91Ó°ÊÓ

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

ii) Mass of the rod is,$M=3.0\text{ k²µ}.$$M=3.0\text{ k²µ}.$

iii) Distance,$d=1.0\text{″¾}.$

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

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-

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

where,Iis moment of inertia, m, Mare 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.

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,

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

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

$I=\frac{1}{12}\left(3\text{ k²µ}\right){\left(4\text{″¾}\right)}^2+\left(3\text{ k²µ}\right){\left(1\text{″¾}\right)}^2$

$I=7.0\text{ k²µ}.{\text{m}}^2$

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

The rotational K.E of the rod is,

$K.E_{rot}=\frac{1}{2}I{\mathrm{Ó\neg }}^2$

${\mathrm{Ó\neg }}^2=\frac{2\text{ }K.E_{rot}}{I}$

$\mathrm{Ó\neg }=\sqrt{\frac{2\text{ }K.E_{rot}}{I}}$

$\mathrm{Ó\neg }=\sqrt{\frac{2\left(20\text{ J}\right)}{7{\text{ k²µm}}^{\text{2}}}}$

$\mathrm{Ó\neg }=2.4\text{ r²¹»å}/\text{s}$

But,$v=r\mathrm{Ó\neg }.\text{ â¶Ä‰H±ð²Ô³¦±ð,}$linear velocity of rod is,

$v=\left(3\text{″¾}\right)(2.4\text{ r²¹»å}/\text{s})$

$v=7.2\text{″¾/s}$

Therefore, linear speed of the end B of the rod as it passes through the vertical position is.$7.2\text{″¾/s}$

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,

$\begin{array}{c}P.E=mgh\\ =mgd(1−\cos \mathrm{θ})\end{array}$

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

$20\text{ J}=\left(3\text{​ k²µ}\right)(9.8\text{″¾}/{\text{s}}^{\text{2}})\left(1\text{″¾}\right)(1−\cos \mathrm{θ})$

$(1−\cos \mathrm{θ})=0.68$

$\cos \mathrm{θ}=0.32$

$\mathrm{θ}=71.35°$

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