Question

It has been suggested that we should use our power plants to generate energy in the off-hours (such as late at night) and store it for use during the day. One idea put forward is to store the energy in large flywheels. Suppose we want to build such a flywheel in the shape of a hollow cylinder of inner radius 0.490 m and outer radius 1.60 m , using concrete of density 2250 kg/m3.
If, for stability, such a heavy flywheel is limited to 1.40 second for each revolution and has negligible friction at its axle, what must be its length to store 2.40 MJ of energy in its rotational motion?

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? (Solve this part without reworking the entire problem!)

Answer 1

Hollow cylinder is build with concrete of density $\rho$ = 2250 kg/m³

Cylinder's inner radius r₁ = 0.490 m

Cylinder's outer radius r₂ = 1.60 m

Volume of hollow cylinder V = $\pi l(r^{2}_{2}-r^{2}_{1})$

$\rho = \frac{M}{V}$

$M = \rho V$

where M mass of the cylinder

Moment of inertia of Hollow cylinder is given by

$I = \frac{M(r^{2}_{1}+r^{2}_{2})}{2}$

$I = \frac{\rho V(r^{2}_{1}+r^{2}_{2})}{2}$ ...............................1)

Time period T = 1.40s

$T = \frac{2\pi }{\omega }$

$\omega = \frac{2\pi }{T }$ ...........................................................2)

Kinectic energy for rotation of hollow cylinder is given by

$K.E = \frac{1}{2}(I\omega ^{2})$

substituting values of equation 1 and 2 in above equation ,we get

$K.E = \frac{1}{2}((\frac{\rho V(r^{2}_{1}+r^{2}_{2})}{2})(\frac{2\pi }{T}) ^{2})$

Substituting V we get

$K.E = \frac{1}{2}((\frac{\rho (\pi l(r^{2}_{2}-r^{2}_{1}))(r^{2}_{1}+r^{2}_{2})}{2})(\frac{2\pi }{T}) ^{2})$ ...........................................3)

we want to store 2.40 MJ of K.E

putting values of K.E, r₁ , r₂ , T ,we get

$2.40\times 10^{6} = \frac{1}{2}((\frac{(2250) \pi l((1.60)^{4}-(0.490)^{4})}{2})(\frac{2\pi }{1.40}) ^{2})$

$2.40\times 10^{6} = (\frac{(2250) \pi^{3} l((1.60)^{4}-(0.490)^{4})}{(1.40)^{2}})$

2) By doubling spin rate means reducing tme period by 2, T₁ = (T/2)

By seeing equation 3, (T/2) will make 4 times factor in numerator

Thus new length

$l_{1} = \frac{l}{4}$

$l_{1} = \frac{10.38}{4}$