A thin rod of mass m and length l is suspended from one of its ends. It is set into oscillation about a horizontal axis. Its angular speed is ω while passing through its mean position. How high will its centre of mass rise from its lowest position?

Question

A thin rod of mass m and length l is suspended from one of its ends. It is set into oscillation about a horizontal axis. Its angular speed is ω while passing through its mean position. How high will its centre of mass rise from its lowest position? (A) (ω2 l2/2 g) (B) (ω2 l2/3 g) (C) (ω2 l2/g) (D) (ω2 l2/6 g)

Tardigrade Admin · Accepted Answer

(1/2) (m l2/3) ⋅ ω2=m g h (Energy conservation) h=(l2 ω2/6 g)

Q. A thin rod of mass $m$ and length $l$ is suspended from one of its ends. It is set into oscillation about a horizontal axis. Its angular speed is $ω$ while passing through its mean position. How high will its centre of mass rise from its lowest position?

$\frac{ω ^{2} l ^{2}}{2 g}$

$\frac{ω ^{2} l ^{2}}{3 g}$

$\frac{ω ^{2} l ^{2}}{g}$

$\frac{ω ^{2} l ^{2}}{6 g}$

$\frac{1}{2} \frac{m l ^{2}}{3} \cdot ω^{2} = m g h$ (Energy conservation)
$h = \frac{l ^{2} ω ^{2}}{6 g}$

Q. A thin rod of mass m and length l is suspended from one of its ends. It is set into oscillation about a horizontal axis. Its angular speed is ω while passing through its mean position. How high will its centre of mass rise from its lowest position?

2gω2l2​

3gω2l2​

gω2l2​

6gω2l2​