Question

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

Answer 1

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

Answer 2

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