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A rod $\mathrm{P Q}$ of mass $\mathrm{m}$ is kept suspended in horizontal position as shown. Linear mass density of rod varies distance $\mathrm{x}$ from its left end $\mathrm{P}$ as $\mathrm{\lambda=k x}$ ( $\mathrm{k }$ is constant). Left string is now cut. Torque about point $\mathrm{ ' Q '}$ just after the string is cut is -

Option: 1
$\mathrm{\frac{m g l}{2}}$

Option: 2
$\mathrm{\frac{m g l}{3}}$

Option: 3
$\mathrm{\frac{m g l}{4}}$

Option: 4
$\mathrm{\frac{m g l}{6}}$

Answers (1)

best_answer

The $\mathrm{CM}$ of the rod, from $\mathrm{P}$

$\mathrm{ x_{c m}=\frac{\int_0^l(d m) x}{\int_0^l d m} }$

$\mathrm{ x_{c m}=\frac{\int_0^l(\lambda d x) x}{\int_0^l \lambda d x}=\frac{\int_0^l(k x d x) x}{\int_0^l k x d x} }$

$\mathrm{ x_{c m}=\frac{2 l}{3} }$

Hence option 2 is correct.

Posted by

Rishi

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