What is the gravitational force of attraction between a particle and a rod?

  • Thread starter Thread starter tech_guru
  • Start date Start date
  • Tags Tags
    Gravitation
AI Thread Summary
To calculate the gravitational force of attraction between a particle of mass m and a uniform slender rod of mass M and length L, integration is necessary. The distance between the particle and the rod is denoted as D. The process involves dividing the rod into small segments, determining the force contribution from each segment, and integrating these contributions. The mass of each segment is derived from the linear density of the rod, which is M/L. The expected result of the calculation is [2GmM]/[D{L^2 + 4D^2}^0.5].
tech_guru
Messages
1
Reaction score
0
OK...this must b simple...but I don't know how to go 'bout...it needs some integration for sure...I have also attached the diagram with it.

Find the gravitation force of attraction between a particle of mass m and a uniform slender rod of mass M and length L for the orientation given in the diagram.Thed istance between the particle and the rod is D.


The answer should come out to be-

[2GmM]/[D{L^2 + 4 D^2}^0.5]
 

Attachments

  • PHYSICS.JPG
    PHYSICS.JPG
    4.7 KB · Views: 454
Physics news on Phys.org
I can't see the picture, but what you generally do in problems like this is break apart the rod into small pieces (dL), find the contribution of one piece to the force (dF), and then integrate. The mass of this small piece will be (M/L)dl, since M/L is the linear density (I'm assuming that the rod has uniform density).
 
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...
Back
Top