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Dynamics of Uniform Circular Motion
Introduction
v2
 F = mac = m --–
r
Since the acceleration is toward the center, the force on the object must
also be toward the center. This force is applied by an exterior object. In the
case of the Moon, the Earth's gravity applies the force.
An object being swung in a circle creates a common misconception. You feel
a force the rope that is interpreted as Centrifugal (center- fleeing) force. You
must apply a force to keep the object from flying off at a tangent. This force
applied on the object is toward the center of the motion. Newton's Third Law of
Motion: States that the object must apply an equal force on you in the opposite
direction.
If there was a centrifuge force on the object, it would fly outward when
released. Instead of flying straight out when released, it flies of tangential
to the point of release. Consider the old style sling shot.
If an object is being swung on a cord you cannot get a perfect horizontal
circle. There are forces in two directions on the object. Centripetal force
pulls the object towards the center but gravity pulls the object downward. The
tension in the rope is the vector sum of the two forces.
Table of Contents
Problems
| 1. |
What is the tension on the rope and angle
below horizontal if a O .150 kg ball is being swung in a horizontal circle
at 2.00 revolutions per second with a 0.600 m rope?
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| 2. |
Rachel, 45.0 kg, is ice skating on a
frictionless surface. She is tied to a 10.0 m rope and is skating in a
circular motion with 80.0 N of tension on the rope. How many
revolutions per minute does she skate?
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| 3. |
A 200. g ball is swung in a vertical circle
using a 0.500 m rope. Calculate the speed the ball must have at the top to continue
moving in a circle. (a) What is the tension on the rope at the top. (b) If
the ball is moving twice as fast at the bottom than at the top, what is
the tension at the bottom? |
Table of Contents
Answers
| 1. |
What is the tension on the rope and angle
below horizontal if a O .150 kg ball is being swung in a horizontal circle
at 2.00 revolutions per second with a 0.600 m rope?
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|
1 s
------------ = 0.500 s per revolution = Period
2.00 rev
2 r
2 0.600 m
v = ------ = -------------------- = 7.54 m/s
t
0.500 s
mv2
0.150 kg x (7.54 m s -1)2
Fc = -------- = ----------------------------------- =
14.2 N
r
0.600 m
Fg =
mg = 0.150 kg x 9.80 m s -2 = 1.47 N  Frope
= ( (14.2 N)2 + (1.47 N)2 ) 1/2 = 14.3 N
1.47 N
 = Tan
-1 ( ------------ ) = 5.91o
14.2 N
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| 2. |
Rachel, 45.0 kg, is ice skating on a
frictionless surface. She is tied to a 10.0 m rope and is skating in a
circular motion with 80.0 N of tension on the rope. How many
revolutions per minute does she skate?
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|
mv2
r Fc
10.0 m x 80.0 N
Fc = -------- => v = (
-------- ) 1/2 = ( ----------------------- )
1/2 = 4.22 m/s
r
m
45.0 kg
2r
2r
2 10.0 m
v = ------ => t = --------- =
---------------- = 14.9
s This
is the Period, time for 1 revolution.
t
v 4.22 m s
-1
1
rev 60 s
--------- x --------- = 4.03 revolutions per minute = 4.03 rpm
14.9 s 1 min
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| 3. |
A 200. g ball is swung in a vertical circle
using a 0.500 m rope. Calculate the speed the ball must have at the top to continue
moving in a circle. (a) What is the tension on the rope at the top. (b) If
the ball is moving twice as fast at the bottom than at the top, what is
the tension at the bottom?
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|
At the top of the circle, gravity provides exactly enough force to keep
the ball in a vertical circle.
T = 0
Fc = Fg = 0.200 kg x
9.80 m s -2 = 1.96 N
mv2
r Fc
1.0 m x 1.96 N
Fc = -------- => v = (
-------- ) 1/2 = ( ----------------------- )
1/2 = 3.13 m/s
Velocity at top of circle.
r
m
0.200 kg
Velocity at bottom = 2 x 3.13 m/s = 6.26 m/s
mv2
0.200 kg x (3.13 m s -1)2
Fc = -------- = ----------------------------------- =
3.92 N
r
0.500 m
T = Fc + Fg
= 3.92 N + 1.96 N = 5.88 N
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Table of Contents
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