Answer to Question #177165 in Classical Mechanics for Joseph Se

Question

The final answer should be in the format of "tolerance of ± in the third significant digit."

A car is traveling around a circular track of a 730-ft radius. If the magnitude of its total acceleration is 11 ft/sec² at the instant when its speed is 50 mi/hr, determine the rate a_t which the car is changing its speed.

a_t = ± _________ ft/sec²

Write the vector expression for the acceleration a of the mass center G of the simple pendulum in both n-t and x-y coordinates for the instant when θ = 33° if θ˙ = 2.41 rad/sec and θ¨ = 4.445 rad/sec².

a = (_____ e_n + _____ e_t) ft/sec²

a = (_____ i + ______ j) ft/sec²

Answer 1

1)

$a_t=a-\frac{v^2}{r}=10.997~\frac{ft}{s^2},$

2)

$a_n=\frac{v^2}{r}=r\theta'^2=v\theta',$

$a_t=v'=r\theta'',$

$a_n=1.467\cdot 2.41=3.535e_n,$

$a_t=730\cdot 4.445=3244.850e_t,$

$i=-a_tsin\theta-a_ncos\theta=-1770.237,$

$j=a_t cos\theta-a_n sin \theta=2719.435.$