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**Angular velocity:**

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If an object is rotated around its axis, it is able to have a certain angle velocity. The angular speed is expressed as the change in speed of the object’s angle of displacement. The angular velocity determines the angular momentum of the object.

The angular momentum can be defined as the product of the moment of inertia, and the velocity of angular rotation.

**Third Kinematic Equation of Motion:**

The third kinematic equation to determine the object’s (moves in the circular motion) speed of angular motion (at various locations) and angular acceleration and angular displacement as illustrated below.

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Concepts and ReasonThis problem is based upon the concepts of angular velocity, and angular acceleration.

The first step is to calculate the angular speed by using the relationship between angular displacement and angular velocity. The appropriate kinematic expression can be used to calculate the angular acceleration.

FundamentalsThe expression gives the angular velocity of a body.

[katex]\omega = \frac{\theta }{t}[/katex]

Here, [katex]\omega [/katex] is angular velocity, θ is angle swept by body and, t is time interval

An object’s angular displacement is described as

[katex]\theta = n2\pi[/katex]

nhere is the number of rotations.The expression “Angular Acceleration of a Body” is used to describe the acceleration.

[katex]\alpha = \frac{{\partial \omega }}{{\partial t}}[/katex]

Here, [katex]\omega [/katex] is angular velocity, α is angular acceleration and t is time interval

This is the kinematic expression of angular acceleration.

[katex]{\omega _{\rm{f}}}^2 = {\omega _{\rm{i}}}^2 + 2\alpha (\theta )[/katex]

The expression gives the angular velocity of a body.

[katex]\omega = \frac{\theta }{t}[/katex]

Here, [katex]\omega [/katex] is angular velocity, θ is angle swept by body and t is time interval.

The object’s angular displacement is.

[katex]\theta = n2\pi[/katex]

nhere is the number of rotations.Substitute 20 in equation [katex]\theta = n2\pi[/katex]

[katex]\theta = (20)2\pi rad[/katex] [katex]= 40\pi rad[/katex]

Substitute [katex]40\pi rad[/katex] for [katex]\theta [/katex] and [katex]5s[/katex] for [katex]t[/katex] in the above equation.

[katex]\omega = \frac{40\pi rad}{5s}[/katex] [katex] = 8\pi rad/s[/katex]

The kinematic expression of angular acceleration can be found at

[katex]{\omega _{\rm{f}}}^2 = {\omega _{\rm{i}}}^2 + 2\alpha (\theta )[/katex]

Substitute 6 in equation [katex]\theta = n2\pi[/katex] for n

[katex]\theta = (6)2\pi rad [/katex] [katex]= 12\pi rad [/katex]

Substitute 0 rad/s for [katex]\omega{_f}[/katex], [katex]8\pi rad/s[/katex] for [katex]\omega{_i}[/katex] and [katex]12\pi rad[/katex] for [katex]\theta [/katex] in equation

[katex]\omega{_f}{^2} = \omega{_i}{^2} + 2\alpha(\theta)[/katex]

[katex]0rad/s = (8\pi rad/s){^2} + 2 \alpha(12\pi rad)[/katex] [katex]\alpha = -8.38rad/s{^2}[/katex]

Answer:The [katex]8.38rad/s{^2} [/katex] angular acceleration for a salad spinner