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?
Answers
Answer:
Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.
Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net F = ma. For uniform circular motion, the acceleration is the centripetal acceleration—a = ac. Thus, the magnitude of centripetal force Fc is Fc = mac.
By using the expressions for centripetal acceleration ac from
a
c
=
v
2
r
;
a
c
=
r
ω
2
, we get two expressions for the centripetal force Fc in terms of mass, velocity, angular velocity, and radius of curvature:
F
c
=
m
v
2
r
;
F
c
=
m
r
ω
2
.
You may use whichever expression for centripetal force is more convenient. Centripetal force Fc is always perpendicular to the path and pointing to the center of curvature, because ac is perpendicular to the velocity and pointing to the center of curvature.
Note that if you solve the first expression for r, you get
r
=
m
v
2
F
c
.
This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.
Answer:
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