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Circular path
SOLUTION
_ 2
mv

m = 0.25 kg F =
r
r = 0.5 m __ 2
(0.25 kg)(2.0 m/s)

v = 2.0 m/s = 0.5 m
F = ? 2 2
Radius (r ) __
(0.25 kg)(4.0 m /s )
=
0.5 m
Velocity (0.25)(4.0) kg·m 2
1
= _ _ _

×

0.5 s 2 m
Centripetal 2
kg·m
force _
= 2
m·s 2
kg·m
_

= 2 2
s
FIGURE 2.26 Centripetal force on the ball causes it to change = 2 N
direction continuously, or accelerate into a circular path. Without
the unbalanced force acting on it, the ball would continue in a
straight line.
EXAMPLE 2.17
Suppose you make the string in example 2.16 one-half as long, 0.25 m.
What force is now needed? (Answer: 4.0 N)
line motion. If other forces were involved, it would follow some
other path. Nonetheless, the apparent outward force has been 2.9 NEWTON’S LAW OF GRAVITATION
given a name just as if it were a real force. The outward tug is
called a centrifugal force. You know that if you drop an object, it always falls to the floor.
The magnitude of the centripetal force required to keep an You define down as the direction of the object’s movement and
object in a circular path depends on the inertia, or mass, of the up as the opposite direction. Objects fall because of the force of
object and the acceleration of the object, just as you learned in gravity, which accelerates objects at g = 9.8 m/s (32 ft/s ) and
2
2
the second law of motion. The acceleration of an object mov- gives them weight, w = mg.
ing in a circle can be shown by geometry or calculus to be di- Gravity is an attractive force, a pull that exists between
rectly proportional to the square of the speed around the circle all objects in the universe. It is a mutual force that, just like all
2
(v ) and inversely proportional to the radius of the circle (r). other forces, comes in matched pairs. Since Earth attracts you
(A smaller radius requires a greater acceleration.) Therefore, with a certain force, you must attract Earth with an exact op-
the acceleration of an object moving in uniform circular mo- posite force. The magnitude of this force of mutual attraction
tion (a c ) is depends on several variables. These variables were first described
2
v _
a c = by Newton in Principia, his famous book on motion that was

r printed in 1687. Newton had, however, worked out his ideas much
earlier, by the age of 24, along with ideas about his laws of motion
equation 2.10
and the formula for centripetal acceleration. In a biography writ-
The magnitude of the centripetal force of an object with a ten by a friend in 1752, Newton stated that the notion of gravita-
mass (m) that is moving with a velocity (v) in a circular orbit tion came to mind during a time of thinking that “was occasioned
of a radius (r) can be found by substituting equation 2.5 in by the fall of an apple.” He was thinking about why the Moon stays
F = ma, or in orbit around Earth rather than moving off in a straight line as
would be predicted by the first law of motion. Perhaps the same
_ 2
mv

F = force that attracts the Moon toward Earth, he thought, attracts the
r
apple to Earth. Newton developed a theoretical equation for grav-
equation 2.11 itational force that explained not only the motion of the Moon but
also the motion of the whole solar system. Today, this relationship
is known as the universal law of gravitation:
EXAMPLE 2.16 Every object in the universe is attracted to every other
A 0.25 kg ball is attached to the end of a 0.5 m string and moved in a object with a force that is directly proportional to the
horizontal circle at 2.0 m/s. What net force is needed to keep the ball product of their masses and inversely proportional to the
in its circular path? square of the distances between them.
2-25 CHAPTER 2 Motion 49