Answer to Question #137748 in Field Theory for alfraid

Question

A particle moves along an arc of a circle of radius R according to the law l = a sin ωt, where l is the displacement from the initial position measured along the arc, and a and ω are constants. Assuming R = 1.00 m, a = 0.80 m, and ω = 2.00 rad/s, find: (a) the magnitude of the total acceleration of the particle at the points l = 0 and l = ±a; (b) the minimum value of the total acceleration wmin and the corresponding displacement lm.

Answer 1

Solution

Acceleration can be written as

$\omega_n=\frac{a^2 \omega^2 \sin\omega t}{R}$

Case first when $l=0$ then

$\omega=0, \pi$

Magnitude of acceleration

$\omega_n=\frac{a^2 \omega^2 }{R}=\frac{(0.80)^2 \times2^2 }{1}=2.56 \frac{rad}{sec^2}$

In case of $l=-a$ or $+a$ then

$\omega =0$

Magnitude of acceleration

$\omega_n=a \omega^2 =(0.80)\times2^2$

$\omega_n=3.20\frac{rad}{sec^2}$