Charles Kingsley

2022-01-04

A slender rod is 80.0 cm long and has mass 0.120 kg. A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.0500-kg sphere is welded to the other end. The rod, pivoting about a stationary, frictionless axis at its center, is held horizontal and released from rest. What is the linear speed of the 0.0500-kg sphere as it passes through its lowest point?

Janet Young

Step 1
$\mathrm{△}U={m}_{2}g{y}_{2}+{m}_{3}g{y}_{3}=g×\left(0.05×\left(-0.4\right)+0.02×0.4\right)=-0.118N.m$
Determine the system's overall potential change. m2 and m3 are the masses at the rod ends. note the rod centre of mass neither gains nor loses potential.
Step 2
${I}_{1}=2\rho {\int }_{0}^{r}{r}^{2}dr=2×\frac{0.12}{0.8}{\int }_{0}^{0.4}{r}^{2}dr=0.0064kg.{m}^{2}$
Determine the rod's moment of inertia.
Step 3
${I}_{2+3}=\sum m{r}^{2}=\left(0.05+0.02\right)×{0.4}^{2}=0.0112kg.{m}^{2}$
determine the end masses' moment of inertia
Step 4
$\mathrm{△}U=K=\frac{1}{2}\left({I}_{1}+{I}_{2+3}\right){\omega }^{2}$
compare the system's kinetic energy to the change in potential energy.
Step 5
$=0.118=\frac{1}{2}×0.0176{\omega }^{2}$
Calculate velocity at lowest point:
Step 6
$\omega =3.66ra\frac{d}{s};v=r\omega =0.4×3.66=1.46\frac{m}{s}$

Debbie Moore

Explanation:
Given that,
Slender rod
Length of rod $=80cm=0.8m$
Mass of slender rod $=0.12kg$
Sphere Bob at one end
Mass $M1=0.02kg$
Sphere Bod at the other end
Mass $M2=0.05kg$
Linear speed of mass 2 at the lowest point
We need to calculate the change in potential of the complete system. m2 and m3 are the masses at the rod ends. note the rod centre of mass neither gains nor loses potential.
So, at the lowest point,
$\mathrm{△}U=M2\cdot g\cdot y2+M1\cdot g\cdot y1$
Note, at the lowest point, the mass 1 is 40cm (0.4m) form the midpoint, Also, the mass 2 is $-40cm\left(-04m\right)$ from the midpoint
$\mathrm{△}U=M2\cdot g\cdot y2+M1\cdot g\cdot y1$
$\mathrm{△}U=0.05\cdot 9.81\cdot \left(-0.4\right)+0.02\cdot 9.82\cdot 0.4$
$\mathrm{△}U=-0.1962+0.07848$
$\mathrm{△}U=-0.11772Nm$
Now, the moment of inertia of the rod is given as
$I=\sum {r}^{2}dm$
$dm=2pdr$
$I=2p\sum {r}^{2}dr$
$I=2×\frac{0.12}{0.8}\sum {r}^{2}dr$; from

$I=0.3\left[\frac{{0.4}^{3}}{3}-0\right]$,from
$I=0.3×0.02133$
$I=0.0064k\frac{g}{{m}^{2}}$.
calculating of inertia of the end masses.
$I\left(1+2\right)=\sum m{r}^{2}=\left(m1+m2\right){r}^{2}$
$I\left(1+2\right)=\left(0.02+0.05\right){0.4}^{2}$
$I\left(1+2\right)=0.07×{0.4}^{2}$
$I\left(1+2\right)=0.0112k\frac{g}{{m}^{2}}$
Now, the Energy of the masses due to angular velocity is given as
$K.E=\frac{1}{2}\left(I+I\left(1+2\right)\right){w}^{2}$
$K.E=\frac{1}{2}\left(0.0064+0.0112\right){w}^{2}$
$K.E=0.0088{w}^{2}$
Using conservation of energy
The potential energy is equal to the kinetic energy of the system
$K.E=P.E$
$0.0088{w}^{2}=0.11772$
Then, ${w}^{2}$

karton

Step 1
The PE of an object is the amount of energy the object will store because it is at a certain height.
The PE is given as
U=mgh
where m= mass of the object, g= acceleration due to gravity, and h= height.
Step 2
The PE of the system is given by
$U={m}_{1}g{h}_{1}+{m}_{2}g{h}_{2}\phantom{\rule{0ex}{0ex}}=g\left({m}_{1}{h}_{1}+{m}_{2}{h}_{2}\right)$
$=9.8m/{s}^{2}\left(0.05kg×\left(-0.4m\right)+0.02kg×0.4m\right)\phantom{\rule{0ex}{0ex}}=-0.1176Nm$
Again, PE of the system is given by
$\phantom{\rule{0ex}{0ex}}U=KE\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left({I}_{3}+{I}_{12}\right){\omega }^{2}\phantom{\rule{0ex}{0ex}}\omega =\sqrt{\frac{2U}{{I}_{3}+{I}_{12}}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{2U}{\left(2p\sum _{0}^{0.4}{r}^{2}dr\right)+\left(\left({m}_{1}+{m}_{2}\right){r}^{2}\right)}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{2U}{\left(2×\frac{0.12kg}{0.8m}\left[\frac{{r}^{3}}{3}{\right]}_{0}^{0.4}\right)+\left(\left(0.05kg+0.02kg\right)\left(0.4m{\right)}^{2}\right)}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{2×\left(0.1176Nm\right)}{\left(2×\frac{0.12kg}{0.8m}\left[\frac{{r}^{3}}{3}{\right]}_{0}^{0.4}\right)+\left(\left(0.05kg+0.02kg\right)\left(0.4m{\right)}^{2}\right)}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{2×\left(0.1176Nm\right)}{\left(0.0064kg{m}^{2}\right)+\left(0.0112kg{m}^{2}\right)}}\phantom{\rule{0ex}{0ex}}=3.655rad/s$
The linear velocity is given by
$\phantom{\rule{0ex}{0ex}}v=r\omega \phantom{\rule{0ex}{0ex}}=0.4m×3.655rad/\phantom{\rule{0ex}{0ex}}=1.462m/s$
Step 3
Hence, the required velocity is 1.462m/s

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