A solid cylinder of mass `20 kg` and radius `0.12 m` rotating wit initial angula speed of `125 rad//s` is placed lightly (i.e. 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 finl total enrgy 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 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.

Explanation

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 stars rolling is
`t = (r omega_(0))/(3mu_(k) g) = (0.12 xx 125)/(3 xx 0.15 xx 9.8) = 3.4 s`
(B) Initial translational energy `= 0` (`:' u = 0)`
rotational energy `= (1)/(2)I omega_(0)^(2) = (1)/(2) ((1)/(2)mr^(2)) omega_(0)^(2) = (1)/(4) xx 20(0.12)^(2) (125)^(2) = 1125 J`
total energy of the cylinder `= 0 + 1125 = 1125J`
(c) As `alpha = (mu_(k) mgR)/(I) = (mu_(k) mg R)/((1)/(2)mR^(2)) :. alpha = (2mu_(k) g)/(R ) = (2xx0.15 xx 9.8)/(0.12) = 24.5 rad//s^(2)`
Rolling begins, when `v = r omega`
Now `omega = omega_(0) - alpha t = 125 - 24.5 xx 3.4 = 41.7 rad//s`
`v = r omega = 0.12 xx 41.7 = 5.004 m//s`
`:.` final translational energy `= (1)/(2)mv^(2) = (1)/(2) xx 20(5)^(2) = 250 J`
final rotational energy `= (1)/(2)I omega^(2) = (1)/(4)mr^(2) omega^(2) = (1)/(4) xx 20 (0.12)^(2) (41.7)^(2) = 125J`
final total energy `= 250 + 125 = 375J`
(d) No. Loss of energy `= 1125 - 375 = 750 J`
(e) Work done by friction for translational motion `= + 250 J`
Taking, `theta = omega_(0) t = (1)/(2) alpha t^(2) = 125 xx 3.4 - (1)/(2) (24.5)(3.4)^(2) = 425 - 141.61 = 283.39 radian`
Work done by friction against rotational motion `= - mu_(k) mg r xx theta`
`= - 0.15 xx 20 xx 9.8 xx 0.12 xx 283.39 = - 1000J`
Thus net work done by friction on the body `= 250 - 1000 = - 750 J`
This accounts for the loss of energy in (d) above.