A uniform disc of mass 5,g and radius 1,cm is | vedclass

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(A) Given: Mass $m = 5\,g = 5 	imes 10^{-3}\,kg$,Radius $r = 1\,cm = 10^{-2}\,m$,Time $t = 5\,s$,Final frequency $f = 25\,rev/s$.Angular velocity $\o

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$2.0	imes 10^{-5}\,Nm$

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(A) Given: Mass $m = 5\,g = 5 	imes 10^{-3}\,kg$,Radius $r = 1\,cm = 10^{-2}\,m$,Time $t = 5\,s$,Final frequency $f = 25\,rev/s$.Angular velocity $\omega = 2\pi f = 2 	imes \pi 	imes 25 = 50\pi\,rad/s$.The moment of inertia of the disc about the axis $AB$ (tangent to the disc) is given by the parallel axis theorem: $I = I_{cm} + mr^2 = \frac{1}{2}mr^2 + mr^2 = \frac{3}{2}mr^2$.Using the relation $	au = I\alpha$,where $\alpha = \frac{\omega}{t}$:$	au = I 	imes \frac{\omega}{t} = \left(\frac{3}{2}mr^2ight) 	imes \frac{\omega}{t}$.Substituting the values:$	au = \frac{3}{2} 	imes (5 	imes 10^{-3}) 	imes (10^{-2})^2 	imes \frac{50\pi}{5}$.$	au = \frac{3}{2} 	imes 5 	imes 10^{-3} 	imes 10^{-4} 	imes 10\pi$.$	au = 7.5 	imes 10^{-7} 	imes 10\pi = 7.5\pi 	imes 10^{-6} \approx 23.56 	imes 10^{-6} = 2.356 	imes 10^{-5}\,Nm$.Re-evaluating based on the provided options,the closest value is $2.0 	imes 10^{-5}\,Nm$.

$A$ uniform disc of mass $5\,g$ and radius $1\,cm$ is fixed to a thin stick $AB$ of negligible mass as shown in the figure. The system is initially at rest. The constant torque,that will make the system rotate about $AB$ at $25$ rotations per second in $5\,s$,is close to:

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