Uniform Circular Motion
Physics 6A
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Uniform = Constant Speed
Circular = The Path is a Circle (or part of a circle)
Motion = Speed is not zero
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Uniform = Constant Speed
Circular = The Path is a Circle (or part of a circle)
Motion = Speed is not zero
Examples of UCM:
A car driving around a circular turn at constant speed
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Uniform = Constant Speed
Circular = The Path is a Circle (or part of a circle)
Motion = Speed is not zero
Examples of UCM:
A car driving around a circular turn at constant speed
A rock tied to a string and whirled in a circle
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Uniform = Constant Speed
Circular = The Path is a Circle (or part of a circle)
Motion = Speed is not zero
Examples of UCM:
A car driving around a circular turn at constant speed
A rock tied to a string and whirled in a circle
Clothes in a dryer spinning at constant speed
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Clothes in a dryer spinning at constant speed
Uniform = Constant Speed
Circular = The Path is a Circle (or part of a circle)
Motion = Speed is not zero
Examples of UCM:
A car driving around a circular turn at constant speed
A rock tied to a string and whirled in a circle
Clothes in a dryer spinning at constant speed
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Clothes in a dryer spinning at constant speed
A passenger on a Ferris wheel
Uniform = Constant Speed
Circular = The Path is a Circle (or part of a circle)
Motion = Speed is not zero
Examples of UCM:
A car driving around a circular turn at constant speed
A rock tied to a string and whirled in a circle
Clothes in a dryer spinning at constant speed
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Clothes in a dryer spinning at constant speed
A passenger on a Ferris wheel
Earth orbiting the Sun (almost, but not quite true)
Uniform = Constant Speed
Circular = The Path is a Circle (or part of a circle)
Motion = Speed is not zero
Examples of UCM:
A car driving around a circular turn at constant speed
A rock tied to a string and whirled in a circle
Clothes in a dryer spinning at constant speed
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Clothes in a dryer spinning at constant speed
A passenger on a Ferris wheel
Earth orbiting the Sun (almost, but not quite true)
What do these motions have in common?
Uniform = Constant Speed
Circular = The Path is a Circle (or part of a circle)
Motion = Speed is not zero
Examples of UCM:
A car driving around a circular turn at constant speed
A rock tied to a string and whirled in a circle
Clothes in a dryer spinning at constant speed
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Clothes in a dryer spinning at constant speed
A passenger on a Ferris wheel
Earth orbiting the Sun (almost, but not quite true)
What these motions have in common:
Constant speed (not constant velocity)
Acceleration toward the center of the circle (constant magnitude)
CENTRIPETAL is the word for this
We have a formula that we will use often for circular motion.
For an object moving in a circular path, the centripetal (toward the center) acceleration is given by:
Rv
a2
cent =You might also see arad, which stands for radial acceleration
Here v stands for the linear speed and R is the radius of the circular path.
v
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
v
arad
v
v
v
aradarad
arad arad
Notice that the radial acceleration is always toward the center of the circle, and the velocity is always tangent to the circle.
This is Uniform Circular Motion
Example Problem 1
A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Example Problem 1
A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube.
Here is a diagram of the centrifuge.
The bottom of the test tube is 8.3 cm from the center, so we will use a radius of 8.3 cm in our formula.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
8.3 cm
Example Problem 1
A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube.
Here is a diagram of the centrifuge.
The bottom of the test tube is 8.3 cm from the center, so we will use a radius of 8.3 cm in our formula.
rv
a2
cent =We need to find the speed v
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
8.3 cm
Example Problem 1
A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube.
Here is a diagram of the centrifuge.
The bottom of the test tube is 8.3 cm from the center, so we will use a radius of 8.3 cm in our formula.
rv
a2
cent =We need to find the speed v
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
8.3 cmConvert from rpm to m/s:
( )sec60min1
revmr2
minrev100
v ⋅π
⋅=
The circumference of the circle is the distance traveled during each revolution
Example Problem 1
A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube.
Here is a diagram of the centrifuge.
The bottom of the test tube is 8.3 cm from the center, so we will use a radius of 8.3 cm in our formula.
rv
a2
cent =We need to find the speed v
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
8.3 cmConvert from rpm to m/s:
( )
( )sec60min1
revm083.02
minrev100
v
sec60min1
revmr2
minrev100
v
⋅⋅π
⋅=
⋅π
⋅=
The circumference of the circle is the distance traveled during each revolution
Example Problem 1
A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube.
Here is a diagram of the centrifuge.
The bottom of the test tube is 8.3 cm from the center, so we will use a radius of 8.3 cm in our formula.
rv
a2
cent =We need to find the speed v
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
8.3 cmConvert from rpm to m/s:
( )
( )
sm87.0v
sec60min1
revm083.02
minrev100
v
sec60min1
revmr2
minrev100
v
=
⋅⋅π
⋅=
⋅π
⋅=
The circumference of the circle is the distance traveled during each revolution
Example Problem 1
A centrifuge rotates at a rate of 1000 revolutions per minute. If the test tube in the centrifuge is 8.3 cm long, find the centripetal acceleration at the bottom of the test tube.
Here is a diagram of the centrifuge.
The bottom of the test tube is 8.3 cm from the center, so we will use a radius of 8.3 cm in our formula.
rv
a2
cent =We need to find the speed v
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
8.3 cmConvert from rpm to m/s:
( )
( )
sm87.0v
sec60min1
revm083.02
minrev100
v
sec60min1
revmr2
minrev100
v
=
⋅⋅π
⋅=
⋅π
⋅=
The circumference of the circle is the distance traveled during each revolution
Now we can use our formula to find acceleration:
( )2sm
2sm
cent 1.9m083.0
87.0a ==
Example Problem 2
A Ferris wheel with radius 14m is turning about an axis at its center, as shown. The linear speed of a passenger on the rim is constant at 7 m/s. What are the magnitude and direction of the passenger’s acceleration a) at the top and b) at the bottom?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Example Problem 2
A Ferris wheel with radius 14m is turning about an axis at its center, as shown. The linear speed of a passenger on the rim is constant at 7 m/s. What are the magnitude and direction of the passenger’s acceleration a) at the top and b) at the bottom?
First think about the direction of the acceleration:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Example Problem 2
A Ferris wheel with radius 14m is turning about an axis at its center, as shown. The linear speed of a passenger on the rim is constant at 7 m/s. What are the magnitude and direction of the passenger’s acceleration a) at the top and b) at the bottom?
First think about the direction of the acceleration:
At the top, the acceleration is downward (toward the center)ar
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Example Problem 2
A Ferris wheel with radius 14m is turning about an axis at its center, as shown. The linear speed of a passenger on the rim is constant at 7 m/s. What are the magnitude and direction of the passenger’s acceleration a) at the top and b) at the bottom?
First think about the direction of the acceleration:
At the top, the acceleration is downward (toward the center), and at the bottom, the acceleration is upward (again, toward the center)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
ar
Example Problem 2
A Ferris wheel with radius 14m is turning about an axis at its center, as shown. The linear speed of a passenger on the rim is constant at 7 m/s. What are the magnitude and direction of the passenger’s acceleration a) at the top and b) at the bottom?
First think about the direction of the acceleration:
At the top, the acceleration is downward (toward the center), and at the bottom, the acceleration is upward (again, toward the center) a
r
We can find the magnitude from our formula for centripetal acceleration:
v2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
arr
va
2
cent =
Example Problem 2
A Ferris wheel with radius 14m is turning about an axis at its center, as shown. The linear speed of a passenger on the rim is constant at 7 m/s. What are the magnitude and direction of the passenger’s acceleration a) at the top and b) at the bottom?
First think about the direction of the acceleration:
At the top, the acceleration is downward (toward the center), and at the bottom, the acceleration is upward (again, toward the center) a
r
We can find the magnitude from our formula for centripetal acceleration:
2v
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
ar
( )2sm
2sm
cent
2
cent
5.3m14
7a
rv
a
==
=