Answer:
Here,
m = 20 kg, r = 0.12 m ; \[{{\omega }_{0}}=125\,rad/s\], \[{{\mu }_{k}}=0.15\]
(a) As
discussed already in Q. 30, Page 5/114, time after which the cylinder starts
rolling is
\[t=\frac{r\,{{\omega
}_{0}}}{3{{\mu }_{k}}g}=\frac{0.12\times 125}{3\times 0.15\times 9.8}=3.4s\]
(b)
Initial translational energy =0 \[(\because
\,\,\,u=0)\]
rotational
energy = \[=\frac{1}{2}I\omega _{0}^{2}=\frac{1}{2}\left( \frac{1}{2}m{{r}^{2}}
\right)\omega _{0}^{2}=\frac{1}{4}\times 20{{\left( 0.12 \right)}^{2}}{{\left(
125 \right)}^{2}}=1125J\]
total
energy of the cylinder = 0 + 1125 = 1125 J
(c)
As \[\alpha =\frac{{{\mu }_{k}}mg\,R}{I}=\frac{{{\mu
}_{k}}mg\,R}{\frac{1}{2}m{{R}^{2}}}\] \[\therefore
\] \[\alpha =\frac{2{{\mu }_{k}}g}{R}=\frac{2\times 0.15\times
9.8}{0.12}=24.5\,rad/{{s}^{2}}\]
Rolling
begins, when \[\upsilon =r\omega \]
Now
\[\omega ={{\omega }_{0}}-\alpha
t=125-24.5\times 3.4=41.7\,rad/s\]
\[\upsilon
=r\omega =0.12\times 41.7=5.004\,m/s\]
\[\therefore
\] final translational energy = \[\frac{1}{2}m{{\upsilon
}^{2}}=\frac{1}{2}\times 20{{\left( 5 \right)}^{2}}=250J\]
final
rotational energy = \[\frac{1}{2}I{{\omega }^{2}}=\frac{1}{4}m{{r}^{2}}{{\omega
}^{2}}=\frac{1}{4}\times 20{{\left( 0.12 \right)}^{2}}{{\left( 41.7
\right)}^{2}}=125J\]
final
total energy = 250 + 125 = 375 J
(d)
No; Loss of energy =1125 ? 375= 750 J
(e)
Work done by friction for translational motion = + 250 J
Taking,
\[\theta ={{\omega }_{0}}t-\frac{1}{2}\alpha {{t}^{2}}=125\times
3.4-\frac{1}{2}\left( 24.5 \right){{\left( 3.4
\right)}^{2}}=425-141.61=283.39radian\]
Work
done by friction against rotational motion = \[-{{\mu }_{k}}mg\,r\times \theta
\]
\[=-0.\text{15}\times
\text{2}0\times \text{9}.\text{8}\times 0.\text{12}\times
\text{283}.\text{39}=-\text{1}000\text{ J}\]
Thus
net work done by friction on the body = 250 - 1000 = - 750 J
This
accounts for the loss of energy in (d) above.
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