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xz_007 [3.2K]
3 years ago
13

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/s 2 .

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

Engineering
1 answer:
Neporo4naja [7]3 years ago
3 0

Answer:

Velocity components

V_r = -16.28 m/s

V_z = -22.8 m/s

V_q = 0 m/s

For Acceleration components;

a_r = -4.07m/s^2

a_z = -5.70m/s^2

a_q = 0m/s^2

Explanation:

We are given:

Speed v_o = 28 m/s

Acceleration a_o= 7 m/s^2

We first need to find the radial position r of washer in x-y plane.

Therefore

r = \sqrt{300^2 + 400^2}

r = 500 mm

To find length along direction OA we have:

L = \sqrt{500^2 + 700^2}L = 860 mm

Therefore, the radial and vertical components of velocity will be given as:

V_r = V_o*cos(Q)

V_z = V_o*sin(Q)

Where Q is the angle between OA and vector r.

Therefore,

V_r = 28 * \frac{r}{L} = > 28 * \frac{500}{860}

V_r = -16.28 m/s

• V_z = 28 * \frac{700}{860} = -22.8

• V_q = 0 m/s

The radial and vertical components of acceleration will be:

a_r = a_o*cos(Q)

a_z = a_o*sin(Q)

Therefore we have:

• a_r = 7* \frac{500}{860} = -4.07m/s^2

• a_z = 7 * \frac{700}{860} = -5.70 m/s^2

• a_q = 0 m/s^2

Note : image is missing, so I attached it

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