The small washer is sliding down the cord OA. When it is at the midpoint, its speed is 28 m>s and its acceleration is 7 m>s2. Express the velocity and acceleration of the washer at this point in terms of its cylindrical components.

Velocity components

V_r = -16.28 m/s

V_z = -22.8 m/s

V_q = 0 m/s

Acceleration components

a_r = -4.07 m/s^2

a_z = -5.70 m/s^2

a_q = 0 m/s^2

Explanation:

Given:

- The diagram missing is attached.

- The speed of the washer v_o = 28 m/s

- The acceleration of washer a_o = 7 m/s^2

Find:

Express the velocity and acceleration of the washer at this point in terms of its cylindrical components.

Solution:

- From figure, compute the radial position r of washer in x-y plane:

                            r = sqrt ( 300^2 + 400^2 )

                            r = 500 mm

- From figure, compute the Length L of along the OA direction:

                            L = sqrt ( 500^2 + 700^2 )

                            r = 860 mm

- The radial and vertical components of velocity are:

                            V_r = V_o*cos(Q)

                            V_z = V_o*sin(Q)

Where, Q is the angle between OA and vector r.

                            V_r = 28*r/L = 28*500 / 860

                            V_r = -16.28 m/s

                            V_z = 28*700/L = 28*700 / 860

                            V_z = -22.8 m/s

                            V_q = 0 m/s

- The radial and vertical components of acceleration are:

                            a_r = a_o*cos(Q)

                            a_z = a_o*sin(Q)

Where, Q is the angle between OA and vector r.

                            a_r = 7*r/L = 7*500 / 860

                            a_r = -4.07 m/s^2

                            a_z = 7*700/L = 7*700 / 860

                            a_z = -5.70 m/s^2

                            a_q = 0 m/s^2

i believe the answer is a

1/0.14nm = 10^10/1.4 a lot