Physics 6A

Rotational Motion

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First some quick geometry review:

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θr

x = rθWe need this formula for arc length to see the connection between rotational motion and linear motion.

We will also need to be able to convert from revolutions to radians.

There are 2π radians in one complete revolution.

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θr

x = rθDefinitions of angular velocity and angular acceleration are analogous to what we had for linear motion.

Angular Velocity = ω = t

Angular Acceleration = α = t

This is the Greek letter omega (not w)

This is the Greek letter alpha (looks kinda like a fish)

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For Campus Learning Assistance Services at UCSB

θr

x = rθDefinitions of angular velocity and angular acceleration are analogous to what we had for linear motion.

Angular Velocity = ω = t

Angular Acceleration = α = t

This is the Greek letter omega (not w)

Example: A centrifuge starts from rest and spins for 7 seconds until it reaches 1000 rpm.

Find the final angular velocity and the angular acceleration (assume constant).

This is the Greek letter alpha (looks kinda like a fish)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

θr

x = rθDefinitions of angular velocity and angular acceleration are analogous to what we had for linear motion.

Angular Velocity = ω = t

Angular Acceleration = α = t

This is the Greek letter omega (not w)

Example: A centrifuge starts from rest and spins for 7 seconds until it reaches 1000 rpm.

Find the final angular velocity and the angular acceleration (assume constant).

rpm stands for “revolutions per minute” – we can treat the first part of this problem just like a unit conversion:

secrad7.104

sec60

min1

rev1

rad2

min1

rev1000

Standard units for angular velocity are radians per second

This is the Greek letter alpha (looks kinda like a fish)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

θr

x = rθDefinitions of angular velocity and angular acceleration are analogous to what we had for linear motion.

Angular Velocity = ω = t

Angular Acceleration = α = t

This is the Greek letter omega (not w)

Example: A centrifuge starts from rest and spins for 7 seconds until it reaches 1000 rpm.

Find the final angular velocity and the angular acceleration (assume constant).

rpm stands for “revolutions per minute” – we can treat the first part of this problem just like a unit conversion:

secrad7.104

sec60

min1

rev1

rad2

min1

rev1000

Standard units for angular velocity are radians per second

Now we can use the definition of angular acceleration:

2secradsec

rad

15sec7

7.104

t

Standard units for angular acceleration are radians per second squared.

This is the Greek letter alpha (looks kinda like a fish)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

We already know how to deal with linear motion.

We have formulas for kinematics, forces, energy and momentum.

We will find similar formulas for rotational motion. Actually, we already know the formulas – they are the same as the linear ones we already know!

All you have to do is translate the variables.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

We already know how to deal with linear motion.

We have formulas for kinematics, forces, energy and momentum.

We will find similar formulas for rotational motion. Actually, we already know the formulas – they are the same as the linear ones we already know!

All you have to do is translate the variables.

We have already seen one case:

x = rθ This translates between distance (linear) and angle (rotational)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

We already know how to deal with linear motion.

We have formulas for kinematics, forces, energy and momentum.

We will find similar formulas for rotational motion. Actually, we already know the formulas – they are the same as the linear ones we already know!

All you have to do is translate the variables.

We have already seen one case:

x = rθ This translates between distance (linear) and angle (rotational)

Here are the other variables:

v = rω linear velocity relates to angular velocity

atan = rα linear acceleration relates to angular acceleration

Notice a pattern here?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

We already know how to deal with linear motion.

We have formulas for kinematics, forces, energy and momentum.

We will find similar formulas for rotational motion. Actually, we already know the formulas – they are the same as the linear ones we already know!

All you have to do is translate the variables.

We have already seen one case:

x = rθ This translates between distance (linear) and angle (rotational)

Here are the other variables:

v = rω linear velocity relates to angular velocity

atan = rα linear acceleration relates to angular acceleration

Multiply the angular quantity by the radius to get the corresponding linear quantity.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas.

Linear Motion (constant a) Rotational Motion (constant α)

x=x0+v0t+½at2 θ=θ0+ω0t+½αt2

v=v0+at ω=ω0+αt

v2=v02+2a(x-x0) ω2=ω0

2+2α(θ-θ0)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas.

Linear Motion (constant a) Rotational Motion (constant α)

x=x0+v0t+½at2 θ=θ0+ω0t+½αt2

v=v0+at ω=ω0+αt

v2=v02+2a(x-x0) ω2=ω0

2+2α(θ-θ0)

Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.

Find the total number of revolutions that the wheels make during the 25 second interval.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas.

Linear Motion (constant a) Rotational Motion (constant α)

x=x0+v0t+½at2 θ=θ0+ω0t+½αt2

v=v0+at ω=ω0+αt

v2=v02+2a(x-x0) ω2=ω0

2+2α(θ-θ0)

Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.

Find the total number of revolutions that the wheels make during the 25 second interval.

We basically have two options on how to proceed. We can switch to angular variables right away, or we can do the corresponding problem in linear variables and translate at the end.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas.

Linear Motion (constant a) Rotational Motion (constant α)

x=x0+v0t+½at2 θ=θ0+ω0t+½αt2

v=v0+at ω=ω0+αt

v2=v02+2a(x-x0) ω2=ω0

2+2α(θ-θ0)

Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.

Find the total number of revolutions that the wheels make during the 25 second interval.

Switching to angular variables right away:

Convert to angular velocity:

srads

m

9.42m35.0

15

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas.

Linear Motion (constant a) Rotational Motion (constant α)

x=x0+v0t+½at2 θ=θ0+ω0t+½αt2

v=v0+at ω=ω0+αt

v2=v02+2a(x-x0) ω2=ω0

2+2α(θ-θ0)

Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.

Find the total number of revolutions that the wheels make during the 25 second interval.

Switching to angular variables right away:

Convert to angular velocity:

srads

m

9.42m35.0

15

Find angular acceleration:

2srads

rad

7.1s25

9.42

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas.

Linear Motion (constant a) Rotational Motion (constant α)

x=x0+v0t+½at2 θ=θ0+ω0t+½αt2

v=v0+at ω=ω0+αt

v2=v02+2a(x-x0) ω2=ω0

2+2α(θ-θ0)

Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.

Find the total number of revolutions that the wheels make during the 25 second interval.

Switching to angular variables right away:

Convert to angular velocity:

srads

m

9.42m35.0

15

Find angular acceleration:

2srads

rad

7.1s25

9.42

Use a kinematics equation:

rad25.531)s25)(7.1(

tt

2

srad

21

221

00

2

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas.

Linear Motion (constant a) Rotational Motion (constant α)

x=x0+v0t+½at2 θ=θ0+ω0t+½αt2

v=v0+at ω=ω0+αt

v2=v02+2a(x-x0) ω2=ω0

2+2α(θ-θ0)

Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.

Find the total number of revolutions that the wheels make during the 25 second interval.

Switching to angular variables right away:

Convert to angular velocity:

srads

m

9.42m35.0

15

Find angular acceleration:

2srads

rad

7.1s25

9.42

Use a kinematics equation:

rad25.531)s25)(7.1(

tt

2

srad

21

221

00

2

Convert to revolutions:

rev6.842

rad25.531

revrad

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas.

Linear Motion (constant a) Rotational Motion (constant α)

x=x0+v0t+½at2 θ=θ0+ω0t+½αt2

v=v0+at ω=ω0+αt

v2=v02+2a(x-x0) ω2=ω0

2+2α(θ-θ0)

Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant.

Find the total number of revolutions that the wheels make during the 25 second interval.

This time do the linear problem first:

Find linear acceleration:

2sms

m

6.0s25

15a

Use a kinematics equation:

Convert to revolutions:

rev3.85)35.0(2

m5.187

revm

m5.187)s25)(6.0(x

attvxx

2

sm

21

221

00

2

(we did some rounding off)

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For Campus Learning Assistance Services at UCSB

Some other major topics for linear motion are Energy, Forces and Momentum.

All of these have analogues for rotational motion as well.

Any moving object will have Kinetic Energy. This applies to rotating objects. Here’s a Formula:

221

rotational IK We know that ω is angular velocity. Comparing with the formula for linear kinetic energy, what do you think I is?

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221

rotational IK We know that ω is angular velocity. Comparing with the formula for linear kinetic energy, what do you think I is?

The I in our formula takes the place of m (mass) in the linear formula. We call it Moment of Inertia (or rotational inertia). It plays the same role in rotational motion that mass plays in linear motion (I quantifies how difficult it is to produce an angular acceleration, just like mass relates to linear acceleration).

The value for I will depend on the shape of your object, but the basic rule of thumb is that the farther the mass is from the axis of rotation, the larger the inertia.

Page 306 in your book has a table of formulas for different shapes.

These are all based on the formula for the moment of inertia of a point particle. You will not have to derive them, just know how to use them.

2particle mRI

Some other major topics for linear motion are Energy, Forces and Momentum.

All of these have analogues for rotational motion as well.

Any moving object will have Kinetic Energy. This applies to rotating objects. Here’s a Formula:

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example

A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.

a) How fast is each sphere moving when it reaches the bottom of the hill?

b) Which sphere will reach the bottom first, the hollow one or the solid one?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

h

θ

We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.

Example

A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.

a) How fast is each sphere moving when it reaches the bottom of the hill?

b) Which sphere will reach the bottom first, the hollow one or the solid one?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

h

θ

We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.

GravRotLin

TopBottom

UKK

EE

Example

A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.

a) How fast is each sphere moving when it reaches the bottom of the hill?

b) Which sphere will reach the bottom first, the hollow one or the solid one?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

h

θ

We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.

mghImv

UKK

EE

2212

21

GravRotLin

TopBottom

we can replace ω with v/r so everything is in terms of the desired unknown

Example

A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.

a) How fast is each sphere moving when it reaches the bottom of the hill?

b) Which sphere will reach the bottom first, the hollow one or the solid one?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

h

θ

We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.

mghImv

mghImv

UKK

EE

2

rv

212

21

2212

21

GravRotLin

TopBottom

we can replace ω with v/r so everything is in terms of the desired unknown

Example

A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.

a) How fast is each sphere moving when it reaches the bottom of the hill?

b) Which sphere will reach the bottom first, the hollow one or the solid one?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

h

θ

We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.

mghImv

mghImv

UKK

EE

2

rv

212

21

2212

21

GravRotLin

TopBottom

we can replace ω with v/r so everything is in terms of the desired unknown

At this point we can substitute the formula for each shape (from table 9.2 on page 279)

Example

A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.

a) How fast is each sphere moving when it reaches the bottom of the hill?

b) Which sphere will reach the bottom first, the hollow one or the solid one?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

h

θ

We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.

mghImv

mghImv

UKK

EE

2

rv

212

21

2212

21

GravRotLin

TopBottom

we can replace ω with v/r so everything is in terms of the desired unknown

At this point we can substitute the formula for each shape (from table 9.2 on page 279)

ghv

mghmv

mghmrmv

710

2107

2

rv2

52

212

21

Solid Sphere

Example

A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.

a) How fast is each sphere moving when it reaches the bottom of the hill?

b) Which sphere will reach the bottom first, the hollow one or the solid one?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

h

θ

We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.

we can replace ω with v/r so everything is in terms of the desired unknown

At this point we can substitute the formula for each shape (from table 9.2 on page 279)

ghv

mghmv

mghmrmv

710

2107

2

rv2

52

212

21

Solid Sphere

ghv

mghmv

mghmrmv

56

265

2

rv2

32

212

21

Hollow Sphere

mghImv

mghImv

UKK

EE

2

rv

212

21

2212

21

GravRotLin

TopBottom

Example

A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.

a) How fast is each sphere moving when it reaches the bottom of the hill?

b) Which sphere will reach the bottom first, the hollow one or the solid one?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

h

θ

We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom.

we can replace ω with v/r so everything is in terms of the desired unknown

At this point we can substitute the formula for each shape (from table 9.2 on page 279)

ghv

mghmv

mghmrmv

710

2107

2

rv2

52

212

21

Solid Sphere

ghv

mghmv

mghmrmv

56

265

2

rv2

32

212

21

Hollow SphereThe solid sphere is faster because its moment of inertia is smaller.

It reaches the bottom first.

mghImv

mghImv

UKK

EE

2

rv

212

21

2212

21

GravRotLin

TopBottom

Example

A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h.

a) How fast is each sphere moving when it reaches the bottom of the hill?

b) Which sphere will reach the bottom first, the hollow one or the solid one?