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What is the force of gravitational attraction between a thin uniform rod of mass M and length l and a particle of mass m lying in the same line as the rod at a distance a from one end

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- Thread starter -EquinoX-
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What is the force of gravitational attraction between a thin uniform rod of mass M and length l and a particle of mass m lying in the same line as the rod at a distance a from one end

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I dont know where to start this problem

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arildno

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Start by defining where the rod lies, in terms of COORDINATES.

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what do you mean?? well I just tried solving this and is it the integral of GMm/(l-a)^2?

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HallsofIvy

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Since everything there is a constant, it's not much of an integral is it? Also there is no "dx" or "dy" or whatever specifying the variable of integration.what do you mean?? well I just tried solving this and is it the integral of GMm/(l-a)^2?

Arildno said

Start by defining where the rod lies, in terms of COORDINATES.

Since this is a linear problem (the point is in the same line as the rod), you can use a number line rather than Cartesian coordinates.

Set up coordinates so one end of the rod is at 0 and the other is at l. The "test object" will be at -a. Now imagine a tiny piece of that rod between x and x+dx. What is the mass of that piece? The distance from that tiny piece to the test mass is a+ x (for "infinitesmal" dx you can treat that as the single distance for the whole piece- that's the whole point of this method). What is the force that tiny piece on the test mass? Now "add" that up (integrate it from x= 0 to x= l) to find the force of the entire rod on the test mass.

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Did you miss typed this?? Should this be force?? So the answer is the integral from 0 to 1 of Gmm/(a+x)^2

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arildno

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Sigh.

Not even the modelling of a straight line segment does it seem you are able to accomplish.

I'll do this problem for you; pay CLOSE attention to the structure of the solution procedure:

1. Modelling of a straight line segment:

We have only need of a single coordinate variable (or ordinate, strictly speaking) to describe the linie segment:

We let one end point of the rod lie at x=0, whereas the other end point lies at x=L. (L is the length of the rod)

Thus, the rod is modelled by the interval [itex]0\leq{x}\leq{L}[/itex], where any point ON the rod is assigned its own ordinate number "x", lying between 0 and L, where the interpretation of its specific x-value is the distance between the point on the rod and the end-point that has been assigned x-value 0.

2. Positioning of separate point mass:

We let the point mass be placed a distance "a" to the left of the end point at x=0, i.e, the point mass has the position x=-a

3. Uniform density:

The mass density at any point on the rod is M/L.

Therefore, a portion of the rod with length I has mass (M/L)*I.

In particular, an infinitesemal portion of length dx has (infinitesemal) mass dM=(M/L)*dx

4. Gravitational attraction on point mass:

This is quite simply the sum of the attractions it experiences from each point of the rod (each with a location between 0 and L). Thus, summing over the attractions of these mass points (individually called dF)will give us the total force F:

[tex]F=\int_{rod}dF=\int_{rod}\frac{GmdM}{(x-(-a))^{2}}=\int_{rod}\frac{GmMdx}{L(x+a)^{2}}=\frac{GMm}{L}\int_{0}^{L}\frac{dx}{(a+x)^{2}}=\frac{GMm}{L}(-\frac{1}{a+L}+\frac{1}{a})=\frac{GMm}{L}\frac{L}{(a+L)a}=\frac{GMm}{a(a+L)}[/tex]

A somewhat interesting observation can be made of this:

If a>>L, then the gravitational force acting upon the mass point is the same as if the rod's mass was concentrated at its center of mass, whereas if L<<a, then the gravitational force acting upon the mass point is the same as if the rod's mass was concentrated at the geometric mean between "a" and "L".

Not even the modelling of a straight line segment does it seem you are able to accomplish.

I'll do this problem for you; pay CLOSE attention to the structure of the solution procedure:

1. Modelling of a straight line segment:

We have only need of a single coordinate variable (or ordinate, strictly speaking) to describe the linie segment:

We let one end point of the rod lie at x=0, whereas the other end point lies at x=L. (L is the length of the rod)

Thus, the rod is modelled by the interval [itex]0\leq{x}\leq{L}[/itex], where any point ON the rod is assigned its own ordinate number "x", lying between 0 and L, where the interpretation of its specific x-value is the distance between the point on the rod and the end-point that has been assigned x-value 0.

2. Positioning of separate point mass:

We let the point mass be placed a distance "a" to the left of the end point at x=0, i.e, the point mass has the position x=-a

3. Uniform density:

The mass density at any point on the rod is M/L.

Therefore, a portion of the rod with length I has mass (M/L)*I.

In particular, an infinitesemal portion of length dx has (infinitesemal) mass dM=(M/L)*dx

4. Gravitational attraction on point mass:

This is quite simply the sum of the attractions it experiences from each point of the rod (each with a location between 0 and L). Thus, summing over the attractions of these mass points (individually called dF)will give us the total force F:

[tex]F=\int_{rod}dF=\int_{rod}\frac{GmdM}{(x-(-a))^{2}}=\int_{rod}\frac{GmMdx}{L(x+a)^{2}}=\frac{GMm}{L}\int_{0}^{L}\frac{dx}{(a+x)^{2}}=\frac{GMm}{L}(-\frac{1}{a+L}+\frac{1}{a})=\frac{GMm}{L}\frac{L}{(a+L)a}=\frac{GMm}{a(a+L)}[/tex]

A somewhat interesting observation can be made of this:

If a>>L, then the gravitational force acting upon the mass point is the same as if the rod's mass was concentrated at its center of mass, whereas if L<<a, then the gravitational force acting upon the mass point is the same as if the rod's mass was concentrated at the geometric mean between "a" and "L".

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if you put a in the right of the rod then we would get the same answer right??

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arildno

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The MAGNITUDE of the force acting upon the mass point would certainly be the same, but what would be the DIRECTION of this force?

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