Answer:
The magnitude of angular acceleration of the motor is [tex]20\ rad/s^2[/tex].
Explanation:
Given that,
Initial angular velocity, [tex]\omega_i=90\ rad/s[/tex]
Final angular velocity, [tex]\omega_f=30\ rad/s[/tex]
Angular displacement, [tex]\theta=180\ rad[/tex]
Let [tex]\alpha[/tex] is the magnitude of the angular acceleration of the motor. It can be calculated using third equation of rotational kinematics as :
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{\omega_f^2-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{30^2-90^2}{2\times 180}\\\\\alpha =-20\ rad/s^2[/tex]
So, the magnitude of angular acceleration of the motor is [tex]20\ rad/s^2[/tex].
The magnitude of the angular acceleration of the motor is [tex]20 \;\rm rad/s^{2}[/tex].
Given data:
The initial angular speed of the motor is, [tex]\omega_{1} = 90 \;\rm rad/s[/tex].
The final angular speed of the motor is, [tex]\omega_{2} = 30 \;\rm rad/s[/tex].
The angle turned by the motor drum is, [tex]\theta =180 \;\rm rad[/tex].
Here we can use the concept and fundamentals of rotational motion. We can apply the third rotational equation of motion to obtain the angular acceleration of the motor as,
[tex]\omega^{2}_{2}=\omega^{2}_{1}+2 \alpha \theta\\\\90^{2}=30^{2}+(2 \times \alpha \times 180)\\\\\alpha = \dfrac{8100-900}{2 \times 180}\\\\\alpha =20 \;\rm rad/s^{2}[/tex]
Thus, we can conclude that the magnitude of the angular acceleration of the motor is [tex]20 \;\rm rad/s^{2}[/tex].
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