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System of Particles and Rotational Motion

Centre of mass

A thin uniform rod of length l and mass m is swinging freely a

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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) $\frac{1}{3} \frac{l ^{2} ω ^{2}}{g}$
(b) $\frac{1}{6} \frac{l ω}{g}$
(c) $\frac{1}{2} \frac{l ^{2} ω ^{2}}{g}$
(d) $\frac{1}{6} \frac{l ^{2} ω ^{2}}{g}$

Views: 5,386 students

Updated on: Oct 25, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2009

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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)

\frac{1}{3} \frac{l ^{2} ω ^{2}}{g}

(b)

\frac{1}{6} \frac{l ω}{g}

(c)

\frac{1}{2} \frac{l ^{2} ω ^{2}}{g}

(d)

\frac{1}{6} \frac{l ^{2} ω ^{2}}{g}

8 mins ago

Discuss this question LIVE

8 mins ago

Text solutionVerified

Exp. (d)
By conservation of energy,
Initial total energy

=

Final total energy

TE_{i} = TE_{f}

\frac{1}{2} l ω^{2} = m g h

\frac{1}{2} \times \frac{1}{3} m l^{2} ω^{2} = m g h

h = \frac{1}{6} \frac{l ^{2} ω ^{2}}{g}

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Question Text	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) $\frac{1}{3} \frac{l ^{2} ω ^{2}}{g}$ (b) $\frac{1}{6} \frac{l ω}{g}$ (c) $\frac{1}{2} \frac{l ^{2} ω ^{2}}{g}$ (d) $\frac{1}{6} \frac{l ^{2} ω ^{2}}{g}$
Updated On	Oct 25, 2023
Topic	System of Particles and Rotational Motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 17
Upvotes	1905
Avg. Video Duration	7 min

Text solutionVerified

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Text solutionVerified

Filo tutor solutions (17)

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