A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is :

Question

A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is : (A) (3g/2 l) sin θ (B) (2g/3l) sin θ (C) (3g/2l) cos θ (D) (2g/2l) sin θ

Tardigrade Admin · Accepted Answer

Torque at angle θ τ = Mg sin θ . <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/be30392dae547abd3b5eebaae8cafb18-.png" /> (l/2) τ = l α l α = Mg sin θ (l/2) ∴ l = (M l2/3) (Ml2/3) .α = Mg sin θ (l/2) (l α/3) = g ( sin θ/2) α = (3g sin θ/2l)

$\frac{3 g}{2 l} sin θ$

$\frac{2 g}{3 l} sin θ$

$\frac{3 g}{2 l} cos θ$

$\frac{2 g}{2 l} sin θ$

Torque at angle $θ$
$τ = M g sin θ .$

$\frac{l}{2}$
$τ = l α$
$l α = M g sin θ \frac{l}{2}$
$∴ l = \frac{M l ^{2}}{3}$
$\frac{M l ^{2}}{3} . α = M g sin θ \frac{l}{2}$
$\frac{l α}{3} = g \frac{s i n θ}{2}$
$α = \frac{3 g s i n θ}{2 l}$

2l3g​sinθ

3l2g​sinθ

2l3g​cosθ

2l2g​sinθ

Torque at angle θ τ=Mgsinθ. 2l​ τ=lα lα=Mgsinθ2l​ ∴l=3Ml2​ 3Ml2​.α=Mgsinθ2l​ 3lα​=g2sinθ​ α=2l3gsinθ​

$\frac{3 g}{2 l} sin θ$

$\frac{2 g}{3 l} sin θ$

$\frac{3 g}{2 l} cos θ$

$\frac{2 g}{2 l} sin θ$

Torque at angle $θ$
$τ = M g sin θ .$

$\frac{l}{2}$
$τ = l α$
$l α = M g sin θ \frac{l}{2}$
$∴ l = \frac{M l ^{2}}{3}$
$\frac{M l ^{2}}{3} . α = M g sin θ \frac{l}{2}$
$\frac{l α}{3} = g \frac{s i n θ}{2}$
$α = \frac{3 g s i n θ}{2 l}$