A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure. Each part subtends a right angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of 18 km/h on the track.
(a) Find the normal force between the road and the cycle just before and just after the cycle crosses C. (b) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed. ? Take g=10 m/s².
Answers
(a). Refers to the attachment for the answer.
Just before C,
mgcos45° = N + mv²/r
⇒ N = mgcos45° - mv²/r
⇒ N = m (gcos45° - v²/r)
⇒ N = 100(10/√2 - 25/100)
∴ N = 100(10/√2 - 1/4)
∴ N = 100 × 6.8217
⇒ N = 682.17 N.
Similarly, After Crossing C,
N = mg + mv²/r
⇒ N = 732 N.
(b). Let the minimum coefficient of friction be µ.
At C, Magnitudes of Frictional force is equal to = 707 N,
Normal forces just before and after C are 682 N and 732 N respectively.
So coefficient of friction µ at these points = F/N
= 707/682
= 1.037 N
and 707/732 =0.96
Though the lower value of µ is 0.96 but if we take this value just before C this will make available frictional force less than 707 N while the tangential component of weight remains 707 N.
Thus it will skid and its speed won't be constant. So minimum coefficient of friction should be µ =1.037.
Hope it helps.
