Calculate the translational speed of a cylinder

AI Thread Summary
To calculate the translational speed of a cylinder rolling down an 11 m incline, one must consider both potential energy and kinetic energy. The potential energy at the top converts into kinetic energy at the bottom, which includes both translational and rotational components. The relevant equation combines these energies: mgh = 1/2 mv^2 + 1/2 Iω^2, where I is the moment of inertia and ω is the angular velocity related to translational speed. The moment of inertia for a cylinder is 2/5Mr^2, and substituting ω = v/r allows for solving the equation. Despite attempts, the correct speed calculation remains elusive, with a reported incorrect result of 12.4 m/s.
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Homework Statement


Calculate the translational speed of a cylinder when it reaches the foot of an incline 11 m high. Assume it starts from rest and rolls without slipping.


Homework Equations





The Attempt at a Solution


How do you find the speed when you only have the height of the incline?
 
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Bones said:

Homework Statement


Calculate the translational speed of a cylinder when it reaches the foot of an incline 11 m high. Assume it starts from rest and rolls without slipping.

How do you find the speed when you only have the height of the incline?

The height gives you potential energy.

You get KE at the bottom, but be careful you also have rotational KE in the cylinder since it rolled without slipping.
 
I am not sure what the equation is. The only one I found was v=sqrt 10/7gH which was not right.
 
Bones said:
I am not sure what the equation is. The only one I found was v=sqrt 10/7gH which was not right.

Show your equations and perhaps we can see where you are going wrong.
 
There's this one: 1/2Mv^2+1/2Icm(omega)^2+Mgy
But I am not sure where to get all this information from just having the height of the incline.
 
Bones said:
There's this one: 1/2Mv^2+1/2Icm(omega)^2+Mgy
But I am not sure where to get all this information from just having the height of the incline.

Ok. You have your potential energy. And at the bottom what's happened to that PE? It's become KE for this problem. So whjat you have then is

m*g*h = Σ KE = 1/2 m*v2 + 1/2*I*ω2

But what is ω ? Happily ω = v/r

Why happily? Because the moment of inertia also will have a term that relates to r.

Why don't you look up the moment of a cylinder and solve the rest of the problem?
 
The moment of inertial for a cylinder is 2/5Mr^2, but I am still not sure how to solve this. I keep getting 12.4m/s which is not correct.
 
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