# Thread: Rotation Bodies - rolling cylinders

1. ## Rotation Bodies - rolling cylinders

Every year at Coopers Hill in Gloucestershire (UK) there is a cheese rolling festival where people race down the hill after a rolling disc of Gloucestershire cheese, the winner getting to keep the cheese. (No, I'm not kidding look here. It seems there will even be an event in Whistler, BC next year if you are really interested.)

Given that the moment of inertia of a disc of mass m and radius r about its axis of symmetry is , what is the speed of the cheese's centre of mass after it has descended a vertical distance of 17.0 m from the top of the hill? Assume that the cheese rolls without slipping.
[Acceleration due to gravity, g=9.81 ms-2]

2. Originally Posted by jddery
Every year at Coopers Hill in Gloucestershire (UK) there is a cheese rolling festival where people race down the hill after a rolling disc of Gloucestershire cheese, the winner getting to keep the cheese. (No, I'm not kidding look here. It seems there will even be an event in Whistler, BC next year if you are really interested.)

Given that the moment of inertia of a disc of mass m and radius r about its axis of symmetry is , what is the speed of the cheese's centre of mass after it has descended a vertical distance of 17.0 m from the top of the hill? Assume that the cheese rolls without slipping.
[Acceleration due to gravity, g=9.81 ms-2]

use energy principles ...

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2$

remember that if the disk rolls w/o slipping, $v = r\omega$

3. hey how would we get the answer of this question with out knowing the radius?

4. Originally Posted by jddery
hey how would we get the answer of this question with out knowing the radius?
the radius is not necessary ... do the algebra and see why for yourself.

5. I based this question off of the time dilation

and since they are saying its at rest
would I assume that v^2/c^2 is zero?
and simply the equation is 2.2 micro seconds/ 1?

And

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