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(a) The wheel's average angular acceleration is 1.076 rad/s².
(b) The angular displacement of the flywheel during this acceleration is 17,455.6 rad.
(c) The torque required to accelerate the flywheel is 35.23 Nm.
The average angular acceleration of the flywheel is calculated as follows;
[tex]\alpha = \frac{\Delta \omega}{t} \\\\\omega_f = 1850\frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} = 193.76 \ rad/s\\\\\alpha = \frac{\omega _f - \omega_i}{t} \\\\\alpha = \frac{193.76 -0}{3\min\times 60 \ s} \\\\\alpha = 1.076 \ rad/s^2[/tex]
The angular displacement of the flywheel during this acceleration is calculated as follows;
[tex]\omega_f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega_f^2 = 0 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2}{2\alpha} \\\\\theta = \frac{(193.76)^2}{2(1.076)} \\\\\theta = 17,445.6 \ rad[/tex]
[tex]I = \frac{1}{2} mr^2\\\\I = \frac{1}{2} \times 155 \times 0.65^2\\\\I = 32.74 \ kgm^2[/tex]
The torque required to accelerate the flywheel is calculated as follows;
[tex]\tau = I \alpha\\\\\tau = 32.74 \times 1.076\\\\\tau = 35.23 \ Nm[/tex]
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