Answer to Question #197181 in Physics for gibson

Question

1.A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of

radius 80 m, he applies brakes and reduces his speed at the rate of 0.5 m/s2. What is the

magnitude and direction of the resultant acceleration of the cyclist on the circular turn?

2. A rider of a bicycle goes around a circular path of radius 50m at 15m/s. the mass of the

rider is 60kg and that of a bicycle is 25kg,

(a) What is the centripetal force keeping the bicycle going around the circular path?

(b) Calculate the normal force acting on the bicycle.

(c) What is the coefficient of friction which the bicycle tires experiences with road?

Answer 1

1) Let's first find the centripetal acceleration of the cyclist:

$$a_c=\dfrac{v^2}{r},$$ $$a_c=\dfrac{(27\ \dfrac{km}{h}\cdot\dfrac{1000\ m}{1\ km}\cdot\dfrac{1\ h}{3600\ s})^2}{80\ m}=0.7\ \dfrac{m}{s^2}.$$

We can find the magnitude of the resultant acceleration of the cyclist from the Pythagorean theorem:

$$a=\sqrt{a_c^2+a_t^2}=\sqrt{(0.7\ \dfrac{m}{s^2})^2+(-0.5\ \dfrac{m}{s^2})^2}=0.86\ \dfrac{m}{s^2}.$$

We can find the direction of the resultant acceleration of the cyclist from the geometry:

$$\theta=tan^{-1}(\dfrac{a_c}{a_t}),$$ $$\theta=tan^{-1}(\dfrac{0.7\ \dfrac{m}{s^2}}{-0.5\ \dfrac{m}{s^2}})=-54.5^{\circ}.$$

Answer:

$a=0.86\ \dfrac{m}{s^2},$ $54.5^{\circ}$ away from the direction of the motion.

2)

(a)

$$F_c=ma_c=\dfrac{mv^2}{r},$$ $$F_c=\dfrac{85\ kg\cdot(15\ \dfrac{m}{s})^2}{50\ m}=382.5\ N.$$

(b)

$$F_N=mg=85\ kg\cdot9.8\ \dfrac{m}{s^2}=833\ N.$$

(c)

$$F_c=F_{fr},$$ $$F_c=\mu F_N,$$ $$\mu=\dfrac{F_c}{F_N}=\dfrac{382.5\ N}{833\ N}=0.46$$