Question

A particle of mass m is placed at a distance d from one end of a uniform rod with length L and mass M as shown in the figure . Find the magnitude of the gravitational force on the particle due to the rod​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Length of rod = L
• Distance from the rod = d.
• Mass of rod = M.
• Mass of the body = m.

$\huge{\bold{\underline{Explanation:-}}}$

#refer the attachment for figure.

As we can see that the distance between small portion of rod and mass "m" is "x".

Now,

Applying gravitational Force for the smaller part.

Therefore,

$\large{\bold{dF = \dfrac{G M dm}{x^2}}}$

As we know,

$\large{\boxed{dm = \dfrac{m}{l} \times dx}}$

Substituting in the above equation,

$\large{dF = \dfrac{GM}{x^2} \times \dfrac{m}{l} \times dx}$

Rearranging,

$\large{dF = \dfrac{GMm}{l} \times \left[ \dfrac{dx}{x^2} \right]}$

Now, integrating,

$\large{ \displaystyle\int dF = \displaystyle\int \dfrac{GMm}{l} \times \left[ \dfrac{dx}{x^2} \right]}$

Applying limits

$\large{ \displaystyle\int^F_0 dF = \displaystyle\int^{(d + l)}_d \dfrac{GMm}{l} \times \left[ \dfrac{dx}{x^2} \right]}$

Note:-

Here the limits have been taken as d and (d +L) because r is the initial distance from the mass "m" and (d + L) is the final distance from the mass "m".

Solving,

$\large{F = \dfrac{GMm}{l} \displaystyle\int^{d+l}_d \left[ \dfrac{dx}{x^2} \right]}$

Integrating,

$\large{F = \dfrac{GMm}{l} \left[ \dfrac{x^{-2 + 1}}{-2 +1} \right]^{d + l}_{d}}$

$\large{F = \dfrac{GMm}{l} \times \left[ \dfrac{x^{-1}}{-1} \right]^{d+l}_d}$

It becomes,

$\large{F = \dfrac{GMm}{l} \times \left[ \dfrac{1}{-x} \right]^{d+l}_{l}}$

Substituting the values of limits,

$\large{F = \dfrac{GMm}{l} \times \left[ - \dfrac{1}{d+l} + \dfrac{1}{d} \right]}$

$\large{F = \dfrac{GMm}{l} \times \left[ \dfrac{d + l - d}{d(l+d)} \right]}$

Final result comes as,

$\huge{\boxed{\boxed{F = \dfrac{GMm}{l} \left[ \dfrac{ l}{d(l+d)} \right]}}}$

Answer 2

Explanation:

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