Download Section 3: Circular Motion

Section 3: Circular Motion
Student Objectives:
•
Solve problem involving circular motion using Newton’s Laws.
•
Recognize the different between tangential and centripetal components of velocity and
what each means for an object moving in a circle or along a circular path.
•
Rotate coordinate systems to follow an object on a circular path, keeping a tangential
axis along the direction of motion and a centripetal axis perpendicular to the motion.
Concepts:
There are three mathematical quantities that will be of primary interest to us as we analyze
the motion of objects in circles. These three quantities are speed, acceleration and force. The
speed of an object moving in a circle is given by the following equation.
The acceleration of an object moving in a circle can be determined by either two of the
following equations.
The equation on the right (above) is derived from the equation on the left by the
substitution of the expression for speed.
The net force (Fnet) acting upon an object moving in circular motion is directed inwards.
While there may by more than one force acting upon the object, the vector sum of all of them
should add up to the net force. In general, the inward force is larger than the outward force (if
any) such that the outward force cancels and the unbalanced force is in the direction of the
center of the circle. The net force is related to the acceleration of the object (as is always the
case) and is thus given by the following three equations:
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The equations in the middle (above) and on the right (above) are derived from the equation
on the left by the substitution of the expressions for acceleration.
Worked Examples:
1. A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m.
Determine the acceleration and the net force acting upon the car.
To determine the acceleration of the car, use the equation a = v2 / R. The
solution is as follows:
a = v2 / R
a = (10.0 m/s)2 / (25.0 m)
a = (100 m2/s2) / (25.0 m)
a = 4 m/s2
To determine the net force acting upon the car, use the equation Fnet = m•a.
The solution is as follows.
Fnet = m • a
Fnet = (900 kg) • (4 m/s2)
Fnet = 3600 N
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2. An object attached to the end of a string swings in a vertical circle (R = 4.2 m). At an
instant when the angle the object makes with respect to the horizontal axis is 26o the
speed of the object is 9.6 m/s and the tension in the string is 72 N. What is the mass of
the object?
m
r
g
26
o
r
v
There are two forces acting on the ball, gravity and the tension in the string.
We can draw a free body diagram for the ball at this point on its circular path.
T
c
r
mg
26o
t
Notice how instead of drawing an x and y axis as we would have for previous
problems instead we have drawn a system of coordinates that is more
appropriate for a problem involving circular motion. The two axis are the
centripetal (c) and tangential (t) axis here. The two forces that act on the ball
are represented as well. Note that the tension force will always act along the
centripetal direction since it will always be directed back towards the center
of the circle. As the angle of the ball changes, the direction (in our coordinate
system, which is not stationary) of gravity would change.
To find the mass of the object, let’s consider writing Newton’s Law for the
centripetal direction.
∑F
c
( )
= ma c = T + mg sin 26 o
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For circular motion we can use, ac = v2 / R to give,
v2
m
= T + mg sin 26 o
R
( )
We can rearrange this equation to have the terms with the mass on one side
and the terms without mass on the other side.
m
v2
− mg sin 26 o = T
R
( )
v2

m  − g sin 26 o  = T
R

( )
Everything in the brackets is a known quantity, thus we can solve for the mass
of the object using,
m=
T
v
o 
 − g sin 26 
R

2
( )
Plugging in the values given,
m ≈ 4.08 kg.
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.
Homework:
1. A 0.5 kg ball on a string is twirled around in a circle perpendicular to the ground. The
ball moves at a constant velocity of 20 m/s around a circle with a radius of 3.0 m.
Determine the acceleration and the net force acting upon the ball.
2. The Cajun Cliffhanger at Great America is a ride in which occupants line the perimeter of
a cylinder and spin in a circle at a high rate of turning. When the cylinder begins spinning
very rapidly, the floor is removed from under the riders' feet.
a.
What affect does a doubling in speed have upon the centripetal force? Explain.
b. Determine the centripetal force acting upon a 40-kg child who makes 10
revolutions around the Cliffhanger in 29.3 seconds. The radius of the barrel is
2.90 meters.
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