Minimum velocity of the particle isdisplaystyle 4sqrt-frac-gL-3-displaystyle 2sqrt-frac-gL-3-displaystyle sqrt-frac-gL-3-displaystyle 3sqrt-frac-gL-3-

Question

Toppr Admin · Accepted Answer

The maximum and minimum velocities will be at A and B respectively.
Work-energy theorem, $- m g (2 L) = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$
Given: $u : v = 2 : 1$
$- m g (2 L) = \frac{1}{2} m v^{2} - \frac{1}{2} m (2 v)^{2}$
$⟹ v = 2 \sqrt{\frac{g L}{3}}$

Minimum velocity of the particle is4√gL32√gL3√gL33√gL3

The maximum and minimum velocities will be at A and B respectively.Work-energy theorem, −mg(2L)=12mv2−12mu2Given: u:v=2:1−mg(2L)=12mv2−12m(2v)2⟹v=2√gL3

Minimum velocity of the particle is

$4 \sqrt{\frac{g L}{3}}$
$2 \sqrt{\frac{g L}{3}}$
$\sqrt{\frac{g L}{3}}$
$3 \sqrt{\frac{g L}{3}}$

The maximum and minimum velocities will be at A and B respectively.
Work-energy theorem, $- m g (2 L) = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$
Given: $u : v = 2 : 1$
$- m g (2 L) = \frac{1}{2} m v^{2} - \frac{1}{2} m (2 v)^{2}$
$⟹ v = 2 \sqrt{\frac{g L}{3}}$