- #1

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- 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?

##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?