Suppose we want to build a flywheel in the shape of a hollow cylinder of inner r
ID: 2156705 • Letter: S
Question
Suppose we want to build a flywheel in the shape of a hollow cylinder of inner radius 0.500 m and outer radius of 1.50m, using concrete of density 2.20x10^3kg/m^3. a) if, for stability, such a heavy flywheel is limited to 1.75 second for each revolution and has negligible friction at its axle, what must be its length to store 2.5MJ of energy in its rotational motion? b) Suppose that by strengthening the frame you could safely double the flywheel's rate of spin. What length of flywheel would you need in that case? ... Please please help...be so kind to show me steps so I can understand how u get to the solution. thank you!Explanation / Answer
Rotational kinetic energy is ½I2 and I for a cylinder like this is (m/2)(r12 + r22).
We know m = L(r22 - r12) so the full equation, which we need to solve for L, is
K = ¼L(r22 - r12)(r12 + r22)2 so
L = 4K/(r22 - r12)(r12 + r22)2
= (1.0 * 107 J)/((2.25 m2 - 0.25 m2)(2.25 m2 + 0.25 m2)()(2.2 * 103 kg/m3)(2/1.75 s)2)
= 22.4 m and to do the second part, just note that kinetic energy is proportional to the square of , so when you double that you need only one fourth as much mass, so divide that length by 4.