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Kinematics

Problem 1.40

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

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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