A problem by John Marvin Macaraeg

Level 3

A particle travels around the limaçon r = 3 + 2 c o s θ r=3+2cos\theta with a constant angular velocity of 2 r a d s e c 2\frac{rad}{sec} , that is, d θ d t = 2 \frac{d\theta}{dt}=2 . If r r is expressed in feet, find v v and a a at the instant when θ = 1 2 π \theta=\frac{1}{2}\pi .


The answer is 20.

John Marvin Macaraeg by John Marvin Macaraeg , 24, USA

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1 solution

Andre Bonhomme
Dec 18, 2013

Since d θ d t \frac{d\theta}{dt} is constant, we have d 2 θ d t 2 = 0 \frac{d^2\theta}{dt^2}=0 , and from the given equation d r d t = − 2 sin ⁡ θ d θ d t = − 4 sin ⁡ θ , \frac{dr}{dt}=-2\sin\theta\frac{d\theta}{dt}=-4\sin\theta, d 2 r d t 2 = − 4 cos ⁡ θ d θ d t = − 8 cos ⁡ θ . \frac{d^2r}{dt^2}=-4\cos\theta\frac{d\theta}{dt}=-8\cos\theta. Substituting in (1) and setting θ = 1 2 π \theta=\frac{1}{2}\pi , we obtain v r = − 4 sin ⁡ θ = − 4 , v_{r}=-4\sin\theta=-4, v θ = ( 3 + 2 cos ⁡ θ ) 2 = 6. v_{\theta}=(3+2\cos\theta)2=6. Hence, v = v r 2 + v θ 2 = ( − 4 ) 2 + ( 6 ) 2 = 7.21 f t s e c . v=\sqrt{v_{r}^{2}+v_{\theta}^{2}}=\sqrt{(-4)^2+(6)^2}=7.21 \frac{ft}{sec}. Substituting in (4) and setting θ = 1 2 π \theta=\frac{1}{2}\pi , we obtain a r = − 8 cos ⁡ θ − 4 ( 3 + 2 cos ⁡ θ ) = − 12 , a_{r}=-8\cos\theta-4(3+2\cos\theta)=-12, a θ = 4 ( − 4 sin ⁡ θ ) = − 16. a_{\theta}=4(-4\sin\theta)=-16. Hence, a = a r 2 + a θ 2 = ( − 12 ) 2 + ( − 16 ) 2 = 20 f t s e c . a=\sqrt{a_{r}^{2}+a_{\theta}^{2}}=\sqrt{(-12)^2+(-16)^2}=20 \frac{ft}{sec}.

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