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]
[Specify your answer in units of 'm/s' (without quotes)]
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?
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