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Example: force required to turn on a bike
Question: A bicyclist is making a turn along an arc of a circle
with radius 20 m, at a speed of 5 m/s. If the combined mass of
the cyclist plus the bike is 60 kg, how great a static friction force
must the road be able to exert on the tires.
Solution: Taking the magnitudes of both sides of Newton’s
second law gives
|F|= |ma|
= m|a| .
Substituting |a|=|v|
2
/r gives
|F|=m|v|
2
/r
˜
80 N
(rounded off to one sig fig).
Example: Don’t hug the center line on a curve!
Question: You’re driving on a mountain road with a steep drop
on your right. When making a left turn, is it safer to hug the
center line or to stay closer to the outside of the road.
Solution: You want whichever choice involves the least accel-
eration, because that will require the least force and entail the
least risk of exceeding the maximum force of static friction.
Assuming the curve is an arc of a circle and your speed is
constant, your car is performing uniform circular motion, with
|a|=|v|
2
/r. The dependence on the square of the speed shows
that driving slowly is the main safety measure you can take, but
for any given speed you also want to have the largest possible
value of r. Even though your instinct is to keep away from that
scary precipice, you are actually less likely to skid if you keep
toward the outside, because then you are describing a larger
circle.
Example: acceleration related to radius and period of rotation
Question: How can the equation for the acceleration in uniform
circular motion be rewritten in terms of the radius of the circle
and the period, T, of the motion, i.e. the time required to go
around once.
Solution: The period can be related to the speed as follows:
|v|=
circumference
T
= 2
p
r/T .
Substituting into the equation|a|=|v|
2
/r gives
|a|=
4
p
2
r
T
2
.
Example: a clothes dryer
Question: My clothes dryer has a drum with an inside radius of
35 cm, and it spins at 48 revolutions per minute. What is the
acceleration of the clothes inside.
Solution: We can solve this by finding the period and plugging in
to the result of the previous example. If it makes 48 revolutions in
one minute, then the period is 1/48 of a minute, or 1.25 s. To get
an acceleration in mks units, we must convert the radius to 0.35
m. Plugging in, the result is 8.8 m/s
2
.
Section 9.2Uniform Circular Motion