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  • A thick rope of mass '  ' and length ' <img alt="10 \mathrm{~m}" src="h

A thick rope of mass ' 1\mathrm{Kg} ' and length ' 10 \mathrm{~m} ' is coiled into a roll of radius ' 10 \mathrm{~cm} '. If this rope is set into the rolling motion while keeping its other end fixed. If the initial velocity given to the centre of mass is ' 1 \mathrm{~m} / \mathrm{s} '. Find the vertical component of the linear momentum just after its release.

Option: 1

0.005 \mathrm{~kg}-\mathrm{m}-{s}^{-1}


Option: 2

-0.015 \mathrm{~kg}-\mathrm{m}-{s}^{-1}


Option: 3

+0.015 \mathrm{~kg}-\mathrm{m}-{s}^{-1}


Option: 4

-0.005 \mathrm{~kg}-\mathrm{m}-{s}^{-1}


Answers (1)

best_answer

Let the distance travelled by the rope is ' x ' then mass of the roll as a function of distance travelled by the roll will be:-

m(x)=M\left(1-\frac{x}{L}\right)

and its radius as a function of distance travelled is given by -

r(x)=R \sqrt{1-\frac{x}{L}}

The vertical component of the velocity of centre of mass of the rope is given by:-

    v_y=\frac{d}{d t}\left[\frac{m(x) r(x)}{M}\right]

    \begin{aligned} & v_y=\frac{d}{d t}\left[\frac{M\left(1-\frac{x}{L}\right) \times R\left(1-\frac{x}{L}\right)^{1 / 2}}{M}\right] \\ & v_y=R\left[\frac{d}{d t}\left[1-\frac{x}{L}\right]^{3 / 2}\right] \end{aligned}

\begin{aligned} & v_y=-\frac{3}{2} \frac{R v_0}{L} \\ & p_y=M v_y=-\frac{3}{2} \frac{M R v_0}{L} \\ & p_y=-\frac{3}{2}\left[\frac{1 \times 0.1 \times 1}{10}\right]=-0.015 \mathrm{~kg}-\mathrm{m}-\mathrm{s}^{-1} \end{aligned}

 

 

Posted by

Devendra Khairwa

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