2014: Problems 12-13

Topic: Dynamics
Concepts: Air resistance


Solution:

Since the helicopter quickly reaches terminal velocity vtv_t, from kinematics, the time TT of flight is given by

T=hvth1T=\frac{h}{v_t} \propto h^1

Thus, α=1\alpha=1 so the answer is E.

Topic: Other
Concepts: Dimensional analysis


Solution:

From the previous problem, we found T=khrβρδWωT=khr^{\beta}\rho^{\delta}W^{\omega}. Taking the dimensions of both sides,

[T]=[h][r]β[ρ]δ[W]ω[T]=[h][r]^{\beta}[\rho]^{\delta}[W]^{\omega}

Using the fact that [ρ]=M/L3[\rho]=M/L^3 and [W]=ML/T2[W]=ML/T^2,

T=L1Lβ(ML3)δ(MLT2)ωT=L^1L^{\beta}\left(\frac{M}{L^3}\right)^{\delta}\left(\frac{ML}{T^2}\right)^{\omega}

Counting powers of TT:

1=2ω1=-2\omega
ω=1/2\omega=-1/2

Counting powers of MM:

0=δ+ω0=\delta+\omega
δ=ω=1/2\delta=-\omega=1/2

Counting powers of LL:

0=1+β3δ+ω0=1+\beta-3\delta+\omega
β=3δω1=4ω1=21=1\beta=3\delta-\omega-1=-4\omega-1=2-1=1

Since β=1\beta=1, the answer is D.

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