A uniform circular disc placed on a rough hor | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

$A$ flywheel rotates about an axis. Due to friction at the axis,it experiences an angular retardation proportional to its angular velocity. If its angular velocity falls to half while it makes $n$ rotations,how many more rotations will it make before coming to rest?

Four spheres,each of mass $M$ and diameter $2a$,are placed at the corners of a square of side $b$. The moment of inertia of this system about an axis along one of the sides of the square is:

$A$ particle executes uniform circular motion with angular momentum $L$. Its rotational kinetic energy becomes half,when the angular frequency is doubled. Its new angular momentum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module