Ideal Fluids in Motion — Question 8

Section 14-7 · Physics pages 2186–2187

Question 8 A bend requires acceleration even at constant speed A parcel follows a smooth circular arc of radius R = 0.50 m at constant speed v = 3.0 m/s. The flow is steady, and the arc turns the parcel through 90◦. The fluid density is 1000 kg/m3. Ignore gravity in the plane of the bend.⃗ a center Quarter-circle parcel path; acceleration points toward its center. (a) Find its acceleration magnitude and the time to traverse the arc. (b) What net force per unit volume is needed, and in which direction? Explain why equal inlet and outlet speeds do not imply zero net force on a parcel.
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Question 8 - Solution A bend requires acceleration even at constant speed⃗ a center (a) Derivation and calculation an = v2/R = 3.02/.500 = 18.0 m/s2 inward. s = R∆θ = Rπ/2 ⇒t = s/v = π(.500) 2(3.0) = .261799 s . (b) Result and interpretation fnet = ρan = (1000)(18.0) = 18000 N/m3 inward. |vf| = |vi|, vf̸ = vi ⇒∆p̸ = 0. A transverse pressure gradient can supply the centripetal force. Equal speeds constrain kinetic energy, not momentum direction. Physical check The force is perpendicular to the instantaneous velocity and can turn the parcel without changing its speed.

Question 8

A bend requires acceleration even at constant speed

A parcel follows a smooth circular arc of radius R=0.50mR=0.50\,\mathrm{m} at constant speed v=3.0m/sv=3.0\,\mathrm{m/s}. The flow is steady, and the arc turns the parcel through 90∘90^\circ. The fluid density is 1000kg/m31000\,\mathrm{kg/m^3}. Ignore gravity in the plane of the bend.

Diagram shown in the original SVG worksheet.

  1. Find its acceleration magnitude and the time to traverse the arc.

  2. What net force per unit volume is needed, and in which direction? Explain why equal inlet and outlet speeds do not imply zero net force on a parcel.

Question 8 - Solution

A bend requires acceleration even at constant speed

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

an=v2/R=3.02/.500=18.0m/s2inward.\boxed{a_n=v^2/R=3.0^2/.500=18.0\,\mathrm{m/s^2}}\quad\text{inward}.s=RΔθ=Rπ/2⇒t=s/v=π(.500)2(3.0)=.261799s.s=R\Delta\theta=R\pi/2\Rightarrow\boxed{t=s/v=\frac{\pi(.500)}{2(3.0)}=.261799\,\mathrm s}.

(b) Result and interpretation

fnet=ρan=(1000)(18.0)=18000N/m3inward.\boxed{f_{\mathrm{net}}=\rho a_n=(1000)(18.0)=18000\,\mathrm{N/m^3}}\quad\text{inward}.|𝒗f|=|𝒗i|,𝒗f≠𝒗i⇒Δ𝒑≠0.|\mathbf v_f|=|\mathbf v_i|,\qquad\mathbf v_f\ne\mathbf v_i\Rightarrow\Delta\mathbf p\ne0.A transverse pressure gradient can supply the centripetal force. Equal speeds constrain kinetic energy, not momentum direction.

Physical check

The force is perpendicular to the instantaneous velocity and can turn the parcel without changing its speed.