Kinetic Energy of a Cylinder Rolling Without Slipping

In summary, the kinetic energy of a cylinder rolling without slipping can be represented by the equation KE_total = KE_translational + KE_rotational, where KE_translational = 1/2 * m * v_cm^2 and KE_rotational = 1/2 * I * w^2. This equation accounts for the combined translational and rotational motion of the cylinder, taking into consideration the different speeds of different parts of the cylinder. It is generally applicable to any rigid body motion and may involve calculus in its explanation.
  • #1
57
7
Homework Statement
This question is not a homework problem, just something I was wondering about when doing rolling without slipping problems. How can the kinetic energy of something rolling down an incline be represented as the simple addition of the translational and rotational kinetic energies? (Better explanation of my question below)
Relevant Equations
##KE_{total} = KE_{translational} + KE_{rotational}##

##KE_{total} = \frac {1} {2} mv_{cm}^2 + \frac1 2 I \omega^2##
Given that there is a cylinder rolling without slipping down an incline, the method I was taught to represent the KE of the cylinder was:

##KE_{total} = KE_{translational} + KE_{rotational}##
##KE_{total} = \frac {1} {2} mv_{cm}^2 + \frac1 2 I \omega^2## Where "cm" is the center of mass, and "I" was about the cylinder's center of mass

I am confused about how the equation above can represent the kinetic energy of the cylinder. The kinetic energy of the cylinder is the kinetic energy of each of its individual particles added together, so I can't really see how that equation above accounts for all that.

I was told that rolling without slipping combines translational and rotational motion, hence why the translational and rotational kinetic energy equations are added together. But, isn't that neglecting that different parts of the cylinder have different speeds? The very bottom of the cylinder at the point of contact has a tangential velocity of 0 while at the very top of the cylinder, it has a much larger velocity. How does the equation account for the kinetic energies of all the particle's KE?
 
Physics news on Phys.org
  • #2
Perhaps you could visualize that cylinder as an array of bars all meeting at their individual middle point.
Each of those imaginary bars would rotate while simultaneously its center of mass translates along a straight line.
In order to start or stop that combined movement, you need energy for the rotation part and for the translation part.
 
  • Like
Likes uSee2
  • #3
uSee2 said:
Homework Statement: This question is not a homework problem, just something I was wondering about when doing rolling without slipping problems. How can the kinetic energy of something rolling down an incline be represented as the simple addition of the translational and rotational kinetic energies? (Better explanation of my question below)
Relevant Equations: ##KE_{total} = KE_{translational} + KE_{rotational}##

##KE_{total} = \frac {1} {2} mv_{cm}^2 + \frac1 2 I \omega^2##

Given that there is a cylinder rolling without slipping down an incline, the method I was taught to represent the KE of the cylinder was:

##KE_{total} = KE_{translational} + KE_{rotational}##
##KE_{total} = \frac {1} {2} mv_{cm}^2 + \frac1 2 I \omega^2## Where "cm" is the center of mass, and "I" was about the cylinder's center of mass

I am confused about how the equation above can represent the kinetic energy of the cylinder. The kinetic energy of the cylinder is the kinetic energy of each of its individual particles added together, so I can't really see how that equation above accounts for all that.

I was told that rolling without slipping combines translational and rotational motion, hence why the translational and rotational kinetic energy equations are added together. But, isn't that neglecting that different parts of the cylinder have different speeds? The very bottom of the cylinder at the point of contact has a tangential velocity of 0 while at the very top of the cylinder, it has a much larger velocity. How does the equation account for the kinetic energies of all the particle's KE?
You need a calculation to show that the addition of translational and rotational KE equates to the total. I'm not sure there is a clever argument to justify it without calculation.

I'm sure that calculation must be online somewhere.

Note that it's generally true for any rigid body motion.
 
  • Like
Likes uSee2
  • #4
Lnewqban said:
Perhaps you could visualize that cylinder as an array of bars all meeting at their individual middle point.
Each of those imaginary bars would rotate while simultaneously its center of mass translates along a straight line.
In order to start or stop that combined movement, you need energy for the rotation part and for the translation part.
Thank you for answering! I do see that in the case of an object spinning while moving through space, that would be true since the tangential velocity at a distance R away from the center of mass would be constant at any point. But when an object is rolling without slipping, the bottom is at rest while the top is moving faster. So any point at an equal distance of R would have varying speeds.
 
  • #5
PeroK said:
You need a calculation to show that the addition of translational and rotational KE equates to the total. I'm not sure there is a clever argument to justify it without calculation.

I'm sure that calculation must be online somewhere.

Note that it's generally true for any rigid body motion.
Is the equation just something that works for rigid body motion? I can't exactly find the explanation online somewhere, but do you think that the explanation would involve calculus? If it does, I'm fine with just accepting that the equation "just works" since I still have no knowledge of calculus yet.
 
  • #6
uSee2 said:
Thank you for answering! I do see that in the case of an object spinning while moving through space, that would be true since the tangential velocity at a distance R away from the center of mass would be constant at any point. But when an object is rolling without slipping, the bottom is at rest while the top is moving faster. So any point at an equal distance of R would have varying speeds.
The result should be the same by both methods.
The velocity of the top particle is double the velocity of the center or axis of the cylinder, so the circular movement of that particle respect to the contact point (v=0) is harder to stop.

The next bar is in contact with the surface, but our bar is lifting the bottom end via rotation; therefore, its potential energy remains constant (while the cylinder is rolling on horizontal surface).

Kinetic energy wheel.png
 
  • #9
For an intuitive argument, consider a cart moving at constant velocity v. Inside the cart is a flywheel on a horizontal axis normal to the cart's motion.
To a person in the cart, it takes a certain amount of energy to get the wheel spinning at some rate. To a bystander, would you expect it to take the same amount of energy?

For an algebraic argument, consider an assemblage ##(m_i)## of particles with mass centre at ##\vec r##: ##\Sigma m_i\vec r_i=\vec r\Sigma m_i=\vec r\cdot m##, so ##\Sigma m_i\vec{ \dot r_i}=\vec {\dot r}\cdot m##.
##KE=\frac 12\Sigma m_i\vec{ \dot r_i}^2=\frac 12\Sigma m_i((\vec{ \dot r_i}-\vec {\dot r})^2+2\vec{ \dot r_i}\cdot\vec {\dot r}-\vec {\dot r}^2)##
##=\frac 12\Sigma m_i(\vec{ \dot r_i}-\vec {\dot r})^2+\frac 12\Sigma m_i
2\vec{ \dot r_i}\cdot\vec {\dot r}-\frac 12\Sigma m_i\vec {\dot r}^2##
##=\frac 12\Sigma m_i(\vec{ \dot r_i}-\vec {\dot r})^2+(m
\vec{ \dot r})\cdot\vec {\dot r}-\frac 12 m\vec {\dot r}^2
##
##=\frac 12\Sigma m_i(\vec{ \dot r_i}-\vec {\dot r})^2+\frac 12 m\vec {\dot r}^2
##
 
  • Like
Likes PeroK and uSee2
  • #10
Lnewqban said:
After reading it, wouldn't that mean that the equation definitely wouldn't work? Since if the axis of rotation was indeed the point of contact, that would mean that there is no translational motion at all. So it would just be:

##KE_{total} = \frac {1} {2} I \omega^2## where ##I## is the moment of inertia about the ground/point of contact.
 
  • #11
uSee2 said:
After reading it, wouldn't that mean that the equation definitely wouldn't work? Since if the axis of rotation was indeed the point of contact, that would mean that there is no translational motion at all. So it would just be:

##KE_{total} = \frac {1} {2} I \omega^2## where ##I## is the moment of inertia about the ground/point of contact.
That is true only instantaneously.
Otherwise, we will have a cylinder constantly rotating about the contact point, and its center the mass will describe a circle of radius similar to the radius of the cylinder.

Wheel rotating.jpg
 
  • Like
Likes uSee2
  • #12
Lnewqban said:
That is true only instantaneously.
Otherwise, we will have a cylinder constantly rotating about the contact point, and its center the mass will describe a circle of radius similar to the radius of the cylinder.

View attachment 326082
I do see that it is rotating, but since the axis of rotation is changing every moment it is as if it were moving translationally. But if it has only rotational kinetic energy instantaneously for every instant, wouldn't that mean that it only has rotational kinetic energy for the entirety of its roll?
 
  • #13
uSee2 said:
I do see that it is rotating, but since the axis of rotation is changing every moment it is as if it were moving translationally. But if it has only rotational kinetic energy instantaneously for every instant, wouldn't that mean that it only has rotational kinetic energy for the entirety of its roll?
You can pick an axis and factor the motion of a rigid object into translation of that axis plus rotation about that axis in many ways. There is no one "fact of the matter" on where the energy "really" is or what the axis of rotation "really" is.

But yes, you can do an energy accounting considering that the entirety of the kinetic energy of a rolling tire is always manifested in rotation about the momentary position of the contact patch if you like.
 
  • #14
uSee2 said:
I do see that it is rotating, but since the axis of rotation is changing every moment it is as if it were moving translationally. But if it has only rotational kinetic energy instantaneously for every instant, wouldn't that mean that it only has rotational kinetic energy for the entirety of its roll?
There is no rotation about the center of mass of the cylinder only.
Even in the case of the last post, the CM, as well as other points, are describing a circle.
The rotational inertia to overcome in that case should still be greater than the rotational inertia of the cylinder about its axle.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html#pax

Pure translation (skidding cylinder on frictionless horizontal surface) requires certain amount of translational kinetic energy.
Translational work = FS = maS

Pure translation (cylinder rotating around frictionless static axle) requires certain amount of rotational kinetic energy.
Rotational work = τθ = Iαθ

A combined movement requires both, which should be added up.
 
  • #15
Lnewqban said:
There is no rotation about the center of mass of the cylinder only.
Even in the case of the last post, the CM, as well as other points, are describing a circle.
The rotational inertia to overcome in that case should still be greater than the rotational inertia of the cylinder about its axle.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html#pax

Pure translation (skidding cylinder on frictionless horizontal surface) requires certain amount of translational kinetic energy.
Translational work = FS = maS

Pure translation (cylinder rotating around frictionless static axle) requires certain amount of rotational kinetic energy.
Rotational work = τθ = Iαθ

A combined movement requires both, which should be added up.
I'm still confused about the idea that there is no rotation about the center of mass. If there was no rotation about the center of mass, wouldn't there be no translational kinetic energy? But it wouldn't move in a circle like in post #11 since the axis of rotation is constantly shifting.
 
  • #16
haruspex said:
For an intuitive argument, consider a cart moving at constant velocity v. Inside the cart is a flywheel on a horizontal axis normal to the cart's motion.
To a person in the cart, it takes a certain amount of energy to get the wheel spinning at some rate. To a bystander, would you expect it to take the same amount of energy?

For an algebraic argument, consider an assemblage ##(m_i)## of particles with mass centre at ##\vec r##: ##\Sigma m_i\vec r_i=\vec r\Sigma m_i=\vec r\cdot m##, so ##\Sigma m_i\vec{ \dot r_i}=\vec {\dot r}\cdot m##.
##KE=\frac 12\Sigma m_i\vec{ \dot r_i}^2=\frac 12\Sigma m_i((\vec{ \dot r_i}-\vec {\dot r})^2+2\vec{ \dot r_i}\cdot\vec {\dot r}-\vec {\dot r}^2)##
##=\frac 12\Sigma m_i(\vec{ \dot r_i}-\vec {\dot r})^2+\frac 12\Sigma m_i
2\vec{ \dot r_i}\cdot\vec {\dot r}-\frac 12\Sigma m_i\vec {\dot r}^2##
##=\frac 12\Sigma m_i(\vec{ \dot r_i}-\vec {\dot r})^2+(m
\vec{ \dot r})\cdot\vec {\dot r}-\frac 12 m\vec {\dot r}^2
##
##=\frac 12\Sigma m_i(\vec{ \dot r_i}-\vec {\dot r})^2+\frac 12 m\vec {\dot r}^2
##
For the first part, it would take the same amount of energy, right? But isn't this different since all parts of the wheel at the edge are moving at the same tangential velocity, while when the cylinder is rolling without slipping the point of contact has a velocity of 0 while the top has twice the velocity of the center of mass?
 
  • #17
Suppose you subdivided the mass ##M## of the cylinder into elemental masses such ##M=\sum_i~dm_i##. Let point ##P## be the point of contact with the surface. This point is instantaneously at rest because there is no slipping.
Let
##\vec R=## the position of the center of mass ##O## of the cylinder relative to ##P##
##\vec r_i=## the position of mass element ##dm_i## relative to ##P##.
##\vec r'_i=## the position of mass element ##dm_i## relative to the center of mass at ##O##.
##\vec v_i=## the speed of mass element ##dm_i## relative to ##P##.
An element ##dm_i## moves with velocity ##\vec \omega\times \vec r'## relative to the center of mass ##O## which moves with velocity ##\vec V_{cm}## relative to point ##P##. Thus the velocity of the element relative to point ##P## is $$\vec v_i=\vec V_{cm}+\vec {\omega}\times \vec r_i'$$ and $$v_i^2=(\vec V_{cm}+\vec {\omega}\times \vec r'_i)\cdot(\vec V_{cm}+\vec {\omega}\times \vec r'_i)=V_{cm}^2+2\vec V_{cm}\cdot(\vec {\omega}\times \vec r'_i)+\omega^2{r'_i}^2.$$The total kinetic energy of the cylinder is $$K=\frac{1}{2}\sum_i dm_i~v_i^2=\frac{1}{2}V_{cm}^2\sum_idm_i+\sum_i dm_i\vec V_{cm}\cdot(\vec {\omega}\times \vec r'_i)+\frac{1}{2}\omega^2\sum_i dm_i{r'_i}^2$$ In the first term $$\sum_i dm_i=M\implies \frac{1}{2}V_{cm}^2\sum_idm_i=\frac{1}{2}MV_{cm}^2.$$In the second term $$\sum_i dm_i\vec V_{cm}\cdot(\vec {\omega}\times \vec {r}'_i)=\sum_i dm_i ~\vec{r}'_i\cdot(\vec V_{cm}\times\vec{\omega})=(\vec V_{cm}\times\vec{\omega})\cdot\sum_i \vec {r}'_i dm_i=0$$because by definition of the center of mass coordinates, ##\sum_i \vec {r}'_i dm_i=0.##

In the third term the definition of the moment of inertia about the center of mass is ##I_{cm}=\sum_i dm_i {r'_i}^2##.

Thus, by adding all the contributions of all elements ##dm_i## to the kinetic energy, we get $$K=\frac{1}{2}MV_{cm}^2+\frac{1}{2}I_{cm}\omega^2.$$
 
  • Like
  • Love
Likes uSee2 and Lnewqban
  • #18
uSee2 said:
For the first part, it would take the same amount of energy, right? But isn't this different since all parts of the wheel at the edge are moving at the same tangential velocity, while when the cylinder is rolling without slipping the point of contact has a velocity of 0 while the top has twice the velocity of the center of mass?
The rolling wheel is just the special case when the bottom-most part of the wheel is moving backwards relative to the centre with the same speed as the centre of the wheel is moving forward: ##v_{bot}=v_{centre}+v_{bot-centre}=v_{centre}-r\omega=0##.
 
  • #19
kuruman said:
Suppose you subdivided the mass ##M## of the cylinder into elemental masses such ##M=\sum_i~dm_i##. Let point ##P## be the point of contact with the surface. This point is instantaneously at rest because there is no slipping.
Let
##\vec R=## the position of the center of mass ##O## of the cylinder relative to ##P##
##\vec r_i=## the position of mass element ##dm_i## relative to ##P##.
##\vec r'_i=## the position of mass element ##dm_i## relative to the center of mass at ##O##.
##\vec v_i=## the speed of mass element ##dm_i## relative to ##P##.
An element ##dm_i## moves with velocity ##\vec \omega\times \vec r'## relative to the center of mass ##O## which moves with velocity ##\vec V_{cm}## relative to point ##P##. Thus the velocity of the element relative to point ##P## is $$\vec v_i=\vec V_{cm}+\vec {\omega}\times \vec r_i'$$ and $$v_i^2=(\vec V_{cm}+\vec {\omega}\times \vec r'_i)\cdot(\vec V_{cm}+\vec {\omega}\times \vec r'_i)=V_{cm}^2+2\vec V_{cm}\cdot(\vec {\omega}\times \vec r'_i)+\omega^2{r'_i}^2.$$The total kinetic energy of the cylinder is $$K=\frac{1}{2}\sum_i dm_i~v_i^2=\frac{1}{2}V_{cm}^2\sum_idm_i+\sum_i dm_i\vec V_{cm}\cdot(\vec {\omega}\times \vec r'_i)+\frac{1}{2}\omega^2\sum_i dm_i{r'_i}^2$$ In the first term $$\sum_i dm_i=M\implies \frac{1}{2}V_{cm}^2\sum_idm_i=\frac{1}{2}MV_{cm}^2.$$In the second term $$\sum_i dm_i\vec V_{cm}\cdot(\vec {\omega}\times \vec {r}'_i)=\sum_i dm_i ~\vec{r}'_i\cdot(\vec V_{cm}\times\vec{\omega})=(\vec V_{cm}\times\vec{\omega})\cdot\sum_i \vec {r}'_i dm_i=0$$because by definition of the center of mass coordinates, ##\sum_i \vec {r}'_i dm_i=0.##

In the third term the definition of the moment of inertia about the center of mass is ##I_{cm}=\sum_i dm_i {r'_i}^2##.

Thus, by adding all the contributions of all elements ##dm_i## to the kinetic energy, we get $$K=\frac{1}{2}MV_{cm}^2+\frac{1}{2}I_{cm}\omega^2.$$
This is perfect! Thank you so much, this clears up a lot and makes a lot of sense. Also, was this derivation in a textbook? I feel that my textbooks should've mentioned this derivation as this seems very critical to understanding, I might have missed something

Also off-topic, but do you have a video of you doing the lesson in this article?

https://www.physicsforums.com/insights/explaining-rolling-motion/

This article was also super helpful as well. I would just like to see the demonstrations/experiments as well.
 
  • Like
Likes Lnewqban and kuruman
  • #20
uSee2 said:
This is perfect! Thank you so much, this clears up a lot and makes a lot of sense. Also, was this derivation in a textbook? I feel that my textbooks should've mentioned this derivation as this seems very critical to understanding, I might have missed something

Also off-topic, but do you have a video of you doing the lesson in this article?

https://www.physicsforums.com/insights/explaining-rolling-motion/

This article was also super helpful as well. I would just like to see the demonstrations/experiments as well.
I am sure you will be able to find this derivation in a standard intermediate mechanics textbook. It's pretty ordinary stuff. Most introductory physics textbooks consider rotations about the center of mass and go from ##\frac{1}{2}\sum_i dm_iv_i^2## to ##\frac{1}{2}I\omega^2## through ##v_i^2 =\omega^2r_i^2## and then tack on the kinetic energy of the CoM but that is a bit of hand waving.

Sorry, there is no video of the lecture on rolling motion.
 
Last edited:
  • #21
uSee2 said:
For the first part, it would take the same amount of energy, right? But isn't this different since all parts of the wheel at the edge are moving at the same tangential velocity, while when the cylinder is rolling without slipping the point of contact has a velocity of 0 while the top has twice the velocity of the center of mass?
The point of doing a calculation is that you answer a question like "how come the equation works". And, yes, every point on a rigid body that is translating and rotating has a different velocity; and, yes, it's not obvious that this simplifies to KE of centre of mass plus rotational KE about the centre of mass. The only way to prove it is to do the calculation. That means ... mathematics!
 

Suggested for: Kinetic Energy of a Cylinder Rolling Without Slipping

Back
Top