question_answer

A solid cylinder of mass 20 kg and radius 0-12 m rotating with initial angular speed of 125 rad/s is placed lightly (Le. without any translational push) on a horizontal table with coefficient of kinetic friction \[{{\mu }_{k}}=0.15\]  between the cylinder and the table. (a) After how long does the cylinder start rolling? (b) What is the initial translational energy, rotational energy and total energy of the cylinder? (c) What is the final (i.e. after rolling begins) translational energy, rotational energy and total energy of the cylinder? (d) Is the final total energy equal to the initial total energy of motion of the cylinder? If not, where does the difference of energy disappear? (e) Account for the loss of total energy of motion in the following way: find the work done by friction on the body for its translational motion. Show that net work done by friction on the body is negative equal in magnitude to the loss of total energy calculated in (d) above.

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.