A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is ω. Its centre of mass rises to a maximum height of

Question

A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is ω. Its centre of mass rises to a maximum height of (A) (1/3) (ℓ2ω2/g)  (B) (1/6) (ℓω/g)  (C) (1/2) (ℓ2ω2/g)  (D) (1/6) (ℓ2ω2/g)

Tardigrade Admin · Accepted Answer

T.Ei=T.Ef (1/2)Iω2=mgh (1/2)×(1/3)mℓ2ω2=mgh ⇒ h=(1/6) (ℓ2ω2/g)

Q. A thin uniform rod of length $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $ω$ . Its centre of mass rises to a maximum height of

$\frac{1}{3} \frac{ℓ ^{2} ω ^{2}}{g}$

$\frac{1}{6} \frac{ℓ ω}{g}$

$\frac{1}{2} \frac{ℓ ^{2} ω ^{2}}{g}$

$\frac{1}{6} \frac{ℓ ^{2} ω ^{2}}{g}$

$T . E_{i} = T . E_{f}$
$\frac{1}{2} I ω^{2} = m g h$
$\frac{1}{2} \times \frac{1}{3} m ℓ^{2} ω^{2} = m g h \Rightarrow h = \frac{1}{6} \frac{ℓ ^{2} ω ^{2}}{g}$

Q. A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is ω. Its centre of mass rises to a maximum height of

31​gℓ2ω2​

61​gℓω​

21​gℓ2ω2​

61​gℓ2ω2​