A solid cylinder having radius R and length L is slipping on a rough horizontal plane. At time t=0 the cylinder has a translational velocity v0=49 m/s, perpendicular to its axis and a rotational velocity v0/4R about the centre. The time taken by the cylinder to start rolling is seconds. (coefficient of kinetic friction K=0.25 and g=9.8 m/s2)
- 15
- 5
- 10
- 7.5
Answer
(2) 5
Solution
[figure]
At t = 0, 0R = v04 < v0, so the lowest point slides forward and kinetic friction f = Kmg acts backward.
Translation: the retardation is a = Kg, so v = v0 - Kgt.
Rotation: = fRI = KmgR 12mR2 = 2 KgR, so R = v04 + 2 Kgt.
Pure rolling starts when v = R:
v0 - Kgt = v04 + 2 Kgt
3 Kgt = 3v04
t = v04 Kg = 494 0.25 9.8 = 5 s
Answer: (2)