A bicycle and rider of mass 100 kg are travelling at the rate of 16 km/h on a level road. A brake is applied to the rear wheel which is 0.9 m in diameter and this is the only resistance acting. How far will the bicycle travel and how many turns will it make before it comes to rest ? The pressure applied on the brake is 100 N and \mu=0.05.
Question : A bicycle and rider of mass 100 kg are travelling at the rat...
Given : m = 100 kg, v = 16 km / h = 4.44 m / s ; D = 0.9 m ; R_{ N }=100 N ; \mu=0.05
Distance travelled by the bicycle before it comes to rest
Let x = Distance travelled (in metres) by the bicycle before it comes to rest.
We know that tangential braking force acting at the point of contact of the brake and wheel,
F_{t}=\mu R_{ N }=0.05 \times 100=5 N …(i)
and work done =F_{t} \times x=5 \times x=5 x N – m
We know that kinetic energy of the bicycle
=\frac{m \cdot v^{2}}{2}=\frac{100(4.44)^{2}}{2}= 986N-m …(ii)
In order to bring the bicycle to rest, the work done against friction must be equal to kinetic energy of the bicycle. Therefore equating equations (i) and (ii)
5x = 986 or x = 986/5 = 197.2 m
Number of revolutions made by the bicycle before it comes to rest
Let N = Required number of revolutions.
We know that distance travelled by the bicycle (x),
197.2=\pi D N=\pi \times 0.9 N=2.83 N\therefore N = 197.2 / 2.83 = 70