Answer to Question #171430 in Mechanical Engineering for edjee

Question

A Stone has an initial velocity of 200 ft per see up to the right at a slope of 4 to 3. The components of acceleration are constant at a_x=-12 ft per second² and a_y=-20 ft per second². Compute the radius of curvature at the start and at the top of the path

Answer 1

$a_x=-12, a_y=-20$

$\vec r''(t)=-12\vec i-20\vec i$

$\vec v_0=200\cdot\dfrac{3}{\sqrt{3^2+4^2}}\vec i+200\cdot\dfrac{4}{\sqrt{3^2+4^2}}\vec j$

$\vec v_0=120\vec i+160\vec j$

$\vec r'(t)=(120-12t)\vec i+(160-20t)\vec j$

$\vec r'(t)\times \vec r''(t)=\begin{vmatrix} \vec i & \vec j & \vec k \\ 120-12t & 160-20t & 0 \\ -12t & -20t & 0 \end{vmatrix}$

$=\vec k((120-12t)(-20t)-(160-20t)(-12t))$

$=-480t\vec k$

$||\vec r'(t)\times \vec r''(t)||=480|t|$

$||\vec r'(t)||=\sqrt{(120-12t)^2+(160-20t)^2}$

$=4\sqrt{34t^2-580t+2500}$

$K(t)=\dfrac{||\vec r'(t)\times \vec r''(t)||}{||\vec r'(t)||^3}$

$=\dfrac{480t}{64(34t^2-580t+2500)^{3/2}}$

$=\dfrac{15t}{2(34t^2-580t+2500)^{3/2}}$

$R(t)=\dfrac{2(34t^2-580t+2500)^{3/2}}{15t}$

At the start of the path: $t=0$

$R(0)=\infin$

At the top of the path: $v_y=0$

$160-20t=0$

$t=8$

$R(8)=\dfrac{2(34(8)^2-580(8)+2500)^{3/2}}{15(8)}=3.6 (ft)$