mechanics question
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mechanics question
1) a body is acted on by a force towards a point. the magnitude of the body is inversely proportional to the square of distance. the path of the body will be?
rahul Posts : 1
Join date : 20080521
Re: mechanics question
F = k/x^2
where k is a constant of proportionality
F= k/x^2 = mvdv/dx
k/x^2 =mvdv/dx
k/x^2dx = mvdv
on integartion u get
v^2 =k/m(1/x)
No we need x as
dx/dt = sqrt(k/m * 1/x)
sqrt(m/k) sqrt(x)/dx = dt
2/3x^3/2sqrt(m/k) = t
now in principle we can write x as a function of time
Now we have
x^3/2 = 3/2sqrt(k/m)*t
on simplifying
x = {3/2 sqrt (k/m)*t}^2/3
now we have x is proportional to t^2/3
That would mean that we can describe the x in terms of a power function
You can easily check the function
x =t^2/3
That would represent the path of x
caution it is defined only for x<0
Hope that helps
where k is a constant of proportionality
F= k/x^2 = mvdv/dx
k/x^2 =mvdv/dx
k/x^2dx = mvdv
on integartion u get
v^2 =k/m(1/x)
No we need x as
dx/dt = sqrt(k/m * 1/x)
sqrt(m/k) sqrt(x)/dx = dt
2/3x^3/2sqrt(m/k) = t
now in principle we can write x as a function of time
Now we have
x^3/2 = 3/2sqrt(k/m)*t
on simplifying
x = {3/2 sqrt (k/m)*t}^2/3
now we have x is proportional to t^2/3
That would mean that we can describe the x in terms of a power function
You can easily check the function
x =t^2/3
That would represent the path of x
caution it is defined only for x<0
Hope that helps
Pinnaka Admin
 Posts : 36
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