The final answer should be in the format of "tolerance of ± in the third significant digit."
1.The piston of the hydraulic cylinder gives pin A a constant velocity v = 3.6 ft/sec in the direction shown for an interval of its motion. For the instant when θ = 67°, determine r˙, r¨, θ˙, and θ¨ where r = line OA. The distance of the line A to line B is 6.8”.
r˙ = _ ft/sec
r¨ = _ ft/sec2
θ˙ = _ rad/sec
θ¨ = _ rad/sec2
r˙=vr=vcosθ=14.0663 fts,\dot{r}=v_r=v cos \theta=14.0663~\frac {ft}{s},r˙=vr=vcosθ=14.0663 sft,
vθ=−vsinθ=−3.3138 fts,v_{\theta}=-v sin \theta=-3.3138~\frac {ft}{s},vθ=−vsinθ=−3.3138 sft,
vθ=rθ˙ ⟹ θ˙=vθr=−0.4873 rads,v_{\theta}=r\dot{\theta}\implies \dot{\theta}=\frac{v_{\theta}}{r}=-0.4873~\frac {rad}{s},vθ=rθ˙⟹θ˙=rvθ=−0.4873 srad,
ar=acosθ=0.3907a fts2,a_r=acos \theta=0.3907a~\frac{ft}{s^2},ar=acosθ=0.3907a s2ft,
aθ=−asinθ=−0.9205 fts2,a_{\theta}=-asin \theta=-0.9205~\frac{ft}{s^2},aθ=−asinθ=−0.9205 s2ft,
ar=r¨−rθ˙2, ⟹ r¨=ar+rθ˙2=0.3907a+1.6147 fts2,a_r=\ddot{r}-r\dot{\theta}^2,\implies \ddot{r}=a_r+r\dot{\theta}^2=0.3907a+1.6147~\frac{ft}{s^2},ar=r¨−rθ˙2,⟹r¨=ar+rθ˙2=0.3907a+1.6147 s2ft,
aθ=rθ¨+2r˙θ˙ ⟹ θ¨=aθ−2r˙θ˙r=1.8807 rads2.a_{\theta}=r\ddot{\theta}+2\dot{r}\dot{\theta}\implies \ddot{\theta}=\frac{a_{\theta}-2\dot{r}\dot{\theta}}{r}=1.8807~\frac{rad}{s^2}.aθ=rθ¨+2r˙θ˙⟹θ¨=raθ−2r˙θ˙=1.8807 s2rad.
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