Rotation Ques 115
- A carpet of mass $M$ made of inextensible material is rolled along its length in the form of a cylinder of radius $R$ and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to $R / 2$.
$(1990,8 M)$
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Answer:
Correct Answer: 115.$v=\sqrt{\frac{14 R g}{3}}$
Solution:
Formula:
- Let $M^{\prime}$ be the mass of unwound carpet. Then,
$$ M^{\prime}=\frac{M}{\pi R^{2}} \pi \left(\frac{R}{2}\right)^{2}=\frac{M}{4} $$
From conservation of mechanical energy :
$$ M g R-M^{\prime} g \frac{R}{2}=\frac{1}{2} \frac{M}{4} v^{2}+\frac{1}{2} I \omega^{2} $$
or $M g R-\frac{M}{4} g \frac{R}{2}=\frac{M v^{2}}{8}+\frac{1}{2} \frac{1}{2} \times \frac{M}{4} \times \frac{R^{2}}{4} \quad \frac{v}{R / 2}$
or $\quad \frac{7}{8} M g R=\frac{3 M v^{2}}{16}$
$\therefore \quad v=\sqrt{\frac{14 R g}{3}}$
BETA
