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3 years ago

11

A Micro –Hydro turbine generator rotor is accelerating uniformly from an angular velocity of 610 rpm to its operating angular ve

locity of 837 rpm. The radius of the rotor is 0.62 m and its rotational acceleration is 5.9 rad/s2 . What is the rotor’s angular velocity (in rad/s) after 3.2 s?

2 answers:

solniwko [45]3 years ago

4 0

Answer:

The rotor's angular velocity is 82.73rad/s

Explanation:

It is a curvilinear movement of a constant radius. If there is uniform angular acceleration, then it is a circular motion with constant acceleration, whose equations are analogous to that of the translational motion.

Calculating the initial velocity of the rotor, V1 in rad/s

V1 = 610rev/minute × 6.28 × 1miute/60secs

V1 = 63.85rad/s

Using kinematic equation to calculate the final velocity of the rotor

Given:

Angular acceleration = 5.9rad/s^2

Time,t= 3.2seconds

V2 = V1 + a × t

V2 = 63.85 + (5.9)× (3.2)

V2 = 63.85 + 18.88

V2 = 82.73rads/s

Zarrin [17]3 years ago

3 0

Answer:

Explanation:

Given:

wo = 610 rpm

To rad/s,

= 610 rpm ×2pi rad/60 s

= 63.88 rad/s

wf = 837 rpm

= 87.65 rad/s

ao = 5.9 rad/s^2

Using equations of circular motion,

wf = wo + aot

87.65 = 63.88 + 5.9×t

t = 4.025 s

Using the same equation,

At t = 3.2s (first 3.2 s of the motion),

wf = 63.88 + 5.9 × 3.2

= 82.76 rad/s

= 82.8 rad/s

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