach of solving the equation of motion, but the solution is in terms of Jacobi elliptic functions. It's just not worth the trouble!

Brief description of elliptic functions here http://en.wikipedia.org/wiki/Pendulum_(mathematics [Broken]) and here http://en.wikipedia.org/wiki/Jacobi_elliptic_functions
 
Last edited by a moderator:
  • #6
121
0
It occurred to me that I may have misunderstood your original approach to solving the problem. If all you need is angular velocity (theta dot), then you can derive an equation for theta dot by using an integrating factor. When you multiply all terms by theta dot, you can integrate the equation explicitly to find theta dot as a function of theta.
 

Related Threads on Dynamics Problem (Find Angular Velocity)

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
0
Views
2K
Replies
3
Views
2K
Replies
1
Views
3K
Replies
30
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
4K
  • Last Post
Replies
1
Views
457
Replies
2
Views
2K
Top