F=ma 2015 solutions

problem number	solution
P01	P01 solution
P02	P02 solution
P03	P03 solution
P04	P04 solution
P05	P05 solution
P06	P06 solution
P07	P07 solution
P08	P08 solution
P09	P09 solution
P10	P10 solution
P11	P11 solution
P12	P12 solution
P13	P13 solution
P14	P14 solution
P15	P15 solution
P16	P16 solution
P17	P17 solution
P18	P18 solution
P19	P19 solution
P20	P20 solution
P21	P21 solution
P22	P22 solution
P23	P23 solution
P24	P24 solution
P25	P25 solution

P17

A flywheel can rotate in order to store kinetic energy. The flywheel is a uniform disk made of a material with a density \(\rho\) and tensile strength \(\sigma\) (measured in Pascals), a radius \(r\), and a thickness \(h\). The flywheel is rotating at the maximum possible angular velocity so that it does not break.

Which of the following expressions correctly gives the maximum kinetic energy per kilogram that can be stored in the flywheel? Assume that \(\alpha\) is a dimensionless constant.

(A) \(\alpha \sqrt{\rho \sigma / r}\)

(B) \(\alpha h \sqrt{\rho \sigma / r}\)

(D) \(\alpha (h/r)(\sigma/\rho)\)

(E) \(\alpha \sigma / \rho\)

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P17 solution

Let \(\lambda\) be the maximum kinetic energy per kilogram that can be stored in the flywheel. If we double the thickness \(h\) of the flywheel, that is equivalent to stacking two flywheels on top of each other. Since each has \(\lambda\) energy per kilogram, the combination also has \(\lambda\) energy per kilogram. Hence, \(\lambda\) is independent of \(h\).

We can use dimensional analysis to find \(\lambda(r, \rho, \sigma)\). First, the dimensions of each variable are:

\[ [\lambda] = \left[ \frac{E}{M} \right] = \frac{ML^2/T^2}{M} = \frac{L^2}{T^2}\]

\[ [r] = L\]

\[ [\rho] = \left[ \frac{M}{V} \right] = \frac{M}{L^3}\]

\[ [\sigma] = \left[ \frac{F}{A} \right] = \frac{ML/T^2}{L^2} = \frac{M}{LT^2}\]

We need 1 power of \(\sigma\) to have the correct number of powers of \(T\):

\[ [\sigma^1] = \frac{M}{LT^2}\]

We need \(-1\) powers of \(\rho\) to cancel out the mass dependence:

\[ \left[ \frac{\sigma}{\rho} \right] = \frac{M}{LT^2} \frac{L^3}{M} = \frac{L^2}{T^2}\]

This has the same dimensions as \(\lambda\) so there is no \(r\) dependence. Thus,

\[ \lambda = \alpha \frac{\sigma}{\rho}\]

where \(\alpha\) is a dimensionless constant, so the answer is \(\boxed{E}\).