Answer:
Velocity components
[tex] V_r = -16.28 m/s [/tex]
[tex] V_z = -22.8 m/s [/tex]
[tex] V_q = 0 m/s [/tex]
For Acceleration components;
[tex] a_r = -4.07m/s^2 [/tex]
[tex] a_z = -5.70m/s^2 [/tex]
[tex] a_q = 0m/s^2 [/tex]
Explanation:
We are given:
[tex] Speed v_o = 28 m/s [/tex]
[tex] Acceleration a_o= 7 m/s^2[/tex]
We first need to find the radial position r of washer in x-y plane.
Therefore
[tex] r = \sqrt{300^2 + 400^2} [/tex]
r = 500 mm
To find length along direction OA we have:
[tex] L = \sqrt{500^2 + 700^2}
L = 860 mm [/tex]
Therefore, the radial and vertical components of velocity will be given as:
[tex] V_r = V_o*cos(Q) [/tex]
[tex] V_z = V_o*sin(Q) [/tex]
Where Q is the angle between OA and vector r.
Therefore,
[tex] V_r = 28 * \frac{r}{L} = > 28 * \frac{500}{860} [/tex]
[tex] V_r = -16.28 m/s [/tex]
• [tex] V_z = 28 * \frac{700}{860} = -22.8 [/tex]
• [tex] V_q = 0 m/s [/tex]
The radial and vertical components of acceleration will be:
[tex] a_r = a_o*cos(Q) [/tex]
[tex] a_z = a_o*sin(Q) [/tex]
Therefore we have:
• [tex] a_r = 7* \frac{500}{860} = -4.07m/s^2 [/tex]
• [tex] a_z = 7 * \frac{700}{860} = -5.70 m/s^2 [/tex]
• [tex] a_q = 0 m/s^2 [/tex]
Note : image is missing, so I attached it
