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Found in: Page 294

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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# In Fig${\mathbf{\text{10-63}}}$., a thin uniform rod (mass, ${\mathbf{\text{3.0 \hspace{0.17em}kg}}}$length${\mathbf{\text{4.0 \hspace{0.17em}m}}}$) rotates freely about a horizontal axis A that is perpendicular to the rod and passes through a point at distance${\mathbf{d}}{\mathbf{=}}{\mathbf{\text{1.0\hspace{0.17em}m}}}$ from the end of the rod. The kinetic energy of the rod as it passes through the vertical position is${\mathbf{\text{20 J}}}$. (a) What is the rotational inertia of the rod about axis A? (b) What is the (linear) speed of the end B of the rod as the rod passes through the vertical position? (c) At what angle u will the rod momentarily stop in its upward swing?

a) Rotational inertia (M.I) of the rod about axis A is.$7.0\text{\hspace{0.17em}}\text{kg}.{\text{m}}^{2}$

b) Linear speed of point B of the rod when the rod is vertical is$7.2\text{\hspace{0.17em}m}/\text{s}$.

c)The angle at which the rod momentarily stops in its upward swing is equal t$71.35°$o.

See the step by step solution

## Step 1: Given

i) Length of the rod is,$L=4.0\text{\hspace{0.17em}}\text{m}.$

ii) Mass of the rod is,$M=3.0\text{\hspace{0.17em}kg}.$$M=3.0\text{\hspace{0.17em}kg}.$

iii) Distance,$d=1.0\text{\hspace{0.17em}m}.$

iv)The K.E. of the rod as when it is vertical is,$K.E=20\text{J}.$

## Step 2: Determining the concept

Find the moment of inertia of the rod about an axis A using the parallel axis theorem. Using the formula for rotational K.E., find the angular speed of the rod. Then, using it’s relation with linear speed, find the linear speed of the end B of the rod as it passes through the vertical position. Using energy conservation, find the angle at which the rod momentarily stops in its upward swing.

The parallel axis theorem is given as-

${\mathbf{I}}{\mathbf{=}}{{\mathbf{I}}}_{\mathrm{com}}{\mathbf{+}}{{\mathbf{MR}}}^{2}$

M.I. ofa uniform rod about its center of mass is,

role="math" localid="1660999291388" ${\mathbf{I}}{\mathbf{=}}\frac{1}{12}{{\mathbf{ML}}}^{2}$

Kinetic energy is given as-

${\mathbf{K}}{\mathbf{.}}{{\mathbf{E}}}_{\mathrm{rot}}{\mathbf{=}}\frac{1}{2}{\mathbf{I}}{\mathbf{}}{{\mathbf{\omega }}}^{2}$

where, Iis moment of inertia, m, M are masses, r is radius, v is velocity, is angular frequency, g is acceleration due to gravity, his height, L is length, K.E. is kinetic energy.

## Step 3: (a) Determining the rotational inertia (M.I) of the rod about axis A

According to the parallel axis theorem,

$I={I}_{com}+M{R}^{2}$

Applying to the given system, the M.I of rod about axis A is,

$I={I}_{com}+M{d}^{2}$

$I=\frac{1}{12}M{L}^{2}+M{d}^{2}$

$I=\frac{1}{12}\left(3\text{\hspace{0.17em}kg}\right){\left(4\text{\hspace{0.17em}m}\right)}^{2}+\left(3\text{\hspace{0.17em}kg}\right){\left(1\text{\hspace{0.17em}m}\right)}^{2}$

$I=7.0\text{\hspace{0.17em}kg}.{\text{m}}^{2}$

Therefore, rotational inertia (M.I.) of the rod about axis A is$7.0\text{\hspace{0.17em}}\text{kg}.{\text{m}}^{2}.$

## Step 4: (b) Determining the linear speed of the end B of the rod as it passes through the vertical position

The rotational K.E of the rod is,

$K.{E}_{rot}=\frac{1}{2}I{\omega }^{2}$

${\omega }^{2}=\frac{2\text{\hspace{0.17em}}K.{E}_{rot}}{I}$

$\omega =\sqrt{\frac{2\text{\hspace{0.17em}}K.{E}_{rot}}{I}}$

$\omega =\sqrt{\frac{2\left(20\text{\hspace{0.17em}J}\right)}{7{\text{\hspace{0.17em}kgm}}^{\text{2}}}}$

$\omega =2.4\text{\hspace{0.17em}rad}/\text{s}$

But,$v=r\omega .\text{\hspace{0.17em}\hspace{0.17em}Hence,}$ linear velocity of rod is,

$v=\left(3\text{\hspace{0.17em}m}\right)\left(2.4\text{\hspace{0.17em}rad}/\text{s}\right)$

$v=7.2\text{\hspace{0.17em}m/s}$

Therefore, linear speed of the end B of the rod as it passes through the vertical position is.$7.2\text{\hspace{0.17em}m/s}$

## Step 4: (b) Determining the angle at which the rod momentarily stops in its upward swing

As per the law of conservation of energy, when the rod momentarily stops it has zero K.E and maximum P.E. Using analogy with simple pendulum, write for maximum P.E as,

$\begin{array}{c}P.E=mgh\\ =mgd\left(1-\mathrm{cos}\theta \right)\end{array}$

But,$P.E=K.{E}_{rot}$

$20\text{\hspace{0.17em}J}=\left(3\text{​\hspace{0.17em}kg}\right)\left(9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\right)\left(1\text{\hspace{0.17em}m}\right)\left(1-\mathrm{cos}\theta \right)$

$\left(1-\mathrm{cos}\theta \right)=0.68$

$\mathrm{cos}\theta =0.32$

$\theta =71.35°$

Hence, the angle at which the rod momentarily stops in its upward swing is equal to.$71.35°$

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