what is the gravitaionalforce between a particle of mass m and a rod of lenght L which are kept at a distance a

Asked by Pallavi Chaturvedi | 6th Dec, 2013, 10:54: PM

Expert Answer:

Given that,
Length of the rod = L
Mass of the particle = M
 
 
 
Let rod AB is placed such that the distance of its near end from the the particle is r.
 
Cosider an elementary portion PQ of the rod, such that OP = x    OQ = x + dx
mass of the portion PQ of the rod = m/L
 
 
 
 
. 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


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P: 11,935
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  0xL , 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:


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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:

Read more: http://www.physicsforums.com
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:
F=roddF=rodGmdM(x(a))2=rodGmMdxL(x+a)2=GMmLL0\fracdx(a+x)2=GMmL(1a+L+1a)=GMmLL(a+L)a=GMma(a+L)


Read more: http://www.physicsforums.com
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:
F=roddF=rodGmdM(x(a))2=rodGmMdxL(x+a)2=GMmLL0\fracdx(a+x)2=GMmL(1a+L+1a)=GMmLL(a+L)a=GMma(a+L)


Read more: http://www.physicsforums.com
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 0x

Answered by Komal Parmar | 9th Dec, 2013, 05:19: PM

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