A machine part is initially rotating at 0.500 rad/s. Its rotation speeds up with constant angular acceleration 2.50 rad/s2. Through what angle has the machine part rotated when its angular speed equals 3.25 rad/s

Given:

Initial angular velocity, wi = 0.5 rad/s

Final angular velocity, wf = 3.25 rad/s

Angular acceleration, ao = 2.5 rad/s^2

Using angular equation of motion,

wf^2 = wi^2 + 2θ * ao

Inputting values,

3.25^2 = 0.5^2 + 2θ * 2.5

θ = 10.3125/5

= 2.0625 rad

But 1 rev = 2pi rad = 360°

1 rev/2pi rad * 2.0625 rad * 360°/1 rev

= 118.17°

= 118.2°

The angle through which the machine part has rotated is 2.0625 rad

Explanation:

Given;

initial angular speed, ωi = 0.500 rad/s

angular acceleration, α = 2.50 rad/s².

Final angular speed, ωf = 3.25 rad/s

Apply kinematic equation;

ωf² = ωi² + 2αθ

where;

θ is the angle through which the machine part has rotated

3.25² = 0.5² + (2 x 2.5) θ

10.5625 = 0.25 + 5θ

5θ = 10.5625 - 0.25

5θ = 10.3125

θ = 10.3125/5

θ = 2.0625 rad

Therefore, the angle through which the machine part has rotated is 2.0625 rad