Answer:
Acceleration will be [tex]\alpha =-1.50rad/sec^2[/tex]
Explanation:
We have given initial angular velocity [tex]\omega _i=37rpm[/tex]
In radian/sec initial angular velocity will be [tex]\omega _i=\frac{2\times \pi 37}{60}=3.873rad/sec[/tex]
Angular velocity after 1.6 sec is 14 rpm
So final angular velocity [tex]\omega _f=\frac{2\times \pi\times 14}{60}=1.465rad/sec[/tex]
Time t = 1.6 sec
We have to find the angular angular acceleration
From first equation of motion we know that
[tex]\omega _f=\omega _+\alpha t[/tex]
[tex]1.465=3.873+\alpha \times 1.6[/tex]
[tex]\alpha =-1.50rad/sec^2[/tex] here negative sign indicates that motion is deaccelerative in nature
The average angular acceleration recorded is -1.5 rad/s²
To solve this problem, we make use of the formula for calculating angular acceleration, with respect to angular speed and time.
Angular Acceleration: This can be defined as the rate of change of angular speed.
Formula
∝ = (ω₂-ω₁)/t........................ Equation 1
where:
From the question,
Given:
Substite these values into equation 1
∝ = (1.47-3.87)/1.6
∝ = -2.4/1.6
∝ = -1.5 rad/s²
Hence, The average angular acceleration recorded is -1.5 rad/s²
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