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: $v_y=0$

$$160-20t=0$$

$$t=8$$

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