Question

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 contact force by the road on the cycle when it is at B and at D. (b) Find the force of friction exerted by the track on the tyres when the cycle is at B, C and D.

Answer 1

Mass of the cycle = 100 kg.

Speed = 18 km/hr. = 5 m/s.

(a). When the Cycle is at B,

Normal Force = mg - mv²/r

⇒ N = m(g - v²/r)

⇒ N = 100(10 - 25/100)

⇒ N = 100/4 × 41

∴ N = 1025 N.

For Point D,

N = mg + mv²/r

⇒ N = 100/4(39)

⇒ N = 975 N.

(b). When the cycle will be at point B and D then the Force of the Friction will be zero, since there is only a vertical force and the body is moving horizontally with a constant speed this means that there will be no force.

Hence, Friction = zero at point B and D.

At Point C,

Friction = mgCos45°

= 100 × 10 × 1/√2

= 707.11  N.

≈ 707.11 N.

Hope it helps.