A thin solid disk of 1 kg is rotating along | vedclass

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(D) Given: Mass $m = 1 \ kg$.Initial angular velocity $\omega_{i} = 1800 \ rpm = 1800 	imes \frac{2 \pi}{60} = 60 \pi \ rad/s$.Final angular velocity

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$40$

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(D) Given: Mass $m = 1 \ kg$.Initial angular velocity $\omega_{i} = 1800 \ rpm = 1800 	imes \frac{2 \pi}{60} = 60 \pi \ rad/s$.Final angular velocity $\omega_{f} = 2100 \ rpm = 2100 	imes \frac{2 \pi}{60} = 70 \pi \ rad/s$.External torque $	au_{ext} = 25 \pi \ Nm$.Time $t = 40 \ s$.Using the equation of rotational motion,$\omega_{f} = \omega_{i} + \alpha t$:$70 \pi = 60 \pi + \alpha(40) \implies 10 \pi = 40 \alpha \implies \alpha = \frac{\pi}{4} \ rad/s^2$.For a disk rotating about its diameter,the moment of inertia is $I = \frac{mR^2}{4}$.Using $	au = I \alpha$:$25 \pi = \left( \frac{1 	imes R^2}{4} ight) 	imes \frac{\pi}{4}$.$25 = \frac{R^2}{16} \implies R^2 = 400 \implies R = 20 \ m$.The diameter of the disk is $D = 2R = 2 	imes 20 = 40 \ m$.

$A$ thin solid disk of $1 \ kg$ is rotating along its diameter axis at the speed of $1800 \ rpm$. By applying an external torque of $25 \pi \ Nm$ for $40 \ s$,the speed increases to $2100 \ rpm$. The diameter of the disk is . . . . . . $m$.

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