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(C) As the cord winds around the shaft,the length of the cord that is free to move decreases. The distance from the particle to the tangency point at

Vedclass · Accepted Answer

$\omega = \frac{\omega_0}{1 - \frac{a	heta}{r_0}}$

Vedclass · Answer

(C) As the cord winds around the shaft,the length of the cord that is free to move decreases. The distance from the particle to the tangency point at any angle $	heta$ is given by $r = r_0 - a	heta$.Since the tension in the cord acts perpendicular to the velocity of the particle,the torque about the center of the shaft is zero. Thus,the angular momentum of the particle about the center of the shaft is conserved.However,a simpler approach is to note that the velocity $v$ of the particle remains constant because the tension force is always perpendicular to the velocity vector.Initially,$v = r_0\omega_0$.At an angle $	heta$,the distance is $r = r_0 - a	heta$,and the angular velocity is $\omega$.Since $v$ is constant,$v = r\omega = (r_0 - a	heta)\omega$.Equating the two expressions for $v$: $r_0\omega_0 = (r_0 - a	heta)\omega$.Solving for $\omega$: $\omega = \frac{r_0\omega_0}{r_0 - a	heta} = \frac{\omega_0}{1 - \frac{a	heta}{r_0}}$.

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