Answer to Question #177169 in Classical Mechanics for Joseph Se

Question

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

A particle moving in the x-y plane has a position vector given by r = 2.26t²i + 1.22t³j, where r is in inches and t is in seconds. Calculate the radius of curvature ρ of the path for the position of the particle when t = 2.0 sec. Sketch the velocity v and the curvature of the path for this particular instant.

ρ = ______ in.

A small particle P starts from point O with a negligible speed and increases its speed to a value v = square root of 2gy, where y is the vertical drop from O. When x = 57 ft, determine the n-component of acceleration of the particle.

y = (x/15)² ft

a_n = _______ ft/sec²

Answer 1

1)

$\rho=\frac{(x'^2+y'^2)^\frac 32}{|x'y''-y'x''|},$

$x'=4.52t,~x''=4.52,$

$y'=3.66t^2,~y''=7.32t,$

$\rho_{t=2}=\frac{((4.52\cdot 2)^2+(3.66\cdot2 ^2)^2)^\frac 32}{|4.52\cdot 2\cdot 7.32\cdot 2-3.66\cdot2^2\cdot 4.52|}=76.979~ft,$

2)

$a_n=v'=(\sqrt{2gy})'=(\sqrt{2g(\frac{x}{15})^2})'=\sqrt{2g}(\frac{x}{15})'=\frac{\sqrt{2g}}{15}=0.969~\frac{ft}{s^2}.$