Physics 6A
Rotational Motion
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First some quick geometry review:
x = rθ
θ
r
We 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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x = rθ
Definitions of angular velocity and angular
acceleration are analogous to what we had
for linear motion.
This is the Greek letter omega (not w)
θ
r
Angular Velocity = ω =
∆θ
∆t
Angular Acceleration = α =
∆ω
∆t
This is the Greek letter alpha
(looks kinda like a fish)
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x = rθ
Definitions of angular velocity and angular
acceleration are analogous to what we had
for linear motion.
This is the Greek letter omega (not w)
θ
r
Angular Velocity = ω =
∆θ
∆t
Angular Acceleration = α =
∆ω
∆t
This is the Greek letter alpha
(looks kinda like a fish)
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).
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x = rθ
Definitions of angular velocity and angular
acceleration are analogous to what we had
for linear motion.
This is the Greek letter omega (not w)
θ
r
Angular Velocity = ω =
∆θ
∆t
Angular Acceleration = α =
∆ω
∆t
This is the Greek letter alpha
(looks kinda like a fish)
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:
1000rev 2πrad 1min
rad
⋅
= 104.7 sec
⋅
1min
1rev 60 sec
Standard units for angular velocity are radians per second
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x = rθ
Definitions of angular velocity and angular
acceleration are analogous to what we had
for linear motion.
This is the Greek letter omega (not w)
θ
Angular Velocity = ω =
r
∆θ
∆t
Angular Acceleration = α =
∆ω
∆t
This is the Greek letter alpha
(looks kinda like a fish)
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:
1000rev 2πrad 1min
rad
⋅
= 104.7 sec
⋅
1min
1rev 60 sec
Standard units for angular velocity are radians per second
Now we can use the definition of angular acceleration:
rad
∆ω 104.7 sec
=
= 15 rad2
α=
sec
∆t
7 sec
Standard units for angular acceleration are radians per second squared.
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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.
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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)
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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?
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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.
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We can use this technique to find angular motion formulas from the analogous
linear motion formulas we already know. Here are some kinematics formulas.
(these are in a table on page 272 of your book)
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=ω02+2α(θ-θ0)
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We can use this technique to find angular motion formulas from the analogous
linear motion formulas we already know. Here are some kinematics formulas.
(these are in a table on page 272 of your book)
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=ω02+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
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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.
(these are in a table on page 272 of your book)
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=ω02+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.
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We can use this technique to find angular motion formulas from the analogous
linear motion formulas we already know. Here are some kinematics formulas.
(these are in a table on page 272 of your book)
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=ω02+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:
15 ms
ω=
= 42.9 rad
s
0.35m
Prepared by Vince Zaccone
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We can use this technique to find angular motion formulas from the analogous
linear motion formulas we already know. Here are some kinematics formulas.
(these are in a table on page 272 of your book)
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=ω02+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:
15 ms
ω=
= 42.9 rad
s
0.35m
Find angular acceleration:
α=
42.9 rad
s
25s
= 1.7 rad
2
s
Prepared by Vince Zaccone
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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.
(these are in a table on page 272 of your book)
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=ω02+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:
15 ms
ω=
= 42.9 rad
s
0.35m
Find angular acceleration:
α=
42.9 rad
s
25s
Use a kinematics equation:
θ = θ0 + ω0 t + 21 αt 2
θ = 21 (1.7 rad
)(25s)2 = 531.25rad
2
s
= 1.7 rad
2
s
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We can use this technique to find angular motion formulas from the analogous
linear motion formulas we already know. Here are some kinematics formulas.
(these are in a table on page 272 of your book)
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=ω02+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:
15 ms
ω=
= 42.9 rad
s
0.35m
Find angular acceleration:
α=
42.9 rad
s
25s
= 1.7 rad
2
s
Use a kinematics equation:
θ = θ0 + ω0 t + 21 αt 2
θ = 21 (1.7 rad
)(25s)2 = 531.25rad
2
s
Convert to revolutions:
531.25rad
= 84.6rev
2π rad
rev
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We can use this technique to find angular motion formulas from the analogous
linear motion formulas we already know. Here are some kinematics formulas.
(these are in a table on page 272 of your book)
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=ω02+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:
a=
15 ms
25s
= 0.6 m2
s
Convert to revolutions:
187.5m
= 85.3rev
m
2π(0.35) rev
Use a kinematics equation:
x = x 0 + v 0 t + 21 at 2
x = 21 (0.6 m2 )(25s)2 = 187.5m
s
(we did some rounding off)
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Some other major topics for linear motion are Energy, Forces and Momentum.
All of these have analogues for rotational motion as well. Forces and Momentum
will be covered in chapter 10. That leaves us with Energy.
Any moving object will have Kinetic Energy. This applies to rotating objects.
Here’s a Formula:
K rotational = 21 I ⋅ ω2
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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Some other major topics for linear motion are Energy, Forces and Momentum.
All of these have analogues for rotational motion as well. Forces and Momentum
will be covered in chapter 10. That leaves us with Energy.
Any moving object will have Kinetic Energy. This applies to rotating objects.
Here’s a Formula:
K rotational = 21 I ⋅ ω2
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 279 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.
Iparticle = mR 2
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Problem 9.48
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?
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Problem 9.48
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?
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.
h
θ
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Problem 9.48
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?
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.
EBottom = ETop
h
θ
KLin + KRot = UGrav
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Problem 9.48
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?
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.
EBottom = ETop
KLin + KRot = UGrav
1
mv 2
2
+ 21 Iω2 = mgh
h
θ
we can replace ω with v/r so everything
is in terms of the desired unknown
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Problem 9.48
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?
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.
EBottom = ETop
KLin + KRot = UGrav
1
mv 2
2
+ 21 Iω2 = mgh
1
mv 2
2
+ 21 I vr
h
θ
we can replace ω with v/r so everything
is in terms of the desired unknown
( )2 = mgh
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Problem 9.48
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?
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.
h
EBottom = ETop
KLin + KRot = UGrav
1
mv 2
2
+ 21 Iω2 = mgh
1
mv 2
2
+ 21 I vr
θ
we can replace ω with v/r so everything
is in terms of the desired unknown
( )2 = mgh
At this point we can substitute the formula for each shape (from table 9.2 on page 279)
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Problem 9.48
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?
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.
h
EBottom = ETop
θ
we can replace ω with v/r so everything
is in terms of the desired unknown
KLin + KRot = UGrav
1
mv 2
2
+ 21 Iω2 = mgh
1
mv 2
2
+ 21 I vr
( )2 = mgh
At this point we can substitute the formula for each shape (from table 9.2 on page 279)
Solid Sphere
(
)( )
2
1
mv 2 + 21 52 mr 2 vr
2
7 mv 2 = mgh
10
v=
= mgh
10
gh
7
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Problem 9.48
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?
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.
h
EBottom = ETop
θ
we can replace ω with v/r so everything
is in terms of the desired unknown
KLin + KRot = UGrav
1
mv 2
2
+ 21 Iω2 = mgh
1
mv 2
2
+ 21 I vr
( )2 = mgh
At this point we can substitute the formula for each shape (from table 9.2 on page 279)
Solid Sphere
(
Hollow Sphere
)( )
2
1
mv 2 + 21 52 mr 2 vr
2
7 mv 2 = mgh
10
v=
10
gh
7
= mgh
1
mv 2
2
5
mv 2
6
v=
+ 21
( mr )( )
2
3
2 v 2
r
= mgh
= mgh
6
gh
5
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Problem 9.48
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?
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.
h
EBottom = ETop
θ
we can replace ω with v/r so everything
is in terms of the desired unknown
KLin + KRot = UGrav
1
mv 2
2
+ 21 Iω2 = mgh
1
mv 2
2
+ 21 I vr
( )2 = mgh
At this point we can substitute the formula for each shape (from table 9.2 on page 279)
Solid Sphere
(
Hollow Sphere
)( )
2
1
mv 2 + 21 52 mr 2 vr
2
7 mv 2 = mgh
10
v=
10
gh
7
= mgh
1
mv 2
2
5
mv 2
6
v=
+ 21
( mr )( )
2
3
= mgh
2 v 2
r
= mgh
The solid sphere is faster because its
moment of inertia is smaller.
It reaches the bottom first.
6
gh
5
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