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Physics / Physics with MasteringPhysics 4 / Chapter 10 / Problem 107GP

Physics with MasteringPhysics | 4th Edition | ISBN: 9780321541635 | Authors: James S. Walker

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A thin, uniform rod of length L and mass M is pivoted

Chapter 10, Problem 107GP

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QUESTION:

A thin, uniform rod of length L and mass M is pivoted about one end, as shown in Figure 10–28. The rod is released from rest in a horizontal position, and allowed to swing downward without friction or air resistance. When the rod is vertical, what are (a) its angular speed \(\omega\) and (b) the tangential speed \(v_\mathrm t\) of its free end?

Equation Transcription:

$\omega{}$

${v}_{t}$

$\omega{}=0$

$L/2$

$x=center\ of\ mass$

${v}_{t}$

$\omega{}$

Text Transcription:

omega

v_t

omega=0

L/2

x=center of mass

v_t

omega

Questions & Answers

QUESTION:

A thin, uniform rod of length L and mass M is pivoted about one end, as shown in Figure 10–28. The rod is released from rest in a horizontal position, and allowed to swing downward without friction or air resistance. When the rod is vertical, what are (a) its angular speed \(\omega\) and (b) the tangential speed \(v_\mathrm t\) of its free end?

Equation Transcription:

$\omega{}$

${v}_{t}$

$\omega{}=0$

$L/2$

$x=center\ of\ mass$

${v}_{t}$

$\omega{}$

Text Transcription:

omega

v_t

omega=0

L/2

x=center of mass

v_t

omega

ANSWER:

a.)

Step 1 of 3

We have to find the angular speed $\omega{}$ of the rod when it is vertical which is released from rest in a horizontal position, and allowed to swing downward without friction or air resistance.

The angular speed can be found by making use of conservation of mechanical energy for the system.

$K+U=E$ (constant )

${U}_{i}+{K}_{i}={K}_{f}+{U}_{f}$

The position of centre of the mass when the rod is vertical corresponds to $y=0,$ so that ${y}_{o}=L/2$ at the start.

Hence,

$0+Mg\ (L/2)=\frac{1}{2}I\ {\omega{}}^{2}$

Where,

$M$ is the mass of uniform rod of

length $L$

$I=$ Moment of Inertia of the rod

pivoted about one end = $\frac{1}{3}$ $M{L}^{2}$

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