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23 December, 10:44

The rotating nozzle sprays a large circular area and turns with the constant angular rate theta overscript dot endscripts = 2.2 rad/s. particles of water move along the tube at the constant rate l overscript dot endscripts = 1.8 m/s relative to the tube. find the magnitudes of the velocity v and acceleration a of a water particle p when the distance l = 0.85 m if the angle beta = 64°.

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  1. 23 December, 14:19
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    Refer to the figure shown below.

    ω = 2.2 rad/s, the constant angular velocity of rotation

    u = 1.8 m/s, the radial velocity of a water particle, P

    v = tangential velocity of P at radius r = 0.85 m from the sprinkler

    P is at angle 64° from the horizontal axis.

    Calculate the tangential velocity of P.

    v = rω = (0.85 m) * (2.2 rad/s) = 1.87 m/s

    Calculate the centripetal acceleration.

    a = v²/r = (1.87 m/s) ² / (0.85 m) = 4.114 m/s².

    This acceleration is directed toward the sprinkler.

    The magnitude of the resultant particle velocity is

    V = √ (v² + u²)

    = √ (1.87² + 1.8²)

    = 2.596 m/s

    Because tan⁻¹ (v/u) = tan⁻¹ (187/1.8) = 46°, the resultant particle velocity is

    64 + 46 = 110° measured counterclocwise from the positive horizontal x-axis.

    Answer:

    The velocity of particle P is 2.6 m/s at an angle of 110° measured counterclockwise from the positive x-axis.

    The acceleration of particle P is 4.1 m/s², and it is directed toward the sprinkler.
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