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DJBruce

Gravitational Force

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So for my mechanics class we have these two problems on our online homework:

 

"An object in the shape of a thin ring has radius A and mass M . A uniform sphere with mass m and very small radius is placed with its center at a distance x to the right of the center of the ring, along a line through the center of the ring, and perpendicular to its plane. What is the magnitude of the gravitational force that the ring-shaped object exerts on the sphere?"

 

and

 

"A thin, uniform rod has length L and mass M. Calculate the magnitude of the gravitational force the rod exerts on a particle with mass m that is at a point x along the axis of the rod a distance from the closest end of the rod."

 

 

For both of these I found the center of mass of each object, and then calculated the gravitational force from that.

 

1) Since the ring is uniform, we know that its center of mass is located at the center of the circle. Since the point object is located a distance x from the center of the circle I found the gravitational force it feels is:

 

[math]F= \frac{GMm}{x^2}[/math]

 

2) Since the rod is uniform we know the center of mass of the rod is located in its center, which is a distance L/2 away from its ends. Knowing the ball is a distance x from the end of the rod I found the gravitational force to be:

 

 

[math]F= \frac{GMm}{(x+\frac{L}{2})^{2}}[/math]

 

I feel like these answers are correct, but the automated grading system says they are not. Can anyone explain what I am doing wrong?

Edited by DJBruce

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For your second problem, all the gravitational force is acting in the x-direction (positive or negative depends on how you define the axes).

Since the rod has uniform mass, it has a linear mass density of \lamda = (M/L).

 

Each small segment of mass dm = \lamda*dl where dl is the length of the small mass segment.

 

They way I understood the problem is that dF=(-GmdM) / (x+n*dl), where lim (n-> infinity) [ n*dl = L].

 

In order to solve this you take the integral from 0 to infinity of (Gm*(M/L)*dl) / (x+n*dl) in terms of n. You can take Gm*M/L outside the integral and you are left to calculate the integral |0infinity [ dl / (x+n*dl)] dn. All the mathematical manipulations are not entirely correct here because I arrived at this result using the idea of first principles. Also it is important to consider dl and x as constants and only n as the variable....

 

I feel that you can prceed from here by yourself. If you solve the integral and multiply with the terms out side (G, m , etc). you should get the right answer.

 

I have not worked on your other question yet. If I solve it I will inform you. Good luck with the rest of your homework

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The gravitational field of an object is generally not Gm/(r^2), where r is the distance from the center of mass. That is only true in some cases. Your result for 1) is pretty obviously wrong, unless you believe that an infinite gravitational force works in the center of a ring (that would give a whole new aspect to a wedding). The 1st question could be solved without integrating (i.e. just from reasoning). I doubt you'll get around these nasty integration signs in the 2nd one.

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Solved the first one mathematically, although it turned out to actually be easier than I had thought... For a uniform ring, you only need to consider the x-direction of the force.

 

So for the contribution of a small unit mass dM, the force dF_x on the mass m can be given by df_x = dF*cos(\theta).

 

They way I am picturing the ring, \theta is the angle between the direction the force dF is acting and the x-axis.

Using basic trigonometry, cos(\theta) = x / (A^2 + x^2)^1/2.

Also we know dF = GmdM / (A^2 + x^2). therefore you get :

 

dF_x = (Gmx / ((A^2 + x^2)^3/2)) dM. I f you integrate the LHS you get F_x, and the right hand side is an integral in terms of dM.

Since everything else in the RHS is a constant, this is basically a very easy integral. Note that this solution is much more mathematically rigorous than the previous solution.

 

Although I generally support strict mathematical practice, as long as you understand the concepts behind the procedure you will be fine.

 

Hope I was helpfull and good luck with the rest of your homework.

 

 

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