Finding the Length and Velocity of a Spring in Circular Motion

  • Thread starter Thread starter Ruskointhehizzy
  • Start date Start date
  • Tags Tags
    Spring
AI Thread Summary
To solve the problem of a ball in circular motion attached to a spring, the spring's extension can be calculated using the equation T_s = kx, where T_s is the tension in the spring, k is the spring constant, and x is the extension. The correct length of the spring during motion is found by adding the relaxed length to the extension x. The velocity of the ball can then be derived from the tension and the new length of the spring, resulting in a velocity of 0.73 m/s. The discussion emphasizes the importance of clearly defining variables and equations to arrive at the correct solutions. Overall, participants found the problem engaging and helpful for their understanding.
Ruskointhehizzy
Messages
8
Reaction score
2

Homework Statement


A ball of mass 0.250 kg is attached to a spring with spring constant 35.0 N/m and relaxed length 0.200 m. The other end of the spring is attached to a vertical pole. The ball is set into circular motion in the horizontal plane with the spring making an angle 25degrees with the pole.
a) find the length of the spring during this motion.
b) find the velocity of the ball during this motion.

The Attempt at a Solution


I have set it up to find v, but am stuck on how to find x. Thank you very much.

IMG_2284.JPG
 

Attachments

  • IMG_2284.JPG
    IMG_2284.JPG
    32.3 KB · Views: 458
  • Like
Likes ephphatha16
Physics news on Phys.org
##T_s = kx## should be the missing equation. Simply the equation of the spring extension.

Some comments what you calculate where would make it easier to follow.
 
  • Like
Likes ephphatha16 and Ruskointhehizzy
mfb said:
##T_s = kx## should be the missing equation. Simply the equation of the spring extension.

Some comments what you calculate where would make it easier to follow.
Okay I thought so but wasn't sure. I will be sure to include comments from here on forth. Thank you very much! I love this site you guys are the best
 
  • Like
Likes mfb
I’d just like to point out, “L+x” is not a final answer to part a). The ‘extension’ x is a variable introduced by you, not given in the problem.

If you use T_s = kx in your vertical equation (on the left) then you can express x in terms of known variables, and then you could calculate the length explicitly.
 
  • Like
Likes ephphatha16
Nathanael said:
I’d just like to point out, “L+x” is not a final answer to part a). The ‘extension’ x is a variable introduced by you, not given in the problem. .
I set Tx to kx and found x using known terms and it gave me .077 which is correct.
using the tension I found and the new length I was able to calculate the velocity at .73m/s which is also correct. Thanks for pointing that out.

thanks for the help everyone! I really enjoyed this problem.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top