A circular object with a mass of 23.7 kg and radius (outer) of 0.7 meters rolls

Question

A circular object with a mass of 23.7 kg and radius (outer) of 0.7 meters rolls without slipping down a 36.5 degree incline. At the top of the incline its center of mass is moving at 5.2 m/s and at the bottom of the incline it is spinning at 25.2 rad/s. If the object makes 10.8 revolutions as it rolls down the incline, what is its moment of inertia about its center in kg m2?

Note: the "c" value found for the moment of inertia may not be physical (greater than one). However, just use whatever value you get for c, even if it's not physical, to calculate your answer.

Explanation / Answer

Length of incline L = 2pi r *n = 2pi*0.7*10.8 = 47.5 m

By energy conservation, mgL sin theta + 0.5mu^2 + 0.5 i w1^2 = 0.5 mv^2 + 0.5 i w2^2

23.7*9.8*47.5* sin 36.5 degree + 0.5*23.7*5.2^2 + 0.5*i*(5.2/0.7)^2 =  0.5*23.7*(25.2*0.7)^2 + 0.5*i*25.2^2

  6562 + 320 + 27.6i = 3687 + 317.5 i

i = [ 6562 + 320-3687]/[317.5-27.6]

= 11.0 kgm^2 answer

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