Question

A track consists of two circular parts ABC and CDE of
equal radius 100 m and joined smoothly as shown in
figure (7-E1). 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. (c) Find the normal force between the
road and the cycle just before and just after the cycle
crosses C. (d) 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/s^2

Answer 1

Answer on Question #58068-Physics-Mechanics-Relativity

A track consists of two circular parts ABC and CDE of equal radius $100\mathrm{m}$ and joined smoothly as shown in figure (7-E1). Each part subtends a right angle at its centre. A cycle weighing $100\mathrm{kg}$ together with the rider travels at a constant speed of $18\mathrm{km/h}$ 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.

(c) Find the normal force between the road and the cycle just before and just after the cycle crosses C.

(d) 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 $g = 10 \, \text{m/s}^2$

Solution

(a) Force diagrams at B and D are shown below.

For B:

$$v = 1 8 \frac {k m}{h} = 5 \frac {m}{s}.$$ $$F _ {c f} = \frac {m v ^ {2}}{R}.$$ $$n = m g - F _ {c f} = 1 0 0 \cdot 1 0 - 1 0 0 \cdot \frac {5 ^ {2}}{1 0 0} = 9 7 5 N$$

For D:

$$n = m g + F _ {c f} = 1 0 0 \cdot 1 0 + 1 0 0 \cdot \frac {5 ^ {2}}{1 0 0} = 1 0 2 5 N$$

(b) At B and D; no friction acts. Because there is no additional tangential force acting on the bicycle.

At C, A component of $mg$ acts along the track. This component is counterbalanced by friction. Thus,

$f = mg\sin 45 = 100\cdot 10\cdot \sin 45 = 707N.$

(c) Force on bicycle just before C is shown in figure.

$$n + F _ {c f} = m g \cos \theta$$ $$n = m g \cos \theta - F _ {c f} = 1 0 0 \cdot 1 0 \cdot \cos 4 5 - 1 0 0 \cdot \frac {5 ^ {2}}{1 0 0} = 6 8 2 N$$

Similarly it can be shown

$$n _ {D} = m g \cos \theta + F _ {c f} = 1 0 0 \cdot 1 0 \cdot \cos 4 5 + 1 0 0 \cdot \frac {5 ^ {2}}{1 0 0} = 7 3 2 N$$

(d) The minimum frictional force is

$$\mu n _ {m i n} = \mu \cdot 6 8 2$$

This force must balance the tangential component of force.

$$\mu \cdot 6 8 2 = 7 0 2 \rightarrow \mu = 1. 0 3 7$$

http://www.AssignmentExpert.com/