A uniform circular disc placed on a rough hor | vedclass

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Question

(A) When the disc is placed on the rough surface,slipping occurs. To oppose the relative motion at the point of contact,kinetic friction $f$ acts in t

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) When the disc is placed on the rough surface,slipping occurs. To oppose the relative motion at the point of contact,kinetic friction $f$ acts in the direction opposite to the velocity $v_0$.For linear motion,the force of friction $f$ provides retardation $a = \frac{f}{m}$.For rotational motion,the torque due to friction about the center of mass is $	au = f \cdot r$. Since the torque acts to oppose the initial angular velocity $\omega_0$,it provides angular retardation $\alpha = \frac{	au}{I} = \frac{f \cdot r}{\frac{1}{2} m r^2} = \frac{2f}{mr}$.If the disc comes to rest after time $t$,both linear velocity and angular velocity become zero simultaneously:$v_0 - at = 0 \implies t = \frac{v_0}{a} = \frac{v_0 m}{f}$$\omega_0 - \alpha t = 0 \implies t = \frac{\omega_0}{\alpha} = \frac{\omega_0 mr}{2f}$Equating the two expressions for time $t$:$\frac{v_0 m}{f} = \frac{\omega_0 mr}{2f}$$\frac{v_0}{\omega_0 r} = \frac{1}{2}$

$A$ uniform circular disc placed on a rough horizontal surface has initially a velocity $v_0$ and an angular velocity $\omega_0$ as shown in the figure. The disc comes to rest after moving some distance in the direction of motion. Then $\frac{v_0}{r\omega_0}$ is

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