Uniform Circular Motion Physics 6A Prepared by Vince Zaccone For Campus Learning Assistance Services...

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Uniform Circular Motion Physics 6A Prepared by Vince Zaccone For Campus Learning Assistance Services at

Transcript of Uniform Circular Motion Physics 6A Prepared by Vince Zaccone For Campus Learning Assistance Services...

Uniform Circular Motion

Physics 6A

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

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

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

A passenger on a Ferris wheel

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

A passenger on a Ferris wheel

Earth orbiting the Sun (almost, but not quite true)

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

A passenger on a Ferris wheel

Earth orbiting the Sun (almost, but not quite true)

What do these motions have in common?

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

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

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

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:

R

va

2

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

arad

v

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

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.

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.

8.3 cm

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

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.

8.3 cm

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.

r

va

2

cent We need to find the speed v

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.

8.3 cm

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.

r

va

2

cent We need to find the speed v

Convert from rpm to m/s:

sec60min1

revmr2

minrev1000

v

The circumference of the circle is the distance traveled during each revolution

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.

8.3 cm

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.

r

va

2

cent We need to find the speed v

Convert from rpm to m/s:

sec60min1

revm083.02

minrev1000

v

sec60min1

revmr2

minrev1000

v

The circumference of the circle is the distance traveled during each revolution

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.

8.3 cm

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.

r

va

2

cent We need to find the speed v

Convert from rpm to m/s:

sm7.8v

sec60min1

revm083.02

minrev1000

v

sec60min1

revmr2

minrev1000

v

The circumference of the circle is the distance traveled during each revolution

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.

8.3 cm

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.

r

va

2

cent We need to find the speed v

Convert from rpm to m/s:

sm7.8v

sec60min1

revm083.02

minrev1000

v

sec60min1

revmr2

minrev1000

v

The circumference of the circle is the distance traveled during each revolution

Now we can use our formula to find acceleration:

2sm

2

sm

cent 910m083.0

7.8a

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?

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)a

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)

a

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)

a

a

We can find the magnitude from our formula for centripetal acceleration:

r

va

2

cent

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)

a

a

We can find the magnitude from our formula for centripetal acceleration:

2s

m

2

sm

cent

2

cent

5.3m14

7a

r

va