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Kinematics

Problem 1.40

A particle moves along an arc of a circle of radius RR according to the law l=asin⁡ωt,l=a \sin \omega t, where ll is the displacement from the initial position measured along the arc, and aa and ω\omega are constants. Assuming R=1.00 m,a=0.80 m,R=1.00 \mathrm{~m}, a=0.80 \mathrm{~m}, and ω=2.00rad/s,\omega=2.00 \mathrm{rad} / \mathrm{s}, find: (a) the magnitude of the total acceleration of the particle at the points l=0l=0 and l=±al=\pm a (b) the minimum value of the total acceleration wmin⁡w_{\min } and the corresponding displacement lml_{m}.

Reveal Answer
 (a) w0=a2ω2/R=2.6 m/s2,wa=aω2=3.2 m/s2 ;  (b) wmin⁡==aω2V1−(R/2a)2‾=2.5 m/s2lm=±a1−R2/2a2=±0.37 m\text { (a) } w_{0}=a^{2} \omega^{2} / R=2.6 \mathrm{~m} / \mathrm{s}^{2}, w_{a}=a \omega^{2}=3.2 \mathrm{~m} / \mathrm{s}^{2} \text { ; } \text { (b) } w_{\min }= =a \omega^{2} V \overline{1-(R / 2 a)^{2}}=2.5 \mathrm{~m} / \mathrm{s}^{2} l_{m}=\pm a \sqrt{1-R^{2} / 2 a^{2}}=\pm 0.37 \mathrm{~m}
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