First slide

Connected bodies

Question

The linear mass density, i.e., mass per unit length of the rope shown in fig. varies 0 to $λ$ from one end
to another. The acceleration of the combined system of figure will be

NA

A

$\frac{F}{[M + (λL / 2)]}$
B

$\frac{F}{(m + λL)}$
C

$\frac{F}{(M + 2 λL)}$
D

$\frac{F}{[(M / 2) + λL]}$

Solution

Consider a small mass dm of the rope at a distance x and x + dx as shown in fig.
Linear mass density at x is given by
$\begin{array}{lll} λ_{x} & = \frac{λ}{L} x \\ ∴ & d_{m} & = λ_{x} dx = \frac{λ}{L} xdx \\ So, & m & = \int_{0}^{L} \frac{λ}{L} xdx = \frac{λ}{L} {[\frac{x^{2}}{2}]}_{0}^{L} = \frac{λL}{2} \end{array}$
If a be the combined acceleration, then
$a = \frac{F}{M + m} = \frac{F}{[M + (λL / 2)]}$ .

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