United States Physics 1 With Calculus - Gravitation

AI Thread Summary
The discussion focuses on calculating the gravitational force on a point mass due to a line mass with uniform density, using the gravitational constant G, mass variables m and M, distance D, and length L. Participants clarify the setup of the Cartesian coordinate system and the cancellation of y-components in the gravitational force calculation. A participant expresses concern about their derived formula, suspecting an error in the dimensions and integration process. Another contributor points out mistakes in the dimensions and suggests a clearer approach to the integration. The final answer is confirmed to be (2GmM)/(D sqrt(D^2 + L^2/4), resolving the initial confusion.
GreenPrint
Messages
1,186
Reaction score
0

Homework Statement



See Attachment

Calculate the force of gravity on the point mass due to the line mass in terms of the gravitational constant G, m, M, D, and L. The line mass has a uniform density.

Homework Equations





The Attempt at a Solution



Ok so apparently I'm suppose to consider the point at were the point mass is to be the origin of the Cartesian coordinate system. To the right of this point is positive. Up from this point is positive. Below this point is negative. If the dotted line is extended through the line mass than the y components of the gravity contributed by the infinitismal masses were cancel each other out directly.

See Attachment

I get (G m M L)/(D sqrt(D^2 + L^2/4) )
in my paper I accidentally put

(G m M L)/(D^2 sqrt(D^2 + L^2/4) )
by accident

for some reason I feel as if this answer is wrong and was wondering if anyone could help me with this problem. I question if I'm really suppose to multiply it by two. It thought that I was suppose to because the horizontal components add together directly.
 

Attachments

  • Capture.PNG
    Capture.PNG
    7.4 KB · Views: 422
  • Untitled.jpg
    Untitled.jpg
    25.5 KB · Views: 438
Physics news on Phys.org
I used λ = M/L for the linear mass density and ended up with a slightly different integral.
If you look carefully at the step where you first use the integral sign, the dimensions are not correct. The dL cancels one of the L's on the bottom, so you have dimensions of GMm/L instead of GMm/L². If you write dF = GM*dm/R² where dm = λ*dL, it is easier to see (I left out the cosine). There is a similar mistake in the very last step, replacing a D with a D² or you would have seen that the dimensions were wrong in the end.

You were wise to integrate 0 to L/2 and multiply by 2. I did -L/2 to L/2 and it was a little more awkward to work out.
 
Is

(2GmM)/(D sqrt(D^2 + L^2/4) )

what you got for the final answer?
 
Without the 2. I never had a 2. Your 2 disappeared when you evaluated at L/2.
 
Right, thanks for the help.
 
Most welcome.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged 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