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Calculus IIChapter 6: 3-Dimensional SpaceSection 6-11

Velocity and Acceleration — Question 3

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Question 3

A particle moves on the circle r→(t)=⟨Rcos(ωt),Rsin(ωt)⟩\vec r(t)=\left\langle R\cos(\omega t),R\sin(\omega t)\right\rangle, where R,ω>0R,\omega>0. Find velocity, acceleration, speed, and show that acceleration points toward the center with magnitude v2/Rv^2/R.

Original worksheet page 1: question and worked solution for 6-11-003
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Question 3 – Solution

Strategy Differentiate twice and compare acceleration with the position vector.

See the diagram in the original worksheet below.

Derivatives v→=Rω⟨−sinωt,cosωt⟩,a→=−Rω2⟨cosωt,sinωt⟩=−ω2r→.\vec v=R\omega\left\langle -\sin\omega t,\cos\omega t\right\rangle,\qquad \vec a=-R\omega^2\left\langle \cos\omega t,\sin\omega t\right\rangle=-\omega^2\vec r. Thus v=Rω\boxed{v=R\omega} and acceleration is radially inward.

Magnitude ∥a→∥=Rω2=(Rω)2R=v2R.\|\vec a\|=R\omega^2=\frac{(R\omega)^2}{R} =\boxed{\frac{v^2}{R}}. Velocity is tangent to the circle and perpendicular to the centripetal acceleration.

Original worksheet page 2: question and worked solution for 6-11-003

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