Question

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°.

Answer 1

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.