Question

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

Answer 1

Answer:

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

Answer 2

The angle this machine part is rotated is equal to 2.0625‬ radian.

Given the following data:

• Initial angular speed = 0.500 rad/s.

• Final angular speed = 3.25 rad/s.

• Angular acceleration = 2.50 $rad/s^2$

To determine the angle this machine part is rotated, we would apply the third equation of rotational motion:

Mathematically, the third equation of rotational motion is given by this formula:

$\omega_f^2 = \omega_i^2 + 2\alpha \theta$

Where:

• $\omega_i$ is the initial angular speed.

• $\omega_f$ is the final angular speed.

• $\alpha$ is the angular acceleration.

• $\theta$ is the angle.

Substituting the given parameters into the formula, we have;

$3.25^2 = 0.500^2 + 2(2.50) \theta\\\\10.5625=0.25+5\theta\\\\5\theta=10.5625-0.25\\\\5\theta=10.3125\\\\\theta=\frac{10.3125}{5}$

Angle = 2.0625‬ rad.

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